Chapter 11: Thermal Properties of Matter

Introduction

Concept Overview

• Temperature measures the hotness or coldness of a body.
• During phase changes like boiling or freezing, temperature remains constant despite heat flow.
• Thermal properties explain why ice feels cold and why temperature stays steady during state changes.

Solved Examples

• Ex 1: Why does temperature remain constant during boiling?
  • Heat energy is used for phase change, not temperature rise.

Practice Set

• Q1: What does temperature measure?
  • A) Energy content B) Hotness or coldness C) Heat flow D) Pressure
• Q2: During freezing, temperature of water
  • A) Increases B) Decreases C) Remains constant D) Fluctuates
• Q3: Why does ice feel cold to touch?
  • A) It has low temperature B) It absorbs heat from hand C) It emits heat D) It has high temperature

Answer Key

• Q1: B - Temperature indicates hotness or coldness, not energy or heat flow.
• Q2: C - Temperature remains constant during freezing as heat is used for phase change.
• Q3: B - Ice absorbs heat from the hand, making it feel cold.

Memory Aids

• Temperature = Hotness/Coldness indicator
• Phase change = Constant temperature despite heat flow
• Ice feels cold because it absorbs heat

Temperature and Heat

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Key Concepts

• Temperature: Relative measure of hotness or coldness (SI unit: Kelvin, °C).
• Heat: Energy transferred due to temperature difference (SI unit: Joule).
• Heat transfer causes changes: temperature rise, expansion, or state change.
• Heat flow direction: from higher to lower temperature.

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transferred due to temperature difference.
• Ex 2: What happens when an object is heated?
  • Temperature may rise, object may expand or change state.

Practice Set

• Q1: Heat is transferred because of
  • A) Pressure difference B) Temperature difference C) Volume difference D) Mass difference
• Q2: Which is a relative measure?
  • A) Heat B) Temperature C) Energy D) Work
• Q3: When an object is heated, which change is NOT possible?
  • A) Temperature rise B) Expansion C) Change of state D) Decrease in temperature

Answer Key

• Q1: B - Heat flows due to temperature difference.
• Q2: B - Temperature is a relative measure of hotness or coldness.
• Q3: D - Heating cannot decrease temperature.

Memory Aids

• Heat = Energy transfer due to temperature difference
• Temperature = Relative hotness/coldness
• Heating effects: Rise in temperature, expansion, or phase change

Measurement of Temperature

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: (t_F - 32)/180 = t_C/100
• Where t_F = temperature in Fahrenheit, t_C = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: t_F = 180/100 × t_C + 32
  • Calculation: t_F = 1.8 × 25 + 32 = 45 + 32 = 77°F
• Ex 2: Convert 95°F to Celsius.
  • Use formula: t_C = 100/180 (t_F - 32)
  • Calculation: t_C = 100/180 (95 - 32) = 100/180 × 63 = 35°C

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: 1.8 × 40 + 32 = 104°F.
• Q3: A) 30°C - Using formula: 100/180 (86 - 32) = 30°C.

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: t_F = 1.8 t_C + 32, t_C = 5/9 (t_F - 32)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: (t_F - 32)/180 = t_C/100
• Where t_F = temperature in Fahrenheit, t_C = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: t_F = 180/100 × t_C + 32
  • Calculation: t_F = 1.8 × 25 + 32 = 45 + 32 = 77°F
• Ex 2: Convert 95°F to Celsius.
  • Use formula: t_C = 100/180 (t_F - 32)
  • Calculation: t_C = 100/180 (95 - 32) = 100/180 × 63 = 35°C

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: 1.8 × 40 + 32 = 104°F.
• Q3: A) 30°C - Using formula: 100/180 (86 - 32) = 30°C.

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: t_F = 1.8 t_C + 32, t_C = 5/9 (t_F - 32)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: (t_F - 32)/180 = t_C/100
• Where t_F = temperature in Fahrenheit, t_C = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: t_F = 180/100 × t_C + 32
  • Calculation: t_F = 1.8 × 25 + 32 = 45 + 32 = 77°F
• Ex 2: Convert 95°F to Celsius.
  • Use formula: t_C = 100/180 (t_F - 32)
  • Calculation: t_C = 100/180 (95 - 32) = 100/180 × 63 = 35°C

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: 1.8 × 40 + 32 = 104°F.
• Q3: A) 30°C - Using formula: 100/180 (86 - 32) = 30°C.

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: t_F = 1.8 t_C + 32, t_C = 5/9 (t_F - 32)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: (t_F - 32)/180 = t_C/100
• Where t_F = temperature in Fahrenheit, t_C = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: t_F = 180/100 × t_C + 32
  • Calculation: t_F = 1.8 × 25 + 32 = 45 + 32 = 77°F
• Ex 2: Convert 95°F to Celsius.
  • Use formula: t_C = 100/180 (t_F - 32)
  • Calculation: t_C = 100/180 (95 - 32) = 100/180 × 63 = 35°C

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: 1.8 × 40 + 32 = 104°F.
• Q3: A) 30°C - Using formula: 100/180 (86 - 32) = 30°C.

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: t_F = 1.8 t_C + 32, t_C = 5/9 (t_F - 32)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: (t_F - 32)/180 = t_C/100
• Where t_F = temperature in Fahrenheit, t_C = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: t_F = 180/100 × t_C + 32
  • Calculation: t_F = 1.8 × 25 + 32 = 45 + 32 = 77°F
• Ex 2: Convert 95°F to Celsius.
  • Use formula: t_C = 100/180 (t_F - 32)
  • Calculation: t_C = 100/180 (95 - 32) = 100/180 × 63 = 35°C

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: 1.8 × 40 + 32 = 104°F.
• Q3: A) 30°C - Using formula: 100/180 (86 - 32) = 30°C.

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: t_F = 1.8 t_C + 32, t_C = 5/9 (t_F - 32)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: (t_F - 32)/180 = t_C/100
• Where t_F = temperature in Fahrenheit, t_C = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: t_F = 180/100 × t_C + 32
  • Calculation: t_F = 1.8 × 25 + 32 = 45 + 32 = 77°F
• Ex 2: Convert 95°F to Celsius.
  • Use formula: t_C = 100/180 (t_F - 32)
  • Calculation: t_C = 100/180 (95 - 32) = 100/180 × 63 = 35°C

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: 1.8 × 40 + 32 = 104°F.
• Q3: A) 30°C - Using formula: 100/180 (86 - 32) = 30°C.

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: t_F = 1.8 t_C + 32, t_C = 5/9 (t_F - 32)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: (t_F - 32)/180 = t_C/100
• Where t_F = temperature in Fahrenheit, t_C = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: t_F = 180/100 × t_C + 32
  • Calculation: t_F = 1.8 × 25 + 32 = 45 + 32 = 77°F
• Ex 2: Convert 95°F to Celsius.
  • Use formula: t_C = 100/180 (t_F - 32)
  • Calculation: t_C = 100/180 (95 - 32) = 100/180 × 63 = 35°C

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: 1.8 × 40 + 32 = 104°F.
• Q3: A) 30°C - Using formula: 100/180 (86 - 32) = 30°C.

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: t_F = 1.8 t_C + 32, t_C = 5/9 (t_F - 32)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: (t_F - 32)/180 = t_C/100
• Where t_F = temperature in Fahrenheit, t_C = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: t_F = 180/100 × t_C + 32
  • Calculation: t_F = 1.8 × 25 + 32 = 45 + 32 = 77°F
• Ex 2: Convert 95°F to Celsius.
  • Use formula: t_C = 100/180 (t_F - 32)
  • Calculation: t_C = 100/180 (95 - 32) = 100/180 × 63 = 35°C

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: 1.8 × 40 + 32 = 104°F.
• Q3: A) 30°C - Using formula: 100/180 (86 - 32) = 30°C.

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: t_F = 1.8 t_C + 32, t_C = 5/9 (t_F - 32)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: (t_F - 32)/180 = t_C/100
• Where t_F = temperature in Fahrenheit, t_C = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: t_F = 180/100 × t_C + 32
  • Calculation: t_F = 1.8 × 25 + 32 = 45 + 32 = 77°F
• Ex 2: Convert 95°F to Celsius.
  • Use formula: t_C = 100/180 (t_F - 32)
  • Calculation: t_C = 100/180 (95 - 32) = 100/180 × 63 = 35°C

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: 1.8 × 40 + 32 = 104°F.
• Q3: A) 30°C - Using formula: 100/180 (86 - 32) = 30°C.

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: t_F = 1.8 t_C + 32, t_C = 5/9 (t_F - 32)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: (t_F - 32)/180 = t_C/100
• Where t_F = temperature in Fahrenheit, t_C = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: t_F = 180/100 × t_C + 32
  • Calculation: t_F = 1.8 × 25 + 32 = 45 + 32 = 77°F
• Ex 2: Convert 95°F to Celsius.
  • Use formula: t_C = 100/180 (t_F - 32)
  • Calculation: t_C = 100/180 (95 - 32) = 100/180 × 63 = 35°C

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: 1.8 × 40 + 32 = 104°F.
• Q3: A) 30°C - Using formula: 100/180 (86 - 32) = 30°C.

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: t_F = 1.8 t_C + 32, t_C = 5/9 (t_F - 32)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: (t_F - 32)/180 = t_C/100
• Where t_F = temperature in Fahrenheit, t_C = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: t_F = 180/100 × t_C + 32
  • Calculation: t_F = 1.8 × 25 + 32 = 45 + 32 = 77°F
• Ex 2: Convert 95°F to Celsius.
  • Use formula: t_C = 100/180 (t_F - 32)
  • Calculation: t_C = 100/180 (95 - 32) = 100/180 × 63 = 35°C

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: 1.8 × 40 + 32 = 104°F.
• Q3: A) 30°C - Using formula: 100/180 (86 - 32) = 30°C.

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: t_F = 1.8 t_C + 32, t_C = 5/9 (t_F - 32)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that change with temperature.
• Commonly used property: volume variation of liquids like mercury or alcohol in glass thermometers.
• Volume of these liquids changes linearly with temperature over a wide range.

Temperature Scales and Fixed Points

• Two fixed reference points define temperature scales: ice point (freezing) and steam point (boiling) of water.
• Ice point = 0°C (32°F), Steam point = 100°C (212°F).
• Common scales: Celsius (°C) and Fahrenheit (°F).

Relation Between Celsius and Fahrenheit

• Formula: \( \frac{t_F - 32}{180} = \frac{t_C}{100} \)
• Where \( t_F \) = temperature in Fahrenheit, \( t_C \) = temperature in Celsius.

Solved Examples

• Ex 1: Convert 25°C to Fahrenheit.
  • Use formula: \( t_F = \frac{180}{100} \times t_C + 32 \)
  • Calculation: \( t_F = 1.8 \times 25 + 32 = 45 + 32 = 77°F \)
• Ex 2: Convert 95°F to Celsius.
  • Use formula: \( t_C = \frac{100}{180} (t_F - 32) \)
  • Calculation: \( t_C = \frac{100}{180} (95 - 32) = \frac{100}{180} \times 63 = 35°C \)

Practice Set

• Q1: What is the freezing point of water in Fahrenheit?
  • A) 0°F
  • B) 32°F
  • C) 100°F
  • D) 212°F
• Q2: Convert 40°C to Fahrenheit.
  • A) 104°F
  • B) 72°F
  • C) 90°F
  • D) 68°F
• Q3: If the temperature is 86°F, what is it in Celsius?
  • A) 30°C
  • B) 32°C
  • C) 28°C
  • D) 25°C

Answer Key

• Q1: B) 32°F - Freezing point of water is 32°F on Fahrenheit scale.
• Q2: A) 104°F - Using formula: \( 1.8 \times 40 + 32 = 104°F \).
• Q3: A) 30°C - Using formula: \( \frac{100}{180} (86 - 32) = 30°C \).

Memory Aids

• Freezing point: 0°C = 32°F
• Boiling point: 100°C = 212°F
• Conversion formula: \( t_F = 1.8 t_C + 32 \), \( t_C = \frac{5}{9} (t_F - 32) \)
• Volume of mercury/alcohol changes linearly with temperature in thermometers.

Thermal Expansion

Concepts

• Thermal expansion: Increase in dimensions due to temperature rise.
• Occurs in solids (length, area, volume), liquids and gases (volume only).
• Gases can expand at constant volume or pressure.
• Expansion depends on material and state (solid, liquid, gas).

Solved Examples

• Ex 1: Calculate length increase of a rod of length 1 m heated by 10°C with linear expansion coefficient α = 1 × 10^-5 °C^-1.
  • Use ΔL = L₀ α ΔT = 1 × 1 × 10^-5 × 10 = 0.0001 m = 0.1 mm

Practice Set

• Q1: Thermal expansion occurs in which states of matter?
  • A. Solids only
  • B. Liquids only
  • C. Solids, liquids, and gases
  • D. Gases only
• Q2: Which dimension expands in liquids on heating?
  • A. Length
  • B. Area
  • C. Volume
  • D. None
• Q3: A gas is heated at constant volume. Which property changes?
  • A. Volume
  • B. Pressure
  • C. Temperature
  • D. Both B and C

Answer Key

• Q1: C - Thermal expansion occurs in solids, liquids, and gases.
• Q2: C - Liquids expand only in volume as they have no fixed shape.
• Q3: D - At constant volume, pressure and temperature change.

Memory Aids

• Thermal expansion = size increase with temperature.
• Solids: length, area, volume expand.
• Liquids & gases: volume expands only.

Concepts

• Thermal expansion: Increase in dimensions due to temperature rise.
• Occurs in solids (length, area, volume), liquids and gases (volume only).
• Gases can expand at constant volume or pressure.
• Expansion depends on material and state (solid, liquid, gas).

Solved Examples

• Ex 1: Calculate length increase of a rod of length 1 m heated by 10°C with linear expansion coefficient α = 1 × 10^-5 °C^-1.
  • Use ΔL = L₀ α ΔT = 1 × 1 × 10^-5 × 10 = 0.0001 m = 0.1 mm

Practice Set

• Q1: Thermal expansion occurs in which states of matter?
  • A. Solids only
  • B. Liquids only
  • C. Solids, liquids, and gases
  • D. Gases only
• Q2: Which dimension expands in liquids on heating?
  • A. Length
  • B. Area
  • C. Volume
  • D. None
• Q3: A gas is heated at constant volume. Which property changes?
  • A. Volume
  • B. Pressure
  • C. Temperature
  • D. Both B and C

Answer Key

• Q1: C - Thermal expansion occurs in solids, liquids, and gases.
• Q2: C - Liquids expand only in volume as they have no fixed shape.
• Q3: D - At constant volume, pressure and temperature change.

Memory Aids

• Thermal expansion = size increase with temperature.
• Solids: length, area, volume expand.
• Liquids & gases: volume expands only.

Concepts

• Thermal expansion: Increase in dimensions due to temperature rise.
• Occurs in solids (length, area, volume), liquids and gases (volume only).
• Gases can expand at constant volume or pressure.
• Expansion depends on material and state (solid, liquid, gas).

Solved Examples

• Ex 1: Calculate length increase of a rod of length 1 m heated by 10°C with linear expansion coefficient α = 1 × 10^-5 °C^-1.
  • Use ΔL = L₀ α ΔT = 1 × 1 × 10^-5 × 10 = 0.0001 m = 0.1 mm

Practice Set

• Q1: Thermal expansion occurs in which states of matter?
  • A. Solids only
  • B. Liquids only
  • C. Solids, liquids, and gases
  • D. Gases only
• Q2: Which dimension expands in liquids on heating?
  • A. Length
  • B. Area
  • C. Volume
  • D. None
• Q3: A gas is heated at constant volume. Which property changes?
  • A. Volume
  • B. Pressure
  • C. Temperature
  • D. Both B and C

Answer Key

• Q1: C - Thermal expansion occurs in solids, liquids, and gases.
• Q2: C - Liquids expand only in volume as they have no fixed shape.
• Q3: D - At constant volume, pressure and temperature change.

Memory Aids

• Thermal expansion = size increase with temperature.
• Solids: length, area, volume expand.
• Liquids & gases: volume expands only.

Concepts

• Thermal expansion: Increase in dimensions due to temperature rise.
• Occurs in solids (length, area, volume), liquids and gases (volume only).
• Gases can expand at constant volume or pressure.
• Expansion depends on material and state (solid, liquid, gas).

Solved Examples

• Ex 1: Calculate length increase of a rod of length 1 m heated by 10°C with linear expansion coefficient α = 1 × 10^-5 °C^-1.
  • Use ΔL = L₀ α ΔT = 1 × 1 × 10^-5 × 10 = 0.0001 m = 0.1 mm

Practice Set

• Q1: Thermal expansion occurs in which states of matter?
  • A. Solids only
  • B. Liquids only
  • C. Solids, liquids, and gases
  • D. Gases only
• Q2: Which dimension expands in liquids on heating?
  • A. Length
  • B. Area
  • C. Volume
  • D. None
• Q3: A gas is heated at constant volume. Which property changes?
  • A. Volume
  • B. Pressure
  • C. Temperature
  • D. Both B and C

Answer Key

• Q1: C - Thermal expansion occurs in solids, liquids, and gases.
• Q2: C - Liquids expand only in volume as they have no fixed shape.
• Q3: D - At constant volume, pressure and temperature change.

Memory Aids

• Thermal expansion = size increase with temperature.
• Solids: length, area, volume expand.
• Liquids & gases: volume expands only.

Concepts

• Thermal expansion: Increase in dimensions due to temperature rise.
• Occurs in solids (length, area, volume), liquids and gases (volume only).
• Gases can expand at constant volume or pressure.
• Expansion depends on material and state (solid, liquid, gas).

Solved Examples

• Ex 1: Calculate length increase of a rod of length 1 m heated by 10°C with linear expansion coefficient α = 1 × 10^-5 °C^-1.
  • Use ΔL = L₀ α ΔT = 1 × 1 × 10^-5 × 10 = 0.0001 m = 0.1 mm

Practice Set

• Q1: Thermal expansion occurs in which states of matter?
  • A. Solids only
  • B. Liquids only
  • C. Solids, liquids, and gases
  • D. Gases only
• Q2: Which dimension expands in liquids on heating?
  • A. Length
  • B. Area
  • C. Volume
  • D. None
• Q3: A gas is heated at constant volume. Which property changes?
  • A. Volume
  • B. Pressure
  • C. Temperature
  • D. Both B and C

Answer Key

• Q1: C - Thermal expansion occurs in solids, liquids, and gases.
• Q2: C - Liquids expand only in volume as they have no fixed shape.
• Q3: D - At constant volume, pressure and temperature change.

Memory Aids

• Thermal expansion = size increase with temperature.
• Solids: length, area, volume expand.
• Liquids & gases: volume expands only.

Concepts

• Thermal expansion: Increase in dimensions due to temperature rise.
• Occurs in solids (length, area, volume), liquids and gases (volume only).
• Gases can expand at constant volume or pressure.
• Expansion depends on material and state (solid, liquid, gas).

Solved Examples

• Ex 1: Calculate length increase of a rod of length 1 m heated by 10°C with linear expansion coefficient α = 1 × 10^-5 °C^-1.
  • Use ΔL = L₀ α ΔT = 1 × 1 × 10^-5 × 10 = 0.0001 m = 0.1 mm

Practice Set

• Q1: Thermal expansion occurs in which states of matter?
  • A. Solids only
  • B. Liquids only
  • C. Solids, liquids, and gases
  • D. Gases only
• Q2: Which dimension expands in liquids on heating?
  • A. Length
  • B. Area
  • C. Volume
  • D. None
• Q3: A gas is heated at constant volume. Which property changes?
  • A. Volume
  • B. Pressure
  • C. Temperature
  • D. Both B and C

Answer Key

• Q1: C - Thermal expansion occurs in solids, liquids, and gases.
• Q2: C - Liquids expand only in volume as they have no fixed shape.
• Q3: D - At constant volume, pressure and temperature change.

Memory Aids

• Thermal expansion = size increase with temperature.
• Solids: length, area, volume expand.
• Liquids & gases: volume expands only.

Concepts

• Thermal expansion: Increase in dimensions due to temperature rise.
• Occurs in solids (length, area, volume), liquids and gases (volume only).
• Gases can expand at constant volume or pressure.
• Expansion depends on material and state (solid, liquid, gas).

Solved Examples

• Ex 1: Calculate length increase of a rod of length 1 m heated by 10°C with linear expansion coefficient α = 1 × 10^-5 °C^-1.
  • Use ΔL = L₀ α ΔT = 1 × 1 × 10^-5 × 10 = 0.0001 m = 0.1 mm

Practice Set

• Q1: Thermal expansion occurs in which states of matter?
  • A. Solids only
  • B. Liquids only
  • C. Solids, liquids, and gases
  • D. Gases only
• Q2: Which dimension expands in liquids on heating?
  • A. Length
  • B. Area
  • C. Volume
  • D. None
• Q3: A gas is heated at constant volume. Which property changes?
  • A. Volume
  • B. Pressure
  • C. Temperature
  • D. Both B and C

Answer Key

• Q1: C - Thermal expansion occurs in solids, liquids, and gases.
• Q2: C - Liquids expand only in volume as they have no fixed shape.
• Q3: D - At constant volume, pressure and temperature change.

Memory Aids

• Thermal expansion = size increase with temperature.
• Solids: length, area, volume expand.
• Liquids & gases: volume expands only.

Concepts

• Thermal expansion: Increase in dimensions due to temperature rise.
• Occurs in solids (length, area, volume), liquids and gases (volume only).
• Gases can expand at constant volume or pressure.
• Expansion depends on material and state (solid, liquid, gas).

Solved Examples

• Ex 1: Calculate length increase of a rod of length 1 m heated by 10°C with linear expansion coefficient α = 1 × 10^-5 °C^-1.
  • Use ΔL = L₀ α ΔT = 1 × 1 × 10^-5 × 10 = 0.0001 m = 0.1 mm

Practice Set

• Q1: Thermal expansion occurs in which states of matter?
  • A. Solids only
  • B. Liquids only
  • C. Solids, liquids, and gases
  • D. Gases only
• Q2: Which dimension expands in liquids on heating?
  • A. Length
  • B. Area
  • C. Volume
  • D. None
• Q3: A gas is heated at constant volume. Which property changes?
  • A. Volume
  • B. Pressure
  • C. Temperature
  • D. Both B and C

Answer Key

• Q1: C - Thermal expansion occurs in solids, liquids, and gases.
• Q2: C - Liquids expand only in volume as they have no fixed shape.
• Q3: D - At constant volume, pressure and temperature change.

Memory Aids

• Thermal expansion = size increase with temperature.
• Solids: length, area, volume expand.
• Liquids & gases: volume expands only.

Concepts

• Thermal expansion: Increase in dimensions due to temperature rise.
• Occurs in solids (length, area, volume), liquids and gases (volume only).
• Gases can expand at constant volume or pressure.
• Expansion depends on material and state (solid, liquid, gas).

Solved Examples

• Ex 1: Calculate length increase of a rod of length 1 m heated by 10°C with linear expansion coefficient \(\alpha = 1 \times 10^{-5} \degree C^{-1}\).
  • Use \(\Delta L = L_0 \alpha \Delta T = 1 \times 1 \times 10^{-5} \times 10 = 0.0001 \text{ m} = 0.1 \text{ mm}\)

Practice Set

• Q1: Thermal expansion occurs in which states of matter?
  • A. Solids only
  • B. Liquids only
  • C. Solids, liquids, and gases
  • D. Gases only
• Q2: Which dimension expands in liquids on heating?
  • A. Length
  • B. Area
  • C. Volume
  • D. None
• Q3: A gas is heated at constant volume. Which property changes?
  • A. Volume
  • B. Pressure
  • C. Temperature
  • D. Both B and C

Answer Key

• Q1: C - Thermal expansion occurs in solids, liquids, and gases.
• Q2: C - Liquids expand only in volume as they have no fixed shape.
• Q3: D - At constant volume, pressure and temperature change.

Memory Aids

• Thermal expansion = size increase with temperature.
• Solids: length, area, volume expand.
• Liquids & gases: volume expands only.

Thermal Expansion in Solids

Types

• Linear expansion: Increase in length ΔL.
• Area expansion: Increase in surface area ΔA.
• Volume expansion: Increase in volume ΔV.

Solved Examples

• Ex 1: Rod length l=2m, α=1.2 × 10^-5 °C^-1, temperature rise ΔT=50°C.
  • New length L_T = L₀(1 + α ΔT) = 2(1 + 1.2 × 10^-5 × 50) = 2.0012 m

Practice Set

• Q1: Linear expansion changes which dimension?
  • A. Length
  • B. Area
  • C. Volume
  • D. Density
• Q2: Area expansion coefficient β relates to increase in?
  • A. Length
  • B. Surface area
  • C. Volume
  • D. Mass
• Q3: Volume expansion coefficient γ is defined as?
  • A. Increase in volume per unit volume per degree rise
  • B. Increase in length per unit length per degree rise
  • C. Increase in area per unit area per degree rise
  • D. None

Answer Key

• Q1: A - Linear expansion changes length.
• Q2: B - Area expansion relates to surface area increase.
• Q3: A - Volume expansion coefficient defines volume increase per unit volume per degree.

Memory Aids

• α: linear expansion coefficient.
• β = 2α, γ = 3α.
• Use L_T = L₀(1 + α ΔT) for length.

Types

• Linear expansion: Increase in length ΔL.
• Area expansion: Increase in surface area ΔA.
• Volume expansion: Increase in volume ΔV.

Solved Examples

• Ex 1: Rod length l=2m, α=1.2 × 10^-5 °C^-1, temperature rise ΔT=50°C.
  • New length L_T = L₀(1 + α ΔT) = 2(1 + 1.2 × 10^-5 × 50) = 2.0012 m

Practice Set

• Q1: Linear expansion changes which dimension?
  • A. Length
  • B. Area
  • C. Volume
  • D. Density
• Q2: Area expansion coefficient β relates to increase in?
  • A. Length
  • B. Surface area
  • C. Volume
  • D. Mass
• Q3: Volume expansion coefficient γ is defined as?
  • A. Increase in volume per unit volume per degree rise
  • B. Increase in length per unit length per degree rise
  • C. Increase in area per unit area per degree rise
  • D. None

Answer Key

• Q1: A - Linear expansion changes length.
• Q2: B - Area expansion relates to surface area increase.
• Q3: A - Volume expansion coefficient defines volume increase per unit volume per degree.

Memory Aids

• α: linear expansion coefficient.
• β = 2α, γ = 3α.
• Use L_T = L₀(1 + α ΔT) for length.

Types

• Linear expansion: Increase in length ΔL.
• Area expansion: Increase in surface area ΔA.
• Volume expansion: Increase in volume ΔV.

Solved Examples

• Ex 1: Rod length l=2m, α=1.2 × 10^-5 °C^-1, temperature rise ΔT=50°C.
  • New length L_T = L₀(1 + α ΔT) = 2(1 + 1.2 × 10^-5 × 50) = 2.0012 m

Practice Set

• Q1: Linear expansion changes which dimension?
  • A. Length
  • B. Area
  • C. Volume
  • D. Density
• Q2: Area expansion coefficient β relates to increase in?
  • A. Length
  • B. Surface area
  • C. Volume
  • D. Mass
• Q3: Volume expansion coefficient γ is defined as?
  • A. Increase in volume per unit volume per degree rise
  • B. Increase in length per unit length per degree rise
  • C. Increase in area per unit area per degree rise
  • D. None

Answer Key

• Q1: A - Linear expansion changes length.
• Q2: B - Area expansion relates to surface area increase.
• Q3: A - Volume expansion coefficient defines volume increase per unit volume per degree.

Memory Aids

• α: linear expansion coefficient.
• β = 2α, γ = 3α.
• Use L_T = L₀(1 + α ΔT) for length.

Types

• Linear expansion: Increase in length \(\Delta L\).
• Area expansion: Increase in surface area \(\Delta A\).
• Volume expansion: Increase in volume \(\Delta V\).

Solved Examples

• Ex 1: Rod length \(l=2m\), \(\alpha=1.2 \times 10^{-5} \degree C^{-1}\), temperature rise \(\Delta T=50\degree C\).
  • New length \(L_T = L_0(1 + \alpha \Delta T) = 2(1 + 1.2 \times 10^{-5} \times 50) = 2.0012 m\)

Practice Set

• Q1: Linear expansion changes which dimension?
  • A. Length
  • B. Area
  • C. Volume
  • D. Density
• Q2: Area expansion coefficient \(\beta\) relates to increase in?
  • A. Length
  • B. Surface area
  • C. Volume
  • D. Mass
• Q3: Volume expansion coefficient \(\gamma\) is defined as?
  • A. Increase in volume per unit volume per degree rise
  • B. Increase in length per unit length per degree rise
  • C. Increase in area per unit area per degree rise
  • D. None

Answer Key

• Q1: A - Linear expansion changes length.
• Q2: B - Area expansion relates to surface area increase.
• Q3: A - Volume expansion coefficient defines volume increase per unit volume per degree.

Memory Aids

• \(\alpha\): linear expansion coefficient.
• \(\beta = 2\alpha\), \(\gamma = 3\alpha\).
• Use \(L_T = L_0(1 + \alpha \Delta T)\) for length.

Coefficients of Thermal Expansion

Definitions & Relations

• Coefficient of linear expansion α = ΔL/(L₀ ΔT).
• Coefficient of area expansion β = ΔA/(A₀ ΔT).
• Coefficient of volume expansion γ = ΔV/(V₀ ΔT).
• Relation: α = β/2 = γ/3.
• Units: °C^-1 or K^-1.

Solved Examples

• Ex 1: Given α = 1 × 10^-5 °C^-1, find β and γ.
  • β = 2α = 2 × 10^-5 °C^-1
  • γ = 3α = 3 × 10^-5 °C^-1

Practice Set

• Q1: If α = 2 × 10^-5 °C^-1, then β = ?
  • A. 1 × 10^-5
  • B. 2 × 10^-5
  • C. 4 × 10^-5
  • D. 6 × 10^-5
• Q2: Relation between coefficients is?
  • A. α = β = γ
  • B. α = β/2 = γ/3
  • C. α = 2β = 3γ
  • D. None
• Q3: Units of α, β, γ are?
  • A. m
  • B. °C^-1
  • C. kg/m³
  • D. N/m²

Answer Key

• Q1: C - β = 2α = 4 × 10^-5 °C^-1.
• Q2: B - α = β/2 = γ/3 is correct relation.
• Q3: B - Units are per degree Celsius or Kelvin.

Memory Aids

• Remember: β = 2α, γ = 3α.
• All coefficients have same units.

Definitions & Relations

• Coefficient of linear expansion α = ΔL/(L₀ ΔT).
• Coefficient of area expansion β = ΔA/(A₀ ΔT).
• Coefficient of volume expansion γ = ΔV/(V₀ ΔT).
• Relation: α = β/2 = γ/3.
• Units: °C^-1 or K^-1.

Solved Examples

• Ex 1: Given α = 1 × 10^-5 °C^-1, find β and γ.
  • β = 2α = 2 × 10^-5 °C^-1
  • γ = 3α = 3 × 10^-5 °C^-1

Practice Set

• Q1: If α = 2 × 10^-5 °C^-1, then β = ?
  • A. 1 × 10^-5
  • B. 2 × 10^-5
  • C. 4 × 10^-5
  • D. 6 × 10^-5
• Q2: Relation between coefficients is?
  • A. α = β = γ
  • B. α = β/2 = γ/3
  • C. α = 2β = 3γ
  • D. None
• Q3: Units of α, β, γ are?
  • A. m
  • B. °C^-1
  • C. kg/m³
  • D. N/m²

Answer Key

• Q1: C - β = 2α = 4 × 10^-5 °C^-1.
• Q2: B - α = β/2 = γ/3 is correct relation.
• Q3: B - Units are per degree Celsius or Kelvin.

Memory Aids

• Remember: β = 2α, γ = 3α.
• All coefficients have same units.

Definitions & Relations

• Coefficient of linear expansion α = ΔL/(L₀ ΔT).
• Coefficient of area expansion β = ΔA/(A₀ ΔT).
• Coefficient of volume expansion γ = ΔV/(V₀ ΔT).
• Relation: α = β/2 = γ/3.
• Units: °C^-1 or K^-1.

Solved Examples

• Ex 1: Given α = 1 × 10^-5 °C^-1, find β and γ.
  • β = 2α = 2 × 10^-5 °C^-1
  • γ = 3α = 3 × 10^-5 °C^-1

Practice Set

• Q1: If α = 2 × 10^-5 °C^-1, then β = ?
  • A. 1 × 10^-5
  • B. 2 × 10^-5
  • C. 4 × 10^-5
  • D. 6 × 10^-5
• Q2: Relation between coefficients is?
  • A. α = β = γ
  • B. α = β/2 = γ/3
  • C. α = 2β = 3γ
  • D. None
• Q3: Units of α, β, γ are?
  • A. m
  • B. °C^-1
  • C. kg/m³
  • D. N/m²

Answer Key

• Q1: C - β = 2α = 4 × 10^-5 °C^-1.
• Q2: B - α = β/2 = γ/3 is correct relation.
• Q3: B - Units are per degree Celsius or Kelvin.

Memory Aids

• Remember: β = 2α, γ = 3α.
• All coefficients have same units.

Definitions & Relations

• Coefficient of linear expansion α = ΔL/(L₀ ΔT).
• Coefficient of area expansion β = ΔA/(A₀ ΔT).
• Coefficient of volume expansion γ = ΔV/(V₀ ΔT).
• Relation: α = β/2 = γ/3.
• Units: °C^-1 or K^-1.

Solved Examples

• Ex 1: Given α = 1 × 10^-5 °C^-1, find β and γ.
  • β = 2α = 2 × 10^-5 °C^-1
  • γ = 3α = 3 × 10^-5 °C^-1

Practice Set

• Q1: If α = 2 × 10^-5 °C^-1, then β = ?
  • A. 1 × 10^-5
  • B. 2 × 10^-5
  • C. 4 × 10^-5
  • D. 6 × 10^-5
• Q2: Relation between coefficients is?
  • A. α = β = γ
  • B. α = β/2 = γ/3
  • C. α = 2β = 3γ
  • D. None
• Q3: Units of α, β, γ are?
  • A. m
  • B. °C^-1
  • C. kg/m³
  • D. N/m²

Answer Key

• Q1: C - β = 2α = 4 × 10^-5 °C^-1.
• Q2: B - α = β/2 = γ/3 is correct relation.
• Q3: B - Units are per degree Celsius or Kelvin.

Memory Aids

• Remember: β = 2α, γ = 3α.
• All coefficients have same units.

Definitions & Relations

• Coefficient of linear expansion α = ΔL/(L₀ ΔT).
• Coefficient of area expansion β = ΔA/(A₀ ΔT).
• Coefficient of volume expansion γ = ΔV/(V₀ ΔT).
• Relation: α = β/2 = γ/3.
• Units: °C^-1 or K^-1.

Solved Examples

• Ex 1: Given α = 1 × 10^-5 °C^-1, find β and γ.
  • β = 2α = 2 × 10^-5 °C^-1
  • γ = 3α = 3 × 10^-5 °C^-1

Practice Set

• Q1: If α = 2 × 10^-5 °C^-1, then β = ?
  • A. 1 × 10^-5
  • B. 2 × 10^-5
  • C. 4 × 10^-5
  • D. 6 × 10^-5
• Q2: Relation between coefficients is?
  • A. α = β = γ
  • B. α = β/2 = γ/3
  • C. α = 2β = 3γ
  • D. None
• Q3: Units of α, β, γ are?
  • A. m
  • B. °C^-1
  • C. kg/m³
  • D. N/m²

Answer Key

• Q1: C - β = 2α = 4 × 10^-5 °C^-1.
• Q2: B - α = β/2 = γ/3 is correct relation.
• Q3: B - Units are per degree Celsius or Kelvin.

Memory Aids

• Remember: β = 2α, γ = 3α.
• All coefficients have same units.

Definitions & Relations

• Coefficient of linear expansion α = ΔL/(L₀ ΔT).
• Coefficient of area expansion β = ΔA/(A₀ ΔT).
• Coefficient of volume expansion γ = ΔV/(V₀ ΔT).
• Relation: α = β/2 = γ/3.
• Units: °C^-1 or K^-1.

Solved Examples

• Ex 1: Given α = 1 × 10^-5 °C^-1, find β and γ.
  • β = 2α = 2 × 10^-5 °C^-1
  • γ = 3α = 3 × 10^-5 °C^-1

Practice Set

• Q1: If α = 2 × 10^-5 °C^-1, then β = ?
  • A. 1 × 10^-5
  • B. 2 × 10^-5
  • C. 4 × 10^-5
  • D. 6 × 10^-5
• Q2: Relation between coefficients is?
  • A. α = β = γ
  • B. α = β/2 = γ/3
  • C. α = 2β = 3γ
  • D. None
• Q3: Units of α, β, γ are?
  • A. m
  • B. °C^-1
  • C. kg/m³
  • D. N/m²

Answer Key

• Q1: C - β = 2α = 4 × 10^-5 °C^-1.
• Q2: B - α = β/2 = γ/3 is correct relation.
• Q3: B - Units are per degree Celsius or Kelvin.

Memory Aids

• Remember: β = 2α, γ = 3α.
• All coefficients have same units.

Definitions & Relations

• Coefficient of linear expansion α = ΔL/(L₀ ΔT).
• Coefficient of area expansion β = ΔA/(A₀ ΔT).
• Coefficient of volume expansion γ = ΔV/(V₀ ΔT).
• Relation: α = β/2 = γ/3.
• Units: °C^-1 or K^-1.

Solved Examples

• Ex 1: Given α = 1 × 10^-5 °C^-1, find β and γ.
  • β = 2α = 2 × 10^-5 °C^-1
  • γ = 3α = 3 × 10^-5 °C^-1

Practice Set

• Q1: If α = 2 × 10^-5 °C^-1, then β = ?
  • A. 1 × 10^-5
  • B. 2 × 10^-5
  • C. 4 × 10^-5
  • D. 6 × 10^-5
• Q2: Relation between coefficients is?
  • A. α = β = γ
  • B. α = β/2 = γ/3
  • C. α = 2β = 3γ
  • D. None
• Q3: Units of α, β, γ are?
  • A. m
  • B. °C^-1
  • C. kg/m³
  • D. N/m²

Answer Key

• Q1: C - β = 2α = 4 × 10^-5 °C^-1.
• Q2: B - α = β/2 = γ/3 is correct relation.
• Q3: B - Units are per degree Celsius or Kelvin.

Memory Aids

• Remember: β = 2α, γ = 3α.
• All coefficients have same units.

Definitions & Relations

• Coefficient of linear expansion \(\alpha = \frac{\Delta L}{L_0 \Delta T}\).
• Coefficient of area expansion \(\beta = \frac{\Delta A}{A_0 \Delta T}\).
• Coefficient of volume expansion \(\gamma = \frac{\Delta V}{V_0 \Delta T}\).
• Relation: \(\alpha = \frac{\beta}{2} = \frac{\gamma}{3}\).
• Units: \(\degree C^{-1}\) or \(K^{-1}\).

Solved Examples

• Ex 1: Given \(\alpha = 1 \times 10^{-5} \degree C^{-1}\), find \(\beta\) and \(\gamma\).
  • \(\beta = 2\alpha = 2 \times 10^{-5} \degree C^{-1}\)
  • \(\gamma = 3\alpha = 3 \times 10^{-5} \degree C^{-1}\)

Practice Set

• Q1: If \(\alpha = 2 \times 10^{-5} \degree C^{-1}\), then \(\beta = ?\)
  • A. \(1 \times 10^{-5}\)
  • B. \(2 \times 10^{-5}\)
  • C. \(4 \times 10^{-5}\)
  • D. \(6 \times 10^{-5}\)
• Q2: Relation between coefficients is?
  • A. \(\alpha = \beta = \gamma\)
  • B. \(\alpha = \frac{\beta}{2} = \frac{\gamma}{3}\)
  • C. \(\alpha = 2\beta = 3\gamma\)
  • D. None
• Q3: Units of \(\alpha, \beta, \gamma\) are?
  • A. m
  • B. \(\degree C^{-1}\)
  • C. kg/m³
  • D. N/m²

Answer Key

• Q1: C - \(\beta = 2\alpha = 4 \times 10^{-5} \degree C^{-1}\).
• Q2: B - \(\alpha = \frac{\beta}{2} = \frac{\gamma}{3}\) is correct relation.
• Q3: B - Units are per degree Celsius or Kelvin.

Memory Aids

• Remember: \(\beta = 2\alpha\), \(\gamma = 3\alpha\).
• All coefficients have same units.

Anomalous Expansion of Water

Key Points

• Water contracts on heating from 0°C to 4°C (negative volume expansion).
• Water expands on heating above 4°C.
• Maximum density at 4°C.
• Explains ice floating and freezing from top.

Solved Examples

• Ex 1: Explain why ice floats on water.
  • Due to anomalous expansion, water is densest at 4°C; ice is less dense and floats.

Practice Set

• Q1: Water has maximum density at?
  • A. 0°C
  • B. 4°C
  • C. 100°C
  • D. -4°C
• Q2: Between 0°C and 4°C, water volume?
  • A. Increases on heating
  • B. Decreases on heating
  • C. Remains constant
  • D. Expands on cooling
• Q3: Anomalous expansion means?
  • A. Normal expansion
  • B. Expansion only in solids
  • C. Contraction on heating in 0-4°C range
  • D. No expansion

Answer Key

• Q1: B - Water is densest at 4°C.
• Q2: B - Water contracts on heating from 0°C to 4°C.
• Q3: C - Anomalous expansion means contraction on heating in 0-4°C.

Memory Aids

• Water densest at 4°C.
• Volume decreases on heating 0-4°C.
• Ice floats due to lower density.

Key Points

• Water contracts on heating from 0°C to 4°C (negative volume expansion).
• Water expands on heating above 4°C.
• Maximum density at 4°C.
• Explains ice floating and freezing from top.

Solved Examples

• Ex 1: Explain why ice floats on water.
  • Due to anomalous expansion, water is densest at 4°C; ice is less dense and floats.

Practice Set

• Q1: Water has maximum density at?
  • A. 0°C
  • B. 4°C
  • C. 100°C
  • D. -4°C
• Q2: Between 0°C and 4°C, water volume?
  • A. Increases on heating
  • B. Decreases on heating
  • C. Remains constant
  • D. Expands on cooling
• Q3: Anomalous expansion means?
  • A. Normal expansion
  • B. Expansion only in solids
  • C. Contraction on heating in 0-4°C range
  • D. No expansion

Answer Key

• Q1: B - Water is densest at 4°C.
• Q2: B - Water contracts on heating from 0°C to 4°C.
• Q3: C - Anomalous expansion means contraction on heating in 0-4°C.

Memory Aids

• Water densest at 4°C.
• Volume decreases on heating 0-4°C.
• Ice floats due to lower density.

Thermal Stress

Concepts & Formulas

• Thermal stress arises when expansion is prevented.
• Thermal strain = ΔL/L = α ΔT.
• Thermal stress = Y × thermal strain = Y α ΔT, where Y is Young's modulus.
• Force developed = Y A α ΔT, A is cross-sectional area.

Solved Examples

• Ex 1: Rod with Y=2 × 10¹¹ Pa, α=1 × 10^-5 °C^-1, ΔT=50°C, A=0.01 m².
  • Thermal stress = 2 × 10¹¹ × 1 × 10^-5 × 50 = 1 × 10⁸ Pa
  • Force = 1 × 10⁸ × 0.01 = 1 × 10⁶ N

Practice Set

• Q1: Thermal stress occurs when?
  • A. Expansion is free
  • B. Expansion is prevented
  • C. Temperature decreases
  • D. No temperature change
• Q2: Thermal strain formula is?
  • A. α ΔT
  • B. ΔL/L
  • C. Both A and B
  • D. None
• Q3: Force developed depends on?
  • A. Young's modulus, area, α, ΔT
  • B. Only temperature
  • C. Only area
  • D. None

Answer Key

• Q1: B - Stress arises when expansion is prevented.
• Q2: C - Thermal strain = α ΔT = ΔL/L.
• Q3: A - Force = Y A α ΔT.

Memory Aids

• Stress = Young's modulus × strain.
• Strain = α ΔT.
• Force = stress × area.

Concepts & Formulas

• Thermal stress arises when expansion is prevented.
• Thermal strain = ΔL/L = α ΔT.
• Thermal stress = Y × thermal strain = Y α ΔT, where Y is Young's modulus.
• Force developed = Y A α ΔT, A is cross-sectional area.

Solved Examples

• Ex 1: Rod with Y=2 × 10¹¹ Pa, α=1 × 10^-5 °C^-1, ΔT=50°C, A=0.01 m².
  • Thermal stress = 2 × 10¹¹ × 1 × 10^-5 × 50 = 1 × 10⁸ Pa
  • Force = 1 × 10⁸ × 0.01 = 1 × 10⁶ N

Practice Set

• Q1: Thermal stress occurs when?
  • A. Expansion is free
  • B. Expansion is prevented
  • C. Temperature decreases
  • D. No temperature change
• Q2: Thermal strain formula is?
  • A. α ΔT
  • B. ΔL/L
  • C. Both A and B
  • D. None
• Q3: Force developed depends on?
  • A. Young's modulus, area, α, ΔT
  • B. Only temperature
  • C. Only area
  • D. None

Answer Key

• Q1: B - Stress arises when expansion is prevented.
• Q2: C - Thermal strain = α ΔT = ΔL/L.
• Q3: A - Force = Y A α ΔT.

Memory Aids

• Stress = Young's modulus × strain.
• Strain = α ΔT.
• Force = stress × area.

Concepts & Formulas

• Thermal stress arises when expansion is prevented.
• Thermal strain = ΔL/L = α ΔT.
• Thermal stress = Y × thermal strain = Y α ΔT, where Y is Young's modulus.
• Force developed = Y A α ΔT, A is cross-sectional area.

Solved Examples

• Ex 1: Rod with Y=2 × 10¹¹ Pa, α=1 × 10^-5 °C^-1, ΔT=50°C, A=0.01 m².
  • Thermal stress = 2 × 10¹¹ × 1 × 10^-5 × 50 = 1 × 10⁸ Pa
  • Force = 1 × 10⁸ × 0.01 = 1 × 10⁶ N

Practice Set

• Q1: Thermal stress occurs when?
  • A. Expansion is free
  • B. Expansion is prevented
  • C. Temperature decreases
  • D. No temperature change
• Q2: Thermal strain formula is?
  • A. α ΔT
  • B. ΔL/L
  • C. Both A and B
  • D. None
• Q3: Force developed depends on?
  • A. Young's modulus, area, α, ΔT
  • B. Only temperature
  • C. Only area
  • D. None

Answer Key

• Q1: B - Stress arises when expansion is prevented.
• Q2: C - Thermal strain = α ΔT = ΔL/L.
• Q3: A - Force = Y A α ΔT.

Memory Aids

• Stress = Young's modulus × strain.
• Strain = α ΔT.
• Force = stress × area.

Concepts & Formulas

• Thermal stress arises when expansion is prevented.
• Thermal strain \(= \frac{\Delta L}{L} = \alpha \Delta T\).
• Thermal stress \(= Y \times \text{thermal strain} = Y \alpha \Delta T\), where \(Y\) is Young's modulus.
• Force developed \(= Y A \alpha \Delta T\), \(A\) is cross-sectional area.

Solved Examples

• Ex 1: Rod with \(Y=2 \times 10^{11} \text{Pa}\), \(\alpha=1 \times 10^{-5} \degree C^{-1}\), \(\Delta T=50\degree C\), \(A=0.01 m^2\).
  • Thermal stress \(= 2 \times 10^{11} \times 1 \times 10^{-5} \times 50 = 1 \times 10^{8} \text{Pa}\)
  • Force \(= 1 \times 10^{8} \times 0.01 = 1 \times 10^{6} \text{N}\)

Practice Set

• Q1: Thermal stress occurs when?
  • A. Expansion is free
  • B. Expansion is prevented
  • C. Temperature decreases
  • D. No temperature change
• Q2: Thermal strain formula is?
  • A. \(\alpha \Delta T\)
  • B. \(\frac{\Delta L}{L}\)
  • C. Both A and B
  • D. None
• Q3: Force developed depends on?
  • A. Young's modulus, area, \(\alpha\), \(\Delta T\)
  • B. Only temperature
  • C. Only area
  • D. None

Answer Key

• Q1: B - Stress arises when expansion is prevented.
• Q2: C - Thermal strain = \(\alpha \Delta T = \frac{\Delta L}{L}\).
• Q3: A - Force = \(Y A \alpha \Delta T\).

Memory Aids

• Stress = Young's modulus × strain.
• Strain = \(\alpha \Delta T\).
• Force = stress × area.

Specific Heat Capacity

Definition and Formula

• Heat capacity (S) = ΔQ/ΔT = ms
• ΔQ: heat supplied to change temperature from T to T + ΔT
• m: mass of the substance
• s: specific heat capacity

Specific Heat Capacity

• Heat absorbed or released per unit mass to change temperature by 1 unit
• Formula: s = (1/m) ΔQ/ΔT = S/m
• Depends on substance nature and temperature

Molar Specific Heat Capacity

• Heat capacity per mole of substance (in gram moles)
• Formula: C = S/μ = (1/μ) ΔQ/ΔT
• Two types:
  • At constant pressure: C_p
  • At constant volume: C_V

Solved Examples

• Ex 1: Calculate specific heat capacity if ΔQ = 500 J, m = 2 kg, ΔT = 5 K.
  • Use s = ΔQ/(m ΔT) = 500/(2 × 5) = 50 J/kg K
• Ex 2: Find molar specific heat capacity if S = 100 J/K, μ = 4 mol.
  • Use C = S/μ = 100/4 = 25 J/mol K

Practice Set

• Q1: Heat capacity S is defined as:
  • A) ΔT/ΔQ
  • B) ΔQ/ΔT
  • C) m × s
  • D) (1/m) ΔQ/ΔT
• Q2: Specific heat capacity s depends on:
  • A) Mass only
  • B) Nature of substance and temperature
  • C) Volume only
  • D) Pressure only
• Q3: Molar specific heat capacity at constant volume is denoted by:
  • A) C_p
  • B) C_V
  • C) s
  • D) S

Answer Key

• Q1: B - Heat capacity is heat supplied per unit temperature change ΔQ/ΔT.
• Q2: B - Specific heat depends on the substance's nature and temperature.
• Q3: B - C_V denotes molar specific heat at constant volume.

Memory Aids

• Heat capacity S = ms
• Specific heat s = S/m
• Molar heat C = S/μ
• C_p at constant pressure, C_V at constant volume

Definition and Formula

• Heat capacity (S) = ΔQ/ΔT = ms
• ΔQ: heat supplied to change temperature from T to T + ΔT
• m: mass of the substance
• s: specific heat capacity

Specific Heat Capacity

• Heat absorbed or released per unit mass to change temperature by 1 unit
• Formula: s = (1/m) ΔQ/ΔT = S/m
• Depends on substance nature and temperature

Molar Specific Heat Capacity

• Heat capacity per mole of substance (in gram moles)
• Formula: C = S/μ = (1/μ) ΔQ/ΔT
• Two types:
  • At constant pressure: C_p
  • At constant volume: C_V

Solved Examples

• Ex 1: Calculate specific heat capacity if ΔQ = 500 J, m = 2 kg, ΔT = 5 K.
  • Use s = ΔQ/(m ΔT) = 500/(2 × 5) = 50 J/kg K
• Ex 2: Find molar specific heat capacity if S = 100 J/K, μ = 4 mol.
  • Use C = S/μ = 100/4 = 25 J/mol K

Practice Set

• Q1: Heat capacity S is defined as:
  • A) ΔT/ΔQ
  • B) ΔQ/ΔT
  • C) m × s
  • D) (1/m) ΔQ/ΔT
• Q2: Specific heat capacity s depends on:
  • A) Mass only
  • B) Nature of substance and temperature
  • C) Volume only
  • D) Pressure only
• Q3: Molar specific heat capacity at constant volume is denoted by:
  • A) C_p
  • B) C_V
  • C) s
  • D) S

Answer Key

• Q1: B - Heat capacity is heat supplied per unit temperature change ΔQ/ΔT.
• Q2: B - Specific heat depends on the substance's nature and temperature.
• Q3: B - C_V denotes molar specific heat at constant volume.

Memory Aids

• Heat capacity S = ms
• Specific heat s = S/m
• Molar heat C = S/μ
• C_p at constant pressure, C_V at constant volume

Definition and Formula

• Heat capacity (S) = ΔQ/ΔT = ms
• ΔQ: heat supplied to change temperature from T to T + ΔT
• m: mass of the substance
• s: specific heat capacity

Specific Heat Capacity

• Heat absorbed or released per unit mass to change temperature by 1 unit
• Formula: s = (1/m) ΔQ/ΔT = S/m
• Depends on substance nature and temperature

Molar Specific Heat Capacity

• Heat capacity per mole of substance (in gram moles)
• Formula: C = S/μ = (1/μ) ΔQ/ΔT
• Two types:
  • At constant pressure: C_p
  • At constant volume: C_V

Solved Examples

• Ex 1: Calculate specific heat capacity if ΔQ = 500 J, m = 2 kg, ΔT = 5 K.
  • Use s = ΔQ/(m ΔT) = 500/(2 × 5) = 50 J/kg K
• Ex 2: Find molar specific heat capacity if S = 100 J/K, μ = 4 mol.
  • Use C = S/μ = 100/4 = 25 J/mol K

Practice Set

• Q1: Heat capacity S is defined as:
  • A) ΔT/ΔQ
  • B) ΔQ/ΔT
  • C) m × s
  • D) (1/m) ΔQ/ΔT
• Q2: Specific heat capacity s depends on:
  • A) Mass only
  • B) Nature of substance and temperature
  • C) Volume only
  • D) Pressure only
• Q3: Molar specific heat capacity at constant volume is denoted by:
  • A) C_p
  • B) C_V
  • C) s
  • D) S

Answer Key

• Q1: B - Heat capacity is heat supplied per unit temperature change ΔQ/ΔT.
• Q2: B - Specific heat depends on the substance's nature and temperature.
• Q3: B - C_V denotes molar specific heat at constant volume.

Memory Aids

• Heat capacity S = ms
• Specific heat s = S/m
• Molar heat C = S/μ
• C_p at constant pressure, C_V at constant volume

Definition and Formula

• Heat capacity (S) = ΔQ/ΔT = ms
• ΔQ: heat supplied to change temperature from T to T + ΔT
• m: mass of the substance
• s: specific heat capacity

Specific Heat Capacity

• Heat absorbed or released per unit mass to change temperature by 1 unit
• Formula: s = (1/m) ΔQ/ΔT = S/m
• Depends on substance nature and temperature

Molar Specific Heat Capacity

• Heat capacity per mole of substance (in gram moles)
• Formula: C = S/μ = (1/μ) ΔQ/ΔT
• Two types:
  • At constant pressure: C_p
  • At constant volume: C_V

Solved Examples

• Ex 1: Calculate specific heat capacity if ΔQ = 500 J, m = 2 kg, ΔT = 5 K.
  • Use s = ΔQ/(m ΔT) = 500/(2 × 5) = 50 J/kg K
• Ex 2: Find molar specific heat capacity if S = 100 J/K, μ = 4 mol.
  • Use C = S/μ = 100/4 = 25 J/mol K

Practice Set

• Q1: Heat capacity S is defined as:
  • A) ΔT/ΔQ
  • B) ΔQ/ΔT
  • C) m × s
  • D) (1/m) ΔQ/ΔT
• Q2: Specific heat capacity s depends on:
  • A) Mass only
  • B) Nature of substance and temperature
  • C) Volume only
  • D) Pressure only
• Q3: Molar specific heat capacity at constant volume is denoted by:
  • A) C_p
  • B) C_V
  • C) s
  • D) S

Answer Key

• Q1: B - Heat capacity is heat supplied per unit temperature change ΔQ/ΔT.
• Q2: B - Specific heat depends on the substance's nature and temperature.
• Q3: B - C_V denotes molar specific heat at constant volume.

Memory Aids

• Heat capacity S = ms
• Specific heat s = S/m
• Molar heat C = S/μ
• C_p at constant pressure, C_V at constant volume

Definition and Formula

• Heat capacity (S) = \( \frac{\Delta Q}{\Delta T} = ms \)
• \( \Delta Q \): heat supplied to change temperature from \( T \) to \( T + \Delta T \)
• \( m \): mass of the substance
• \( s \): specific heat capacity

Specific Heat Capacity

• Heat absorbed or released per unit mass to change temperature by 1 unit
• Formula: \( s = \frac{1}{m} \frac{\Delta Q}{\Delta T} = \frac{S}{m} \)
• Depends on substance nature and temperature

Molar Specific Heat Capacity

• Heat capacity per mole of substance (in gram moles)
• Formula: \( C = \frac{S}{\mu} = \frac{1}{\mu} \frac{\Delta Q}{\Delta T} \)
• Two types:
  • At constant pressure: \( C_p \)
  • At constant volume: \( C_V \)

Solved Examples

• Ex 1: Calculate specific heat capacity if \( \Delta Q = 500 \text{ J} \), \( m = 2 \text{ kg} \), \( \Delta T = 5 \text{ K} \).
  • Use \( s = \frac{\Delta Q}{m \Delta T} = \frac{500}{2 \times 5} = 50 \text{ J/kg K} \)
• Ex 2: Find molar specific heat capacity if \( S = 100 \text{ J/K} \), \( \mu = 4 \text{ mol} \).
  • Use \( C = \frac{S}{\mu} = \frac{100}{4} = 25 \text{ J/mol K} \)

Practice Set

• Q1: Heat capacity \( S \) is defined as:
  • A) \( \frac{\Delta T}{\Delta Q} \)
  • B) \( \frac{\Delta Q}{\Delta T} \)
  • C) \( m \times s \)
  • D) \( \frac{1}{m} \frac{\Delta Q}{\Delta T} \)
• Q2: Specific heat capacity \( s \) depends on:
  • A) Mass only
  • B) Nature of substance and temperature
  • C) Volume only
  • D) Pressure only
• Q3: Molar specific heat capacity at constant volume is denoted by:
  • A) \( C_p \)
  • B) \( C_V \)
  • C) \( s \)
  • D) \( S \)

Answer Key

• Q1: B - Heat capacity is heat supplied per unit temperature change \( \frac{\Delta Q}{\Delta T} \).
• Q2: B - Specific heat depends on the substance's nature and temperature.
• Q3: B - \( C_V \) denotes molar specific heat at constant volume.

Memory Aids

• Heat capacity \( S = ms \)
• Specific heat \( s = \frac{S}{m} \)
• Molar heat \( C = \frac{S}{\mu} \)
• \( C_p \) at constant pressure, \( C_V \) at constant volume

Calorimetry

Concept and Device

• Calorimetry measures specific heat or latent heat of substances.
• A calorimeter is the device used for heat measurement.
• Heat transfer occurs between bodies at different temperatures until thermal equilibrium.

Principle

• Based on the law of conservation of energy.
• Heat lost by hot body = Heat gained by cold body (no heat loss to surroundings).
• Formula: Heat Lost = Heat Gained

Solved Examples

• Ex 1: A 100 g hot metal at 100°C is placed in 200 g water at 25°C. Find final temperature if no heat loss occurs. Use specific heat of metal = 0.4 J/g°C, water = 4.2 J/g°C.
  • Heat lost by metal = Heat gained by water
  • 100 × 0.4 × (100 - T) = 200 × 4.2 × (T - 25)
  • Solve for T: T ≈ 28.6°C

Practice Set

• Q1: What is the principle of calorimetry?
  • A) Heat lost = Heat gained
  • B) Heat lost > Heat gained
  • C) Heat lost < Heat gained
  • D) No heat transfer
• Q2: A 50 g metal at 150°C is dropped into 100 g water at 20°C. Final temperature is 30°C. If specific heat of water is 4.2 J/g°C, find specific heat of metal.
  • A) 0.84 J/g°C
  • B) 0.42 J/g°C
  • C) 1.68 J/g°C
  • D) 0.21 J/g°C
• Q3: In a calorimeter, 200 g of water at 25°C is mixed with 100 g of ice at 0°C. What is the final temperature? (Latent heat of fusion of ice = 336 J/g, specific heat of water = 4.2 J/g°C)
  • A) 0°C
  • B) 5°C
  • C) 10°C
  • D) 15°C

Answer Key

• Q1: A - Principle states heat lost equals heat gained in calorimetry.
• Q2: B - Using heat lost = heat gained, specific heat of metal = (100 × 4.2 × (30-20))/(50 × (150-30)) = 0.42 J/g°C.
• Q3: A - Ice melts absorbing heat, final temperature remains 0°C until all ice melts.

Memory Aids

• Heat lost = Heat gained (No heat loss to surroundings).
• Use formula: m c ΔT for heat transfer.
• Latent heat involved during phase change, no temperature change.

Concept and Device

• Calorimetry measures specific heat or latent heat of substances.
• A calorimeter is the device used for heat measurement.
• Heat transfer occurs between bodies at different temperatures until thermal equilibrium.

Principle

• Based on the law of conservation of energy.
• Heat lost by hot body = Heat gained by cold body (no heat loss to surroundings).
• Formula: Heat Lost = Heat Gained

Solved Examples

• Ex 1: A 100 g hot metal at 100°C is placed in 200 g water at 25°C. Find final temperature if no heat loss occurs. Use specific heat of metal = 0.4 J/g°C, water = 4.2 J/g°C.
  • Heat lost by metal = Heat gained by water
  • 100 × 0.4 × (100 - T) = 200 × 4.2 × (T - 25)
  • Solve for T: T ≈ 28.6°C

Practice Set

• Q1: What is the principle of calorimetry?
  • A) Heat lost = Heat gained
  • B) Heat lost > Heat gained
  • C) Heat lost < Heat gained
  • D) No heat transfer
• Q2: A 50 g metal at 150°C is dropped into 100 g water at 20°C. Final temperature is 30°C. If specific heat of water is 4.2 J/g°C, find specific heat of metal.
  • A) 0.84 J/g°C
  • B) 0.42 J/g°C
  • C) 1.68 J/g°C
  • D) 0.21 J/g°C
• Q3: In a calorimeter, 200 g of water at 25°C is mixed with 100 g of ice at 0°C. What is the final temperature? (Latent heat of fusion of ice = 336 J/g, specific heat of water = 4.2 J/g°C)
  • A) 0°C
  • B) 5°C
  • C) 10°C
  • D) 15°C

Answer Key

• Q1: A - Principle states heat lost equals heat gained in calorimetry.
• Q2: B - Using heat lost = heat gained, specific heat of metal = (100 × 4.2 × (30-20))/(50 × (150-30)) = 0.42 J/g°C.
• Q3: A - Ice melts absorbing heat, final temperature remains 0°C until all ice melts.

Memory Aids

• Heat lost = Heat gained (No heat loss to surroundings).
• Use formula: m c ΔT for heat transfer.
• Latent heat involved during phase change, no temperature change.

Concept and Device

• Calorimetry measures specific heat or latent heat of substances.
• A calorimeter is the device used for heat measurement.
• Heat transfer occurs between bodies at different temperatures until thermal equilibrium.

Principle

• Based on the law of conservation of energy.
• Heat lost by hot body = Heat gained by cold body (no heat loss to surroundings).
• Formula: Heat Lost = Heat Gained

Solved Examples

• Ex 1: A 100 g hot metal at 100°C is placed in 200 g water at 25°C. Find final temperature if no heat loss occurs. Use specific heat of metal = 0.4 J/g°C, water = 4.2 J/g°C.
  • Heat lost by metal = Heat gained by water
  • 100 × 0.4 × (100 - T) = 200 × 4.2 × (T - 25)
  • Solve for T: T ≈ 28.6°C

Practice Set

• Q1: What is the principle of calorimetry?
  • A) Heat lost = Heat gained
  • B) Heat lost > Heat gained
  • C) Heat lost < Heat gained
  • D) No heat transfer
• Q2: A 50 g metal at 150°C is dropped into 100 g water at 20°C. Final temperature is 30°C. If specific heat of water is 4.2 J/g°C, find specific heat of metal.
  • A) 0.84 J/g°C
  • B) 0.42 J/g°C
  • C) 1.68 J/g°C
  • D) 0.21 J/g°C
• Q3: In a calorimeter, 200 g of water at 25°C is mixed with 100 g of ice at 0°C. What is the final temperature? (Latent heat of fusion of ice = 336 J/g, specific heat of water = 4.2 J/g°C)
  • A) 0°C
  • B) 5°C
  • C) 10°C
  • D) 15°C

Answer Key

• Q1: A - Principle states heat lost equals heat gained in calorimetry.
• Q2: B - Using heat lost = heat gained, specific heat of metal = \( \frac{100 \times 4.2 \times (30-20)}{50 \times (150-30)} = 0.42 \) J/g°C.
• Q3: A - Ice melts absorbing heat, final temperature remains 0°C until all ice melts.

Memory Aids

• Heat lost = Heat gained (No heat loss to surroundings).
• Use formula: \( m c \Delta T \) for heat transfer.
• Latent heat involved during phase change, no temperature change.

Concept and Device

• Calorimetry measures specific heat or latent heat of substances.
• A calorimeter is the device used for heat measurement.
• Heat transfer occurs between bodies at different temperatures until thermal equilibrium.

Principle

• Based on the law of conservation of energy.
• Heat lost by hot body = Heat gained by cold body (no heat loss to surroundings).
• Formula: Heat Lost = Heat Gained

Solved Examples

• Ex 1: A 100 g hot metal at 100°C is placed in 200 g water at 25°C. Find final temperature if no heat loss occurs. Use specific heat of metal = 0.4 J/g°C, water = 4.2 J/g°C.
  • Heat lost by metal = Heat gained by water
  • 100 × 0.4 × (100 - T) = 200 × 4.2 × (T - 25)
  • Solve for T: T ≈ 28.6°C

Practice Set

• Q1: What is the principle of calorimetry?
  • A) Heat lost = Heat gained
  • B) Heat lost > Heat gained
  • C) Heat lost < Heat gained
  • D) No heat transfer
• Q2: A 50 g metal at 150°C is dropped into 100 g water at 20°C. Final temperature is 30°C. If specific heat of water is 4.2 J/g°C, find specific heat of metal.
  • A) 0.84 J/g°C
  • B) 0.42 J/g°C
  • C) 1.68 J/g°C
  • D) 0.21 J/g°C
• Q3: In a calorimeter, 200 g of water at 25°C is mixed with 100 g of ice at 0°C. What is the final temperature? (Latent heat of fusion of ice = 336 J/g, specific heat of water = 4.2 J/g°C)
  • A) 0°C
  • B) 5°C
  • C) 10°C
  • D) 15°C

Answer Key

• Q1: A - Principle states heat lost equals heat gained in calorimetry.
• Q2: B - Using heat lost = heat gained, specific heat of metal = \( \frac{100 \times 4.2 \times (30-20)}{50 \times (150-30)} = 0.42 \) J/g°C.
• Q3: A - Ice melts absorbing heat, final temperature remains 0°C until all ice melts.

Memory Aids

• Heat lost = Heat gained (No heat loss to surroundings).
• Use formula: \( m c \Delta T \) for heat transfer.
• Latent heat involved during phase change, no temperature change.

Heat Transfer

Overview

• Heat transfer occurs due to temperature difference between systems.
• Three modes: Conduction, Convection, Radiation.

Solved Examples

• Ex 1: Heat flows from hot to cold due to temperature difference.

Practice Set

• Q1: Which mode of heat transfer requires no medium?
  A. Conduction
  B. Convection
  C. Radiation
  D. None
• Q2: Heat transfer by actual motion of particles occurs in:
  A. Conduction
  B. Convection
  C. Radiation
  D. All
• Q3: Heat transfer due to temperature difference is called:
  A. Work
  B. Heat Transfer
  C. Energy Transfer
  D. Power

Answer Key

• Q1: C - Radiation transfers heat without a medium.
• Q2: B - Convection involves motion of heated particles.
• Q3: B - Heat transfer occurs due to temperature difference.

Memory Aids

• Heat transfer modes: Conduction (solid), Convection (fluid), Radiation (vacuum).
• Heat flows from hot to cold.

Conduction

Key Points

• Heat transfer through a substance without particle motion.
• Heat flow rate: Q ∝ A (T₁ - T₂) t/d
• Formula: Q = K A (T₁ - T₂) t/d, where K is thermal conductivity.
• Rate of heat flow: dQ/dt = -K A dT/dx
• Negative sign indicates heat flows opposite to temperature gradient.

Solved Examples

• Ex 1: Calculate heat flow through rod with area A, thickness d, temperature difference T₁ - T₂, time t, and conductivity K.
  Use Q = K A (T₁ - T₂) t/d.

Practice Set

• Q1: Heat conduction depends on:
  A. Area
  B. Temperature difference
  C. Thickness
  D. All
• Q2: If thickness doubles, heat flow:
  A. Doubles
  B. Halves
  C. Unchanged
  D. Quadruples
• Q3: Rate of heat flow formula is:
  A. dQ/dt = K A dT/dx
  B. dQ/dt = -K A dT/dx
  C. Q = K A (T₁ - T₂) t
  D. None

Answer Key

• Q1: D - All factors affect conduction.
• Q2: B - Heat flow inversely proportional to thickness.
• Q3: B - Correct formula includes negative sign.

Memory Aids

• Conduction formula: Q = K A ΔT t/d.
• Heat flows opposite to temperature gradient.

Key Points

• Heat transfer through a substance without particle motion.
• Heat flow rate: Q ∝ A (T₁ - T₂) t/d
• Formula: Q = K A (T₁ - T₂) t/d, where K is thermal conductivity.
• Rate of heat flow: dQ/dt = -K A dT/dx
• Negative sign indicates heat flows opposite to temperature gradient.

Solved Examples

• Ex 1: Calculate heat flow through rod with area A, thickness d, temperature difference T₁ - T₂, time t, and conductivity K.
  Use Q = K A (T₁ - T₂) t/d.

Practice Set

• Q1: Heat conduction depends on:
  A. Area
  B. Temperature difference
  C. Thickness
  D. All
• Q2: If thickness doubles, heat flow:
  A. Doubles
  B. Halves
  C. Unchanged
  D. Quadruples
• Q3: Rate of heat flow formula is:
  A. dQ/dt = K A dT/dx
  B. dQ/dt = -K A dT/dx
  C. Q = K A (T₁ - T₂) t
  D. None

Answer Key

• Q1: D - All factors affect conduction.
• Q2: B - Heat flow inversely proportional to thickness.
• Q3: B - Correct formula includes negative sign.

Memory Aids

• Conduction formula: Q = K A ΔT t/d.
• Heat flows opposite to temperature gradient.

Key Points

• Heat transfer through a substance without particle motion.
• Heat flow rate: \( Q \propto \frac{A (T_1 - T_2) t}{d} \)
• Formula: \( Q = \frac{K A (T_1 - T_2) t}{d} \), where \(K\) is thermal conductivity.
• Rate of heat flow: \( \frac{dQ}{dt} = -K A \frac{dT}{dx} \)
• Negative sign indicates heat flows opposite to temperature gradient.

Solved Examples

• Ex 1: Calculate heat flow through rod with area \(A\), thickness \(d\), temperature difference \(T_1 - T_2\), time \(t\), and conductivity \(K\).
  Use \( Q = \frac{K A (T_1 - T_2) t}{d} \).

Practice Set

• Q1: Heat conduction depends on:
  A. Area
  B. Temperature difference
  C. Thickness
  D. All
• Q2: If thickness doubles, heat flow:
  A. Doubles
  B. Halves
  C. Unchanged
  D. Quadruples
• Q3: Rate of heat flow formula is:
  A. \( \frac{dQ}{dt} = K A \frac{dT}{dx} \)
  B. \( \frac{dQ}{dt} = -K A \frac{dT}{dx} \)
  C. \( Q = K A (T_1 - T_2) t \)
  D. None

Answer Key

• Q1: D - All factors affect conduction.
• Q2: B - Heat flow inversely proportional to thickness.
• Q3: B - Correct formula includes negative sign.

Memory Aids

• Conduction formula: \( Q = \frac{K A \Delta T t}{d} \).
• Heat flows opposite to temperature gradient.

Key Points

• Heat transfer through a substance without particle motion.
• Heat flow rate: \( Q \propto \frac{A (T_1 - T_2) t}{d} \)
• Formula: \( Q = \frac{K A (T_1 - T_2) t}{d} \), where \(K\) is thermal conductivity.
• Rate of heat flow: \( \frac{dQ}{dt} = -K A \frac{dT}{dx} \)
• Negative sign indicates heat flows opposite to temperature gradient.

Solved Examples

• Ex 1: Calculate heat flow through rod with area \(A\), thickness \(d\), temperature difference \(T_1 - T_2\), time \(t\), and conductivity \(K\).
  Use \( Q = \frac{K A (T_1 - T_2) t}{d} \).

Practice Set

• Q1: Heat conduction depends on:
  A. Area
  B. Temperature difference
  C. Thickness
  D. All
• Q2: If thickness doubles, heat flow:
  A. Doubles
  B. Halves
  C. Unchanged
  D. Quadruples
• Q3: Rate of heat flow formula is:
  A. \( \frac{dQ}{dt} = K A \frac{dT}{dx} \)
  B. \( \frac{dQ}{dt} = -K A \frac{dT}{dx} \)
  C. \( Q = K A (T_1 - T_2) t \)
  D. None

Answer Key

• Q1: D - All factors affect conduction.
• Q2: B - Heat flow inversely proportional to thickness.
• Q3: B - Correct formula includes negative sign.

Memory Aids

• Conduction formula: \( Q = \frac{K A \Delta T t}{d} \).
• Heat flows opposite to temperature gradient.

Convection

Key Points

• Heat transfer by actual motion of heated particles in liquids and gases.
• Causes natural phenomena: trade winds, monsoons, land and sea breezes.

Solved Examples

• Ex 1: Identify convection in atmospheric phenomena like monsoons.

Practice Set

• Q1: Convection occurs in:
  A. Solids
  B. Liquids and gases
  C. Vacuum
  D. None
• Q2: Which causes monsoons?
  A. Conduction
  B. Convection
  C. Radiation
  D. Reflection
• Q3: Heat transfer by particle motion is:
  A. Radiation
  B. Conduction
  C. Convection
  D. None

Answer Key

• Q1: B - Convection occurs in fluids.
• Q2: B - Monsoons caused by convection.
• Q3: C - Particle motion heat transfer is convection.

Memory Aids

• Convection = fluid motion heat transfer.
• Think of wind and water currents.

Radiation

Key Points

• Heat transfer without heating the medium.
• Heat radiations (Q) on a body are partly reflected (Q_r), absorbed (Q_a), and transmitted (Q_t).
• Energy balance: Q = Q_a + Q_r + Q_t
• Coefficients: absorptance a = Q_a/Q, reflectance r = Q_r/Q, transmittance t = Q_t/Q
• Sum: a + r + t = 1
• Perfect bodies: reflector (r=1), absorber (a=1), transmitter (t=1).
• Perfect black body: a=1, r=0, t=0

Solved Examples

• Ex 1: Calculate absorptance if Q_a = 50J and total radiation Q = 100J.
  Use a = Q_a/Q = 0.5.

Practice Set

• Q1: Sum of absorptance, reflectance, transmittance is:
  A. 0
  B. 1
  C. Depends
  D. None
• Q2: A perfect absorber has:
  A. a=1, r=0, t=0
  B. a=0, r=1, t=0
  C. a=0, r=0, t=1
  D. a=1, r=1, t=1
• Q3: Radiation heat transfer requires:
  A. Medium
  B. Vacuum
  C. Solid
  D. Liquid

Answer Key

• Q1: B - Sum of coefficients equals 1.
• Q2: A - Perfect absorber absorbs all radiation.
• Q3: B - Radiation can occur in vacuum.

Memory Aids

• Radiation = no medium needed.
• Sum of a, r, t = 1.
• Black body absorbs all.

Key Points

• Heat transfer without heating the medium.
• Heat radiations (Q) on a body are partly reflected (Q_r), absorbed (Q_a), and transmitted (Q_t).
• Energy balance: Q = Q_a + Q_r + Q_t
• Coefficients: absorptance a = Q_a/Q, reflectance r = Q_r/Q, transmittance t = Q_t/Q
• Sum: a + r + t = 1
• Perfect bodies: reflector (r=1), absorber (a=1), transmitter (t=1).
• Perfect black body: a=1, r=0, t=0

Solved Examples

• Ex 1: Calculate absorptance if Q_a = 50J and total radiation Q = 100J.
  Use a = Q_a/Q = 0.5.

Practice Set

• Q1: Sum of absorptance, reflectance, transmittance is:
  A. 0
  B. 1
  C. Depends
  D. None
• Q2: A perfect absorber has:
  A. a=1, r=0, t=0
  B. a=0, r=1, t=0
  C. a=0, r=0, t=1
  D. a=1, r=1, t=1
• Q3: Radiation heat transfer requires:
  A. Medium
  B. Vacuum
  C. Solid
  D. Liquid

Answer Key

• Q1: B - Sum of coefficients equals 1.
• Q2: A - Perfect absorber absorbs all radiation.
• Q3: B - Radiation can occur in vacuum.

Memory Aids

• Radiation = no medium needed.
• Sum of a, r, t = 1.
• Black body absorbs all.

Key Points

• Heat transfer without heating the medium.
• Heat radiations (Q) on a body are partly reflected (Q_r), absorbed (Q_a), and transmitted (Q_t).
• Energy balance: \( Q = Q_a + Q_r + Q_t \)
• Coefficients: absorptance \(a = \frac{Q_a}{Q}\), reflectance \(r = \frac{Q_r}{Q}\), transmittance \(t = \frac{Q_t}{Q}\)
• Sum: \( a + r + t = 1 \)
• Perfect bodies: reflector (r=1), absorber (a=1), transmitter (t=1).
• Perfect black body: \( a=1, r=0, t=0 \)

Solved Examples

• Ex 1: Calculate absorptance if \( Q_a = 50J \) and total radiation \( Q = 100J \).
  Use \( a = \frac{Q_a}{Q} = 0.5 \).

Practice Set

• Q1: Sum of absorptance, reflectance, transmittance is:
  A. 0
  B. 1
  C. Depends
  D. None
• Q2: A perfect absorber has:
  A. a=1, r=0, t=0
  B. a=0, r=1, t=0
  C. a=0, r=0, t=1
  D. a=1, r=1, t=1
• Q3: Radiation heat transfer requires:
  A. Medium
  B. Vacuum
  C. Solid
  D. Liquid

Answer Key

• Q1: B - Sum of coefficients equals 1.
• Q2: A - Perfect absorber absorbs all radiation.
• Q3: B - Radiation can occur in vacuum.

Memory Aids

• Radiation = no medium needed.
• Sum of a, r, t = 1.
• Black body absorbs all.

Key Points

• Heat transfer without heating the medium.
• Heat radiations (Q) on a body are partly reflected (Q_r), absorbed (Q_a), and transmitted (Q_t).
• Energy balance: \( Q = Q_a + Q_r + Q_t \)
• Coefficients: absorptance \(a = \frac{Q_a}{Q}\), reflectance \(r = \frac{Q_r}{Q}\), transmittance \(t = \frac{Q_t}{Q}\)
• Sum: \( a + r + t = 1 \)
• Perfect bodies: reflector (r=1), absorber (a=1), transmitter (t=1).
• Perfect black body: \( a=1, r=0, t=0 \)

Solved Examples

• Ex 1: Calculate absorptance if \( Q_a = 50J \) and total radiation \( Q = 100J \).
  Use \( a = \frac{Q_a}{Q} = 0.5 \).

Practice Set

• Q1: Sum of absorptance, reflectance, transmittance is:
  A. 0
  B. 1
  C. Depends
  D. None
• Q2: A perfect absorber has:
  A. a=1, r=0, t=0
  B. a=0, r=1, t=0
  C. a=0, r=0, t=1
  D. a=1, r=1, t=1
• Q3: Radiation heat transfer requires:
  A. Medium
  B. Vacuum
  C. Solid
  D. Liquid

Answer Key

• Q1: B - Sum of coefficients equals 1.
• Q2: A - Perfect absorber absorbs all radiation.
• Q3: B - Radiation can occur in vacuum.

Memory Aids

• Radiation = no medium needed.
• Sum of a, r, t = 1.
• Black body absorbs all.

Wien’s Displacement Law

Key Points

• Product of wavelength of max intensity and temperature is constant: λ_m T = b = 2.89 × 10^-3 mK
• As temperature increases, peak wavelength shifts to shorter wavelengths.
• Used to find star temperature: T = b/λ_m

Solved Examples

• Ex 1: Find temperature if peak wavelength λ_m = 500 nm.
  Use T = 2.89 × 10^-3/(500 × 10^-9) = 5780 K.

Practice Set

• Q1: Wien’s constant b value is:
  A. 2.89 × 10^-3 mK
  B. 3.0 × 10⁸ m/s
  C. 5.67 × 10^-8 W/m²K⁴
  D. None
• Q2: Peak wavelength shifts when temperature:
  A. Increases
  B. Decreases
  C. Constant
  D. None
• Q3: Temperature of star with λ_m = 1 μm is:
  A. 2890 K
  B. 5780 K
  C. 1000 K
  D. 2000 K

Answer Key

• Q1: A - Wien’s constant is 2.89 × 10^-3 mK.
• Q2: A - Peak wavelength decreases as temperature increases.
• Q3: A - T = 2.89 × 10^-3/(1 × 10^-6) = 2890 K.

Memory Aids

• Wien’s law: λ_m T = constant.
• Hotter objects peak at shorter wavelengths.

Key Points

• Product of wavelength of max intensity and temperature is constant: λ_m T = b = 2.89 × 10^-3 mK
• As temperature increases, peak wavelength shifts to shorter wavelengths.
• Used to find star temperature: T = b/λ_m

Solved Examples

• Ex 1: Find temperature if peak wavelength λ_m = 500 nm.
  Use T = 2.89 × 10^-3/(500 × 10^-9) = 5780 K.

Practice Set

• Q1: Wien’s constant b value is:
  A. 2.89 × 10^-3 mK
  B. 3.0 × 10⁸ m/s
  C. 5.67 × 10^-8 W/m²K⁴
  D. None
• Q2: Peak wavelength shifts when temperature:
  A. Increases
  B. Decreases
  C. Constant
  D. None
• Q3: Temperature of star with λ_m = 1 μm is:
  A. 2890 K
  B. 5780 K
  C. 1000 K
  D. 2000 K

Answer Key

• Q1: A - Wien’s constant is 2.89 × 10^-3 mK.
• Q2: A - Peak wavelength decreases as temperature increases.
• Q3: A - T = 2.89 × 10^-3/(1 × 10^-6) = 2890 K.

Memory Aids

• Wien’s law: λ_m T = constant.
• Hotter objects peak at shorter wavelengths.

Key Points

• Product of wavelength of max intensity and temperature is constant: λ_m T = b = 2.89 × 10^-3 mK
• As temperature increases, peak wavelength shifts to shorter wavelengths.
• Used to find star temperature: T = b/λ_m

Solved Examples

• Ex 1: Find temperature if peak wavelength λ_m = 500 nm.
  Use T = 2.89 × 10^-3/(500 × 10^-9) = 5780 K.

Practice Set

• Q1: Wien’s constant b value is:
  A. 2.89 × 10^-3 mK
  B. 3.0 × 10⁸ m/s
  C. 5.67 × 10^-8 W/m²K⁴
  D. None
• Q2: Peak wavelength shifts when temperature:
  A. Increases
  B. Decreases
  C. Constant
  D. None
• Q3: Temperature of star with λ_m = 1 μm is:
  A. 2890 K
  B. 5780 K
  C. 1000 K
  D. 2000 K

Answer Key

• Q1: A - Wien’s constant is 2.89 × 10^-3 mK.
• Q2: A - Peak wavelength decreases as temperature increases.
• Q3: A - T = 2.89 × 10^-3/(1 × 10^-6) = 2890 K.

Memory Aids

• Wien’s law: λ_m T = constant.
• Hotter objects peak at shorter wavelengths.

Key Points

• Product of wavelength of max intensity and temperature is constant: λ_m T = b = 2.89 × 10^-3 mK
• As temperature increases, peak wavelength shifts to shorter wavelengths.
• Used to find star temperature: T = b/λ_m

Solved Examples

• Ex 1: Find temperature if peak wavelength λ_m = 500 nm.
  Use T = 2.89 × 10^-3/(500 × 10^-9) = 5780 K.

Practice Set

• Q1: Wien’s constant b value is:
  A. 2.89 × 10^-3 mK
  B. 3.0 × 10⁸ m/s
  C. 5.67 × 10^-8 W/m²K⁴
  D. None
• Q2: Peak wavelength shifts when temperature:
  A. Increases
  B. Decreases
  C. Constant
  D. None
• Q3: Temperature of star with λ_m = 1 μm is:
  A. 2890 K
  B. 5780 K
  C. 1000 K
  D. 2000 K

Answer Key

• Q1: A - Wien’s constant is 2.89 × 10^-3 mK.
• Q2: A - Peak wavelength decreases as temperature increases.
• Q3: A - T = 2.89 × 10^-3/(1 × 10^-6) = 2890 K.

Memory Aids

• Wien’s law: λ_m T = constant.
• Hotter objects peak at shorter wavelengths.

Key Points

• Product of wavelength of max intensity and temperature is constant: λ_m T = b = 2.89 × 10^-3 mK
• As temperature increases, peak wavelength shifts to shorter wavelengths.
• Used to find star temperature: T = b/λ_m

Solved Examples

• Ex 1: Find temperature if peak wavelength λ_m = 500 nm.
  Use T = 2.89 × 10^-3/(500 × 10^-9) = 5780 K.

Practice Set

• Q1: Wien’s constant b value is:
  A. 2.89 × 10^-3 mK
  B. 3.0 × 10⁸ m/s
  C. 5.67 × 10^-8 W/m²K⁴
  D. None
• Q2: Peak wavelength shifts when temperature:
  A. Increases
  B. Decreases
  C. Constant
  D. None
• Q3: Temperature of star with λ_m = 1 μm is:
  A. 2890 K
  B. 5780 K
  C. 1000 K
  D. 2000 K

Answer Key

• Q1: A - Wien’s constant is 2.89 × 10^-3 mK.
• Q2: A - Peak wavelength decreases as temperature increases.
• Q3: A - T = 2.89 × 10^-3/(1 × 10^-6) = 2890 K.

Memory Aids

• Wien’s law: λ_m T = constant.
• Hotter objects peak at shorter wavelengths.

Key Points

• Product of wavelength of max intensity and temperature is constant: \( \lambda_m T = b = 2.89 \times 10^{-3} \text{ mK} \)
• As temperature increases, peak wavelength shifts to shorter wavelengths.
• Used to find star temperature: \( T = \frac{b}{\lambda_m} \)

Solved Examples

• Ex 1: Find temperature if peak wavelength \( \lambda_m = 500 \text{ nm} \).
  Use \( T = \frac{2.89 \times 10^{-3}}{500 \times 10^{-9}} = 5780 \text{ K} \).

Practice Set

• Q1: Wien’s constant \( b \) value is:
  A. \( 2.89 \times 10^{-3} \text{ mK} \)
  B. \( 3.0 \times 10^{8} \text{ m/s} \)
  C. \( 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \)
  D. None
• Q2: Peak wavelength shifts when temperature:
  A. Increases
  B. Decreases
  C. Constant
  D. None
• Q3: Temperature of star with \( \lambda_m = 1 \mu m \) is:
  A. 2890 K
  B. 5780 K
  C. 1000 K
  D. 2000 K

Answer Key

• Q1: A - Wien’s constant is \( 2.89 \times 10^{-3} \text{ mK} \).
• Q2: A - Peak wavelength decreases as temperature increases.
• Q3: A - \( T = \frac{2.89 \times 10^{-3}}{1 \times 10^{-6}} = 2890 K \).

Memory Aids

• Wien’s law: \( \lambda_m T = \text{constant} \).
• Hotter objects peak at shorter wavelengths.

Key Points

• Product of wavelength of max intensity and temperature is constant: \( \lambda_m T = b = 2.89 \times 10^{-3} \text{ mK} \)
• As temperature increases, peak wavelength shifts to shorter wavelengths.
• Used to find star temperature: \( T = \frac{b}{\lambda_m} \)

Solved Examples

• Ex 1: Find temperature if peak wavelength \( \lambda_m = 500 \text{ nm} \).
  Use \( T = \frac{2.89 \times 10^{-3}}{500 \times 10^{-9}} = 5780 \text{ K} \).

Practice Set

• Q1: Wien’s constant \( b \) value is:
  A. \( 2.89 \times 10^{-3} \text{ mK} \)
  B. \( 3.0 \times 10^{8} \text{ m/s} \)
  C. \( 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \)
  D. None
• Q2: Peak wavelength shifts when temperature:
  A. Increases
  B. Decreases
  C. Constant
  D. None
• Q3: Temperature of star with \( \lambda_m = 1 \mu m \) is:
  A. 2890 K
  B. 5780 K
  C. 1000 K
  D. 2000 K

Answer Key

• Q1: A - Wien’s constant is \( 2.89 \times 10^{-3} \text{ mK} \).
• Q2: A - Peak wavelength decreases as temperature increases.
• Q3: A - \( T = \frac{2.89 \times 10^{-3}}{1 \times 10^{-6}} = 2890 K \).

Memory Aids

• Wien’s law: \( \lambda_m T = \text{constant} \).
• Hotter objects peak at shorter wavelengths.

Kirchhoff’s Law

Key Points

• Ratio of emissive power to absorptive power is same for all surfaces at same temperature.
• Equal to emissive power of a perfect black body at that temperature.
• Mathematically: E₁/A₁ = E₂/A₂ = ⋯ = E/A (perfect black body)

Solved Examples

• Ex 1: For two surfaces at same temperature, E₁/A₁ = E₂/A₂.

Practice Set

• Q1: Kirchhoff’s law relates:
  A. Emissive and absorptive powers
  B. Reflectance and transmittance
  C. Temperature and wavelength
  D. None
• Q2: Emissive power ratio is equal to:
  A. Absorptive power ratio
  B. Reflective power ratio
  C. Transmittance ratio
  D. None
• Q3: Kirchhoff’s law applies to:
  A. Different temperatures
  B. Same temperature
  C. Vacuum only
  D. None

Answer Key

• Q1: A - Kirchhoff’s law relates emissive and absorptive powers.
• Q2: A - Ratios of emissive and absorptive powers are equal.
• Q3: B - Applies at same temperature.

Memory Aids

• Emissive/absorptive power ratio constant at given temperature.
• Perfect black body reference.

Key Points

• Ratio of emissive power to absorptive power is same for all surfaces at same temperature.
• Equal to emissive power of a perfect black body at that temperature.
• Mathematically: E₁/A₁ = E₂/A₂ = ⋯ = E/A (perfect black body)

Solved Examples

• Ex 1: For two surfaces at same temperature, E₁/A₁ = E₂/A₂.

Practice Set

• Q1: Kirchhoff’s law relates:
  A. Emissive and absorptive powers
  B. Reflectance and transmittance
  C. Temperature and wavelength
  D. None
• Q2: Emissive power ratio is equal to:
  A. Absorptive power ratio
  B. Reflective power ratio
  C. Transmittance ratio
  D. None
• Q3: Kirchhoff’s law applies to:
  A. Different temperatures
  B. Same temperature
  C. Vacuum only
  D. None

Answer Key

• Q1: A - Kirchhoff’s law relates emissive and absorptive powers.
• Q2: A - Ratios of emissive and absorptive powers are equal.
• Q3: B - Applies at same temperature.

Memory Aids

• Emissive/absorptive power ratio constant at given temperature.
• Perfect black body reference.

Key Points

• Ratio of emissive power to absorptive power is same for all surfaces at same temperature.
• Equal to emissive power of a perfect black body at that temperature.
• Mathematically: \( \frac{E_1}{A_1} = \frac{E_2}{A_2} = \cdots = \frac{E}{A} \) (perfect black body)

Solved Examples

• Ex 1: For two surfaces at same temperature, \( \frac{E_1}{A_1} = \frac{E_2}{A_2} \).

Practice Set

• Q1: Kirchhoff’s law relates:
  A. Emissive and absorptive powers
  B. Reflectance and transmittance
  C. Temperature and wavelength
  D. None
• Q2: Emissive power ratio is equal to:
  A. Absorptive power ratio
  B. Reflective power ratio
  C. Transmittance ratio
  D. None
• Q3: Kirchhoff’s law applies to:
  A. Different temperatures
  B. Same temperature
  C. Vacuum only
  D. None

Answer Key

• Q1: A - Kirchhoff’s law relates emissive and absorptive powers.
• Q2: A - Ratios of emissive and absorptive powers are equal.
• Q3: B - Applies at same temperature.

Memory Aids

• Emissive/absorptive power ratio constant at given temperature.
• Perfect black body reference.

Key Points

• Ratio of emissive power to absorptive power is same for all surfaces at same temperature.
• Equal to emissive power of a perfect black body at that temperature.
• Mathematically: \( \frac{E_1}{A_1} = \frac{E_2}{A_2} = \cdots = \frac{E}{A} \) (perfect black body)

Solved Examples

• Ex 1: For two surfaces at same temperature, \( \frac{E_1}{A_1} = \frac{E_2}{A_2} \).

Practice Set

• Q1: Kirchhoff’s law relates:
  A. Emissive and absorptive powers
  B. Reflectance and transmittance
  C. Temperature and wavelength
  D. None
• Q2: Emissive power ratio is equal to:
  A. Absorptive power ratio
  B. Reflective power ratio
  C. Transmittance ratio
  D. None
• Q3: Kirchhoff’s law applies to:
  A. Different temperatures
  B. Same temperature
  C. Vacuum only
  D. None

Answer Key

• Q1: A - Kirchhoff’s law relates emissive and absorptive powers.
• Q2: A - Ratios of emissive and absorptive powers are equal.
• Q3: B - Applies at same temperature.

Memory Aids

• Emissive/absorptive power ratio constant at given temperature.
• Perfect black body reference.

Stefan’s Law

Key Points

• Radiant energy emitted per unit area by perfect black body E ∝ T⁴.
• Formula: E = σT⁴, σ = 5.67 × 10^-8 W/m²K⁴.
• For ordinary body: E = eσT⁴, where e is emissivity.
• Stefan-Boltzmann law: E = eAσ(T⁴ - T₀⁴)t when surrounded by body at T₀.

Solved Examples

• Ex 1: Calculate radiant energy for body with emissivity e, area A, temperature T, time t.
  Use Q = eAσT⁴t.

Practice Set

• Q1: Stefan’s constant σ value is:
  A. 5.67 × 10^-8 W/m²K⁴
  B. 2.89 × 10^-3 mK
  C. 3.0 × 10⁸ m/s
  D. None
• Q2: Radiant energy E is proportional to:
  A. T
  B. T²
  C. T⁴
  D. T³
• Q3: Stefan-Boltzmann law includes:
  A. T⁴ - T₀⁴
  B. T - T₀
  C. T² - T₀²
  D. None

Answer Key

• Q1: A - Stefan’s constant value is 5.67 × 10^-8 W/m²K⁴.
• Q2: C - Radiant energy proportional to T⁴.
• Q3: A - Stefan-Boltzmann law uses difference of fourth powers.

Memory Aids

• Stefan’s law: E = σT⁴.
• Emissivity e modifies for ordinary bodies.

Key Points

• Radiant energy emitted per unit area by perfect black body E ∝ T⁴.
• Formula: E = σT⁴, σ = 5.67 × 10^-8 W/m²K⁴.
• For ordinary body: E = eσT⁴, where e is emissivity.
• Stefan-Boltzmann law: E = eAσ(T⁴ - T₀⁴)t when surrounded by body at T₀.

Solved Examples

• Ex 1: Calculate radiant energy for body with emissivity e, area A, temperature T, time t.
  Use Q = eAσT⁴t.

Practice Set

• Q1: Stefan’s constant σ value is:
  A. 5.67 × 10^-8 W/m²K⁴
  B. 2.89 × 10^-3 mK
  C. 3.0 × 10⁸ m/s
  D. None
• Q2: Radiant energy E is proportional to:
  A. T
  B. T²
  C. T⁴
  D. T³
• Q3: Stefan-Boltzmann law includes:
  A. T⁴ - T₀⁴
  B. T - T₀
  C. T² - T₀²
  D. None

Answer Key

• Q1: A - Stefan’s constant value is 5.67 × 10^-8 W/m²K⁴.
• Q2: C - Radiant energy proportional to T⁴.
• Q3: A - Stefan-Boltzmann law uses difference of fourth powers.

Memory Aids

• Stefan’s law: E = σT⁴.
• Emissivity e modifies for ordinary bodies.

Key Points

• Radiant energy emitted per unit area by perfect black body \( E \propto T^4 \).
• Formula: \( E = \sigma T^4 \), \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \).
• For ordinary body: \( E = e \sigma T^4 \), where \( e \) is emissivity.
• Stefan-Boltzmann law: \( E = e A \sigma (T^4 - T_0^4) t \) when surrounded by body at \( T_0 \).

Solved Examples

• Ex 1: Calculate radiant energy for body with emissivity \( e \), area \( A \), temperature \( T \), time \( t \).
  Use \( Q = e A \sigma T^4 t \).

Practice Set

• Q1: Stefan’s constant \( \sigma \) value is:
  A. \( 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \)
  B. \( 2.89 \times 10^{-3} \text{ mK} \)
  C. \( 3.0 \times 10^{8} \text{ m/s} \)
  D. None
• Q2: Radiant energy \( E \) is proportional to:
  A. \( T \)
  B. \( T^2 \)
  C. \( T^4 \)
  D. \( T^3 \)
• Q3: Stefan-Boltzmann law includes:
  A. \( T^4 - T_0^4 \)
  B. \( T - T_0 \)
  C. \( T^2 - T_0^2 \)
  D. None

Answer Key

• Q1: A - Stefan’s constant value is \( 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \).
• Q2: C - Radiant energy proportional to \( T^4 \).
• Q3: A - Stefan-Boltzmann law uses difference of fourth powers.

Memory Aids

• Stefan’s law: \( E = \sigma T^4 \).
• Emissivity \( e \) modifies for ordinary bodies.

Key Points

• Radiant energy emitted per unit area by perfect black body \( E \propto T^4 \).
• Formula: \( E = \sigma T^4 \), \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \).
• For ordinary body: \( E = e \sigma T^4 \), where \( e \) is emissivity.
• Stefan-Boltzmann law: \( E = e A \sigma (T^4 - T_0^4) t \) when surrounded by body at \( T_0 \).

Solved Examples

• Ex 1: Calculate radiant energy for body with emissivity \( e \), area \( A \), temperature \( T \), time \( t \).
  Use \( Q = e A \sigma T^4 t \).

Practice Set

• Q1: Stefan’s constant \( \sigma \) value is:
  A. \( 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \)
  B. \( 2.89 \times 10^{-3} \text{ mK} \)
  C. \( 3.0 \times 10^{8} \text{ m/s} \)
  D. None
• Q2: Radiant energy \( E \) is proportional to:
  A. \( T \)
  B. \( T^2 \)
  C. \( T^4 \)
  D. \( T^3 \)
• Q3: Stefan-Boltzmann law includes:
  A. \( T^4 - T_0^4 \)
  B. \( T - T_0 \)
  C. \( T^2 - T_0^2 \)
  D. None

Answer Key

• Q1: A - Stefan’s constant value is \( 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \).
• Q2: C - Radiant energy proportional to \( T^4 \).
• Q3: A - Stefan-Boltzmann law uses difference of fourth powers.

Memory Aids

• Stefan’s law: \( E = \sigma T^4 \).
• Emissivity \( e \) modifies for ordinary bodies.

Ohm’s Law for Thermal Conduction

Key Points

• Thermal current: dQ/dt = (T₁ - T₂)/R_th
• Thermal resistance: R_th = L/(K A)
• Equivalent to Ohm’s law with temperature as potential difference.
• Rate of heat flow: H = ΔT/R_th

Solved Examples

• Ex 1: Calculate thermal current for rod with T₁, T₂, L, K, A.
  Use H = K A (T₁ - T₂)/L.

Practice Set

• Q1: Thermal resistance R_th is:
  A. L/(K A)
  B. K A/L
  C. L K A
  D. None
• Q2: Thermal current is analogous to:
  A. Voltage
  B. Resistance
  C. Electric current
  D. Power
• Q3: Rate of heat flow formula:
  A. H = ΔT/R_th
  B. H = R_thΔT
  C. H = ΔTR_th
  D. None

Answer Key

• Q1: A - Thermal resistance formula.
• Q2: C - Thermal current analogous to electric current.
• Q3: A - Rate of heat flow formula.

Memory Aids

• Thermal resistance R_th = L/(K A).
• Heat flow like current, temperature like voltage.

Key Points

• Thermal current: dQ/dt = (T₁ - T₂)/R_th
• Thermal resistance: R_th = L/(K A)
• Equivalent to Ohm’s law with temperature as potential difference.
• Rate of heat flow: H = ΔT/R_th

Solved Examples

• Ex 1: Calculate thermal current for rod with T₁, T₂, L, K, A.
  Use H = K A (T₁ - T₂)/L.

Practice Set

• Q1: Thermal resistance R_th is:
  A. L/(K A)
  B. K A/L
  C. L K A
  D. None
• Q2: Thermal current is analogous to:
  A. Voltage
  B. Resistance
  C. Electric current
  D. Power
• Q3: Rate of heat flow formula:
  A. H = ΔT/R_th
  B. H = R_thΔT
  C. H = ΔTR_th
  D. None

Answer Key

• Q1: A - Thermal resistance formula.
• Q2: C - Thermal current analogous to electric current.
• Q3: A - Rate of heat flow formula.

Memory Aids

• Thermal resistance R_th = L/(K A).
• Heat flow like current, temperature like voltage.

Key Points

• Thermal current: \( \frac{dQ}{dt} = \frac{T_1 - T_2}{R_{th}} \)
• Thermal resistance: \( R_{th} = \frac{L}{K A} \)
• Equivalent to Ohm’s law with temperature as potential difference.
• Rate of heat flow: \( H = \frac{\Delta T}{R_{th}} \)

Solved Examples

• Ex 1: Calculate thermal current for rod with \( T_1, T_2, L, K, A \).
  Use \( H = \frac{K A (T_1 - T_2)}{L} \).

Practice Set

• Q1: Thermal resistance \( R_{th} \) is:
  A. \( \frac{L}{K A} \)
  B. \( \frac{K A}{L} \)
  C. \( L K A \)
  D. None
• Q2: Thermal current is analogous to:
  A. Voltage
  B. Resistance
  C. Electric current
  D. Power
• Q3: Rate of heat flow formula:
  A. \( H = \frac{\Delta T}{R_{th}} \)
  B. \( H = R_{th} \Delta T \)
  C. \( H = \Delta T R_{th} \)
  D. None

Answer Key

• Q1: A - Thermal resistance formula.
• Q2: C - Thermal current analogous to electric current.
• Q3: A - Rate of heat flow formula.

Memory Aids

• Thermal resistance \( R_{th} = \frac{L}{K A} \).
• Heat flow like current, temperature like voltage.

Key Points

• Thermal current: \( \frac{dQ}{dt} = \frac{T_1 - T_2}{R_{th}} \)
• Thermal resistance: \( R_{th} = \frac{L}{K A} \)
• Equivalent to Ohm’s law with temperature as potential difference.
• Rate of heat flow: \( H = \frac{\Delta T}{R_{th}} \)

Solved Examples

• Ex 1: Calculate thermal current for rod with \( T_1, T_2, L, K, A \).
  Use \( H = \frac{K A (T_1 - T_2)}{L} \).

Practice Set

• Q1: Thermal resistance \( R_{th} \) is:
  A. \( \frac{L}{K A} \)
  B. \( \frac{K A}{L} \)
  C. \( L K A \)
  D. None
• Q2: Thermal current is analogous to:
  A. Voltage
  B. Resistance
  C. Electric current
  D. Power
• Q3: Rate of heat flow formula:
  A. \( H = \frac{\Delta T}{R_{th}} \)
  B. \( H = R_{th} \Delta T \)
  C. \( H = \Delta T R_{th} \)
  D. None

Answer Key

• Q1: A - Thermal resistance formula.
• Q2: C - Thermal current analogous to electric current.
• Q3: A - Rate of heat flow formula.

Memory Aids

• Thermal resistance \( R_{th} = \frac{L}{K A} \).
• Heat flow like current, temperature like voltage.

Thermal Conduction in Slabs

Key Points

• Series arrangement: Slabs stacked one after another.
• Heat flow: Q = A (T₁ - T₂) t/(x₁/K₁ + x₂/K₂ + x₃/K₃)
• Equivalent conductivity: 1/K = (x₁/K₁ + x₂/K₂ + x₃/K₃)/(x₁ + x₂ + x₃)
• Parallel arrangement: Slabs side by side.
• Heat flow: Q = (K₁ A₁ + K₂ A₂ + K₃ A₃) (T₁ - T₂)/x t
• Equivalent conductivity: K = (K₁ A₁ + K₂ A₂ + K₃ A₃)/(A₁ + A₂ + A₃)

Solved Examples

• Ex 1: Calculate heat flow through series slabs using given thickness and conductivity.
• Ex 2: Calculate equivalent conductivity for parallel slabs.

Practice Set

• Q1: In series slabs, total resistance is:
  A. Sum of individual resistances
  B. Product of resistances
  C. Average resistance
  D. None
• Q2: Equivalent conductivity in parallel slabs is:
  A. Weighted average
  B. Sum
  C. Difference
  D. None
• Q3: Heat flow in series slabs depends on:
  A. Total thickness
  B. Individual conductivities
  C. Both
  D. None

Answer Key

• Q1: A - Resistances add in series.
• Q2: A - Weighted average by area.
• Q3: C - Both thickness and conductivity affect heat flow.

Memory Aids

• Series: resistances add, conductivity inverse sum.
• Parallel: conductivity weighted average.

Key Points

• Series arrangement: Slabs stacked one after another.
• Heat flow: Q = A (T₁ - T₂) t/(x₁/K₁ + x₂/K₂ + x₃/K₃)
• Equivalent conductivity: 1/K = (x₁/K₁ + x₂/K₂ + x₃/K₃)/(x₁ + x₂ + x₃)
• Parallel arrangement: Slabs side by side.
• Heat flow: Q = (K₁ A₁ + K₂ A₂ + K₃ A₃) (T₁ - T₂)/x t
• Equivalent conductivity: K = (K₁ A₁ + K₂ A₂ + K₃ A₃)/(A₁ + A₂ + A₃)

Solved Examples

• Ex 1: Calculate heat flow through series slabs using given thickness and conductivity.
• Ex 2: Calculate equivalent conductivity for parallel slabs.

Practice Set

• Q1: In series slabs, total resistance is:
  A. Sum of individual resistances
  B. Product of resistances
  C. Average resistance
  D. None
• Q2: Equivalent conductivity in parallel slabs is:
  A. Weighted average
  B. Sum
  C. Difference
  D. None
• Q3: Heat flow in series slabs depends on:
  A. Total thickness
  B. Individual conductivities
  C. Both
  D. None

Answer Key

• Q1: A - Resistances add in series.
• Q2: A - Weighted average by area.
• Q3: C - Both thickness and conductivity affect heat flow.

Memory Aids

• Series: resistances add, conductivity inverse sum.
• Parallel: conductivity weighted average.

Key Points

• Series arrangement: Slabs stacked one after another.
• Heat flow: \( Q = \frac{A (T_1 - T_2) t}{\frac{x_1}{K_1} + \frac{x_2}{K_2} + \frac{x_3}{K_3}} \)
• Equivalent conductivity: \( \frac{1}{K} = \left( \frac{x_1}{K_1} + \frac{x_2}{K_2} + \frac{x_3}{K_3} \right) \frac{1}{x_1 + x_2 + x_3} \)
• Parallel arrangement: Slabs side by side.
• Heat flow: \( Q = (K_1 A_1 + K_2 A_2 + K_3 A_3) \frac{(T_1 - T_2)}{x} t \)
• Equivalent conductivity: \( K = \frac{K_1 A_1 + K_2 A_2 + K_3 A_3}{A_1 + A_2 + A_3} \)

Solved Examples

• Ex 1: Calculate heat flow through series slabs using given thickness and conductivity.
• Ex 2: Calculate equivalent conductivity for parallel slabs.

Practice Set

• Q1: In series slabs, total resistance is:
  A. Sum of individual resistances
  B. Product of resistances
  C. Average resistance
  D. None
• Q2: Equivalent conductivity in parallel slabs is:
  A. Weighted average
  B. Sum
  C. Difference
  D. None
• Q3: Heat flow in series slabs depends on:
  A. Total thickness
  B. Individual conductivities
  C. Both
  D. None

Answer Key

• Q1: A - Resistances add in series.
• Q2: A - Weighted average by area.
• Q3: C - Both thickness and conductivity affect heat flow.

Memory Aids

• Series: resistances add, conductivity inverse sum.
• Parallel: conductivity weighted average.

Key Points

• Series arrangement: Slabs stacked one after another.
• Heat flow: \( Q = \frac{A (T_1 - T_2) t}{\frac{x_1}{K_1} + \frac{x_2}{K_2} + \frac{x_3}{K_3}} \)
• Equivalent conductivity: \( \frac{1}{K} = \left( \frac{x_1}{K_1} + \frac{x_2}{K_2} + \frac{x_3}{K_3} \right) \frac{1}{x_1 + x_2 + x_3} \)
• Parallel arrangement: Slabs side by side.
• Heat flow: \( Q = (K_1 A_1 + K_2 A_2 + K_3 A_3) \frac{(T_1 - T_2)}{x} t \)
• Equivalent conductivity: \( K = \frac{K_1 A_1 + K_2 A_2 + K_3 A_3}{A_1 + A_2 + A_3} \)

Solved Examples

• Ex 1: Calculate heat flow through series slabs using given thickness and conductivity.
• Ex 2: Calculate equivalent conductivity for parallel slabs.

Practice Set

• Q1: In series slabs, total resistance is:
  A. Sum of individual resistances
  B. Product of resistances
  C. Average resistance
  D. None
• Q2: Equivalent conductivity in parallel slabs is:
  A. Weighted average
  B. Sum
  C. Difference
  D. None
• Q3: Heat flow in series slabs depends on:
  A. Total thickness
  B. Individual conductivities
  C. Both
  D. None

Answer Key

• Q1: A - Resistances add in series.
• Q2: A - Weighted average by area.
• Q3: C - Both thickness and conductivity affect heat flow.

Memory Aids

• Series: resistances add, conductivity inverse sum.
• Parallel: conductivity weighted average.

Ice Layer Growth Time

Key Points

• Time to grow ice layer from thickness x₁ to x₂:
• t = (ρL)/(2KT) (x₂² - x₁²), where T is atmosphere temperature.
• Time intervals ratio for thickness changes: Δt₁ : Δt₂ : Δt₃ = 1 : 3 : 5.

Solved Examples

• Ex 1: Calculate time to grow ice from 0 to x using given parameters.

Practice Set

• Q1: Time to grow ice layer depends on:
  A. Thickness squared difference
  B. Temperature
  C. Thermal conductivity
  D. All
• Q2: Ratio of time intervals for thickness growth is:
  A. 1:2:3
  B. 1:3:5
  C. 2:4:6
  D. 1:1:1
• Q3: Ice growth time formula includes:
  A. ρL
  B. K
  C. T
  D. All

Answer Key

• Q1: D - All factors affect ice growth time.
• Q2: B - Time intervals ratio is 1:3:5.
• Q3: D - Formula includes density, latent heat, conductivity, temperature.

Memory Aids

• Ice growth time ∝ (x₂² - x₁²).
• Time intervals increase as odd numbers.

Key Points

• Time to grow ice layer from thickness x₁ to x₂:
• t = (ρL)/(2KT) (x₂² - x₁²), where T is atmosphere temperature.
• Time intervals ratio for thickness changes: Δt₁ : Δt₂ : Δt₃ = 1 : 3 : 5.

Solved Examples

• Ex 1: Calculate time to grow ice from 0 to x using given parameters.

Practice Set

• Q1: Time to grow ice layer depends on:
  A. Thickness squared difference
  B. Temperature
  C. Thermal conductivity
  D. All
• Q2: Ratio of time intervals for thickness growth is:
  A. 1:2:3
  B. 1:3:5
  C. 2:4:6
  D. 1:1:1
• Q3: Ice growth time formula includes:
  A. ρL
  B. K
  C. T
  D. All

Answer Key

• Q1: D - All factors affect ice growth time.
• Q2: B - Time intervals ratio is 1:3:5.
• Q3: D - Formula includes density, latent heat, conductivity, temperature.

Memory Aids

• Ice growth time ∝ (x₂² - x₁²).
• Time intervals increase as odd numbers.

Key Points

• Time to grow ice layer from thickness \( x_1 \) to \( x_2 \):
• \( t = \frac{\rho L}{2 K T} (x_2^2 - x_1^2) \), where \( T \) is atmosphere temperature.
• Time intervals ratio for thickness changes: \( \Delta t_1 : \Delta t_2 : \Delta t_3 = 1 : 3 : 5 \).

Solved Examples

• Ex 1: Calculate time to grow ice from 0 to \( x \) using given parameters.

Practice Set

• Q1: Time to grow ice layer depends on:
  A. Thickness squared difference
  B. Temperature
  C. Thermal conductivity
  D. All
• Q2: Ratio of time intervals for thickness growth is:
  A. 1:2:3
  B. 1:3:5
  C. 2:4:6
  D. 1:1:1
• Q3: Ice growth time formula includes:
  A. \( \rho L \)
  B. \( K \)
  C. \( T \)
  D. All

Answer Key

• Q1: D - All factors affect ice growth time.
• Q2: B - Time intervals ratio is 1:3:5.
• Q3: D - Formula includes density, latent heat, conductivity, temperature.

Memory Aids

• Ice growth time \( \propto (x_2^2 - x_1^2) \).
• Time intervals increase as odd numbers.

Key Points

• Time to grow ice layer from thickness \( x_1 \) to \( x_2 \):
• \( t = \frac{\rho L}{2 K T} (x_2^2 - x_1^2) \), where \( T \) is atmosphere temperature.
• Time intervals ratio for thickness changes: \( \Delta t_1 : \Delta t_2 : \Delta t_3 = 1 : 3 : 5 \).

Solved Examples

• Ex 1: Calculate time to grow ice from 0 to \( x \) using given parameters.

Practice Set

• Q1: Time to grow ice layer depends on:
  A. Thickness squared difference
  B. Temperature
  C. Thermal conductivity
  D. All
• Q2: Ratio of time intervals for thickness growth is:
  A. 1:2:3
  B. 1:3:5
  C. 2:4:6
  D. 1:1:1
• Q3: Ice growth time formula includes:
  A. \( \rho L \)
  B. \( K \)
  C. \( T \)
  D. All

Answer Key

• Q1: D - All factors affect ice growth time.
• Q2: B - Time intervals ratio is 1:3:5.
• Q3: D - Formula includes density, latent heat, conductivity, temperature.

Memory Aids

• Ice growth time \( \propto (x_2^2 - x_1^2) \).
• Time intervals increase as odd numbers.

Newton's Law of Cooling

Explanation

• Rate of cooling -dQ/dt is proportional to temperature difference (T₂ - T₁) if difference < 40°C.
• Mathematically, dQ/dt = k (T₂ - T₁), where k depends on surface area and nature.
• Temperature decreases exponentially: T = T₀ + (T_i - T₀) e^-kt, k = constant.
• Plot of ln(T - T₀) vs time t is a straight line with slope -k and intercept ln(T_i - T₀).
• Shortcut formula: (T₁ - T₂)/t = k ((T₁ + T₂)/2 - T₀) to avoid calculating e^-kt.

Solved Examples

• Ex 1: Find temperature after time t if T_i = 80° C, T₀ = 30° C, k = 0.1 min^-1.
  • Use T = T₀ + (T_i - T₀) e^-kt
  • Calculate T = 30 + (80 - 30) e^-0.1t
• Ex 2: Given ln(T - T₀) vs t graph slope = -0.05, find k.
  • From slope m = -k, so k = 0.05

Practice Set

• Q1: If initial temperature T_i = 90° C, surrounding T₀ = 30° C, and k = 0.2 min^-1, temperature after 5 min is?
  • A) 40°C
  • B) 50°C
  • C) 60°C
  • D) 70°C
• Q2: The plot of ln(T - T₀) vs time is a straight line with slope -0.1. What is the value of k?
  • A) 0.1
  • B) -0.1
  • C) 1
  • D) -1
• Q3: Using shortcut formula, if T₁ = 80° C, T₂ = 60° C, T₀ = 30° C, and time t = 10 min, find k.
  • A) 0.05
  • B) 0.1
  • C) 0.15
  • D) 0.2

Answer Key

• Q1: B) 50°C - Use T = 30 + 60 e^{-0.2 × 5} = 30 + 60 e^-1 ≈ 50° C.
• Q2: A) 0.1 - Slope m = -k, so k = 0.1.
• Q3: A) 0.05 - Using (T₁ - T₂)/t = k ((T₁ + T₂)/2 - T₀), solve for k.

Memory Aids

• Cooling rate ∝ temperature difference (small difference).
• Temperature formula: T = T₀ + (T_i - T₀) e^-kt.
• Plot ln(T - T₀) vs t is straight line with slope -k.
• Shortcut: (T₁ - T₂)/t = k ((T₁ + T₂)/2 - T₀).

Explanation

• Rate of cooling -dQ/dt is proportional to temperature difference (T₂ - T₁) if difference < 40°C.
• Mathematically, dQ/dt = k (T₂ - T₁), where k depends on surface area and nature.
• Temperature decreases exponentially: T = T₀ + (T_i - T₀) e^-kt, k = constant.
• Plot of ln(T - T₀) vs time t is a straight line with slope -k and intercept ln(T_i - T₀).
• Shortcut formula: (T₁ - T₂)/t = k ((T₁ + T₂)/2 - T₀) to avoid calculating e^-kt.

Solved Examples

• Ex 1: Find temperature after time t if T_i = 80° C, T₀ = 30° C, k = 0.1 min^-1.
  • Use T = T₀ + (T_i - T₀) e^-kt
  • Calculate T = 30 + (80 - 30) e^-0.1t
• Ex 2: Given ln(T - T₀) vs t graph slope = -0.05, find k.
  • From slope m = -k, so k = 0.05

Practice Set

• Q1: If initial temperature T_i = 90° C, surrounding T₀ = 30° C, and k = 0.2 min^-1, temperature after 5 min is?
  • A) 40°C
  • B) 50°C
  • C) 60°C
  • D) 70°C
• Q2: The plot of ln(T - T₀) vs time is a straight line with slope -0.1. What is the value of k?
  • A) 0.1
  • B) -0.1
  • C) 1
  • D) -1
• Q3: Using shortcut formula, if T₁ = 80° C, T₂ = 60° C, T₀ = 30° C, and time t = 10 min, find k.
  • A) 0.05
  • B) 0.1
  • C) 0.15
  • D) 0.2

Answer Key

• Q1: B) 50°C - Use T = 30 + 60 e^{-0.2 × 5} = 30 + 60 e^-1 ≈ 50° C.
• Q2: A) 0.1 - Slope m = -k, so k = 0.1.
• Q3: A) 0.05 - Using (T₁ - T₂)/t = k ((T₁ + T₂)/2 - T₀), solve for k.

Memory Aids

• Cooling rate ∝ temperature difference (small difference).
• Temperature formula: T = T₀ + (T_i - T₀) e^-kt.
• Plot ln(T - T₀) vs t is straight line with slope -k.
• Shortcut: (T₁ - T₂)/t = k ((T₁ + T₂)/2 - T₀).

Explanation

• Rate of cooling -dQ/dt is proportional to temperature difference (T₂ - T₁) if difference < 40°C.
• Mathematically, dQ/dt = k (T₂ - T₁), where k depends on surface area and nature.
• Temperature decreases exponentially: T = T₀ + (T_i - T₀) e^-kt, k = constant.
• Plot of ln(T - T₀) vs time t is a straight line with slope -k and intercept ln(T_i - T₀).
• Shortcut formula: (T₁ - T₂)/t = k ((T₁ + T₂)/2 - T₀) to avoid calculating e^-kt.

Solved Examples

• Ex 1: Find temperature after time t if T_i = 80° C, T₀ = 30° C, k = 0.1 min^-1.
  • Use T = T₀ + (T_i - T₀) e^-kt
  • Calculate T = 30 + (80 - 30) e^-0.1t
• Ex 2: Given ln(T - T₀) vs t graph slope = -0.05, find k.
  • From slope m = -k, so k = 0.05

Practice Set

• Q1: If initial temperature T_i = 90° C, surrounding T₀ = 30° C, and k = 0.2 min^-1, temperature after 5 min is?
  • A) 40°C
  • B) 50°C
  • C) 60°C
  • D) 70°C
• Q2: The plot of ln(T - T₀) vs time is a straight line with slope -0.1. What is the value of k?
  • A) 0.1
  • B) -0.1
  • C) 1
  • D) -1
• Q3: Using shortcut formula, if T₁ = 80° C, T₂ = 60° C, T₀ = 30° C, and time t = 10 min, find k.
  • A) 0.05
  • B) 0.1
  • C) 0.15
  • D) 0.2

Answer Key

• Q1: B) 50°C - Use T = 30 + 60 e^{-0.2 × 5} = 30 + 60 e^-1 ≈ 50° C.
• Q2: A) 0.1 - Slope m = -k, so k = 0.1.
• Q3: A) 0.05 - Using (T₁ - T₂)/t = k ((T₁ + T₂)/2 - T₀), solve for k.

Memory Aids

• Cooling rate ∝ temperature difference (small difference).
• Temperature formula: T = T₀ + (T_i - T₀) e^-kt.
• Plot ln(T - T₀) vs t is straight line with slope -k.
• Shortcut: (T₁ - T₂)/t = k ((T₁ + T₂)/2 - T₀).

Explanation

• Rate of cooling \( -\frac{dQ}{dt} \) is proportional to temperature difference \( (T_2 - T_1) \) if difference < 40°C.
• Mathematically, \( \frac{dQ}{dt} = k (T_2 - T_1) \), where \( k \) depends on surface area and nature.
• Temperature decreases exponentially: \( T = T_0 + (T_i - T_0) e^{-kt} \), \( k \) = constant.
• Plot of \( \ln(T - T_0) \) vs time \( t \) is a straight line with slope \( -k \) and intercept \( \ln(T_i - T_0) \).
• Shortcut formula: \( \frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_0 \right) \) to avoid calculating \( e^{-kt} \).

Solved Examples

• Ex 1: Find temperature after time \( t \) if \( T_i = 80^\circ C \), \( T_0 = 30^\circ C \), \( k = 0.1 \) min\(^{-1}\).
  • Use \( T = T_0 + (T_i - T_0) e^{-kt} \)
  • Calculate \( T = 30 + (80 - 30) e^{-0.1t} \)
• Ex 2: Given \( \ln(T - T_0) \) vs \( t \) graph slope \( = -0.05 \), find \( k \).
  • From slope \( m = -k \), so \( k = 0.05 \)

Practice Set

• Q1: If initial temperature \( T_i = 90^\circ C \), surrounding \( T_0 = 30^\circ C \), and \( k = 0.2 \) min\(^{-1}\), temperature after 5 min is?
  • A) 40°C
  • B) 50°C
  • C) 60°C
  • D) 70°C
• Q2: The plot of \( \ln(T - T_0) \) vs time is a straight line with slope \(-0.1\). What is the value of \( k \)?
  • A) 0.1
  • B) -0.1
  • C) 1
  • D) -1
• Q3: Using shortcut formula, if \( T_1 = 80^\circ C \), \( T_2 = 60^\circ C \), \( T_0 = 30^\circ C \), and time \( t = 10 \) min, find \( k \).
  • A) 0.05
  • B) 0.1
  • C) 0.15
  • D) 0.2

Answer Key

• Q1: B) 50°C - Use \( T = 30 + 60 e^{-0.2 \times 5} = 30 + 60 e^{-1} \approx 50^\circ C \).
• Q2: A) 0.1 - Slope \( m = -k \), so \( k = 0.1 \).
• Q3: A) 0.05 - Using \( \frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_0 \right) \), solve for \( k \).

Memory Aids

• Cooling rate ∝ temperature difference (small difference).
• Temperature formula: \( T = T_0 + (T_i - T_0) e^{-kt} \).
• Plot \( \ln(T - T_0) \) vs \( t \) is straight line with slope \( -k \).
• Shortcut: \( \frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_0 \right) \).