Chapter 11: Thermal Properties of Matter

Introduction

Thermal Properties of Matter

• Temperature measures hotness or coldness of a body.
• During phase changes (e.g., boiling, freezing), temperature remains constant despite heat flow.
• Heat is energy transfer due to temperature difference.
• Thermal properties explain phenomena like why ice feels cold.

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) Mass
• Q2: During freezing, temperature of water
  • A) Increases B) Decreases C) Remains constant D) Fluctuates
• Q3: Ice feels cold because
  • A) It has low temperature B) It absorbs heat from hand C) It emits heat D) It has high heat content

Answer Key

• Q1: B - Temperature indicates hotness or coldness.
• Q2: C - Temperature remains constant during phase change.
• Q3: B - Ice absorbs heat from hand, causing cold sensation.

Memory Aids

• Phase change = constant temperature despite heat flow.
• Temperature = measure of hotness/coldness, not energy.
• Heat = energy transfer due to temperature difference.

Temperature and Heat

Key Concepts

• Temperature: Relative measure of hotness or coldness.
• Heat: Energy transferred due to temperature difference.
• Heating effects: temperature rise, expansion, or state change.
• Units: Temperature (Kelvin, °C), Heat (Joule, J).

Solved Examples

• Ex 1: Define heat.
  • Heat is energy transfer between systems 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 may NOT occur?
  • 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/coldness.
• Q3: D - Heating does not cause temperature decrease.

Memory Aids

• Heat = energy transfer due to \(\Delta T\).
• Temperature = relative hotness, not absolute energy.
• Heating effects: rise in temperature, expansion, or phase change.

Measurement of Temperature

Thermometer and Temperature Measurement

• A thermometer measures temperature using physical properties that vary with temperature.
• Commonly, volume change of liquids like mercury or alcohol is used due to linear variation with temperature.
• Glass thermometers contain such liquids to indicate temperature changes.

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 (Celsius), 32°F (Fahrenheit).
• Steam point: 100°C (Celsius), 212°F (Fahrenheit).
• Common scales: Celsius and Fahrenheit.

Relation Between Celsius and Fahrenheit

• Linear relation: \( \frac{t_F - 32}{180} = \frac{t_C}{100} \)
• Or equivalently: \( t_F = \frac{9}{5} t_C + 32 \)

Solved Examples

Ex 1

• Convert 25°C to Fahrenheit.
• Use formula: \( t_F = \frac{9}{5} \times 25 + 32 = 45 + 32 = 77^{\circ}F \)

Ex 2

• Convert 95°F to Celsius.
• Use formula: \( t_C = \frac{5}{9} (95 - 32) = \frac{5}{9} \times 63 = 35^{\circ}C \)

Practice Set

Q1

What is the boiling point of water in Fahrenheit?

• A) 100°F
• B) 212°F
• C) 180°F
• D) 32°F

Q2

Convert 0°C to Fahrenheit.

• A) 0°F
• B) 32°F
• C) 100°F
• D) 212°F

Q3

If temperature is 68°F, what is it in Celsius?

• A) 20°C
• B) 25°C
• C) 30°C
• D) 15°C

Answer Key

• Q1: B) 212°F — Boiling point of water is 212°F on Fahrenheit scale.
• Q2: B) 32°F — Freezing point of water is 0°C = 32°F.
• Q3: A) 20°C — Using \( t_C = \frac{5}{9} (68 - 32) = 20^{\circ}C \).

Memory Aids

• Use \( t_F = \frac{9}{5} t_C + 32 \) to convert Celsius to Fahrenheit.
• Use \( t_C = \frac{5}{9} (t_F - 32) \) to convert Fahrenheit to Celsius.
• Remember fixed points: 0°C = 32°F (freezing), 100°C = 212°F (boiling).

Ideal Gas Equation and Absolute Temperature

Gas Behaviour and Temperature Scales

• All gases at low density show similar expansion behaviour.
• Variables: Pressure (P), Volume (V), Temperature (T).
• Temperature in Kelvin: \( T = t + 273.15 \), where \( t \) is in °C.
• Boyle's Law (at constant T): \( PV = \text{constant} \).
• Charles' Law (at constant P): \( \frac{V}{T} = \text{constant} \).
• Pressure vs Temperature at constant volume is linear; extrapolates to zero pressure at -273.15°C (absolute zero).
• Absolute zero is the lowest possible temperature; basis of Kelvin scale (0 K = -273.15°C).
• Ideal Gas Equation: \( PV = \mu RT \), where \( \mu \) = moles, \( R = 8.31\,\mathrm{J\,mol^{-1}K^{-1}} \).

Pressure-Temperature Relation for Gases

• Pressure increases linearly with temperature at constant volume.
• Extrapolation of pressure lines for different gases meet at absolute zero (-273.15°C).
• Graph shows pressure vs temperature for gases A, B, C starting near 0 K.
• Pressure is nearly zero at absolute zero; increases with temperature.

Temperature Scales and Conversion

• Kelvin scale starts at absolute zero (0 K = -273.15°C).
• Celsius scale: 0°C (freezing), 100°C (boiling) of water.
• Fahrenheit scale: 32°F (freezing), 212°F (boiling) of water.
• Unit size same for Kelvin and Celsius scales.
• Conversion formula: \( T = t_C + 273.15 \).
• General relation for any scale: \[ \frac{\text{Reading} - \text{Lower fixed point}}{\text{Upper fixed point} - \text{Lower fixed point}} = \text{constant} \]
• Example: \[ \frac{t_C - 0}{100} = \frac{t_F - 32}{180} = \frac{t_R - 0}{80} = \frac{t_K - 273.15}{100} \]

Solved Examples

Ex 1

• Given: \( P = 2\,\mathrm{atm} \), \( V = 5\,\mathrm{L} \), \( T = 300\,\mathrm{K} \), find \( n \) using \( PV = nRT \).
• Use \( R = 0.0821\,\mathrm{L\,atm\,mol^{-1}K^{-1}} \).
• Calculation: \( n = \frac{PV}{RT} = \frac{2 \times 5}{0.0821 \times 300} \approx 0.406\,\mathrm{mol} \).

Ex 2

• Convert 25°C to Kelvin.
• Calculation: \( T = 25 + 273.15 = 298.15\,\mathrm{K} \).

Practice Set

Q1

• At constant temperature, if volume of a gas is halved, pressure will be:
  • A) Halved
  • B) Doubled
  • C) Unchanged
  • D) Quartered

Q2

• What is the temperature in Celsius corresponding to 0 K?
  • A) -273.15°C
  • B) 0°C
  • C) 273.15°C
  • D) -100°C

Q3

• A gas occupies 4 L at 27°C and 1 atm. What volume will it occupy at 127°C and 1 atm?
  • A) 5 L
  • B) 6 L
  • C) 7 L
  • D) 8 L

Answer Key

• Q1: B) Doubled — Boyle's law: \( PV = \text{constant} \), halving volume doubles pressure.
• Q2: A) -273.15°C — 0 K equals absolute zero in Celsius.
• Q3: D) 8 L — Using Charles' law: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), convert temperatures to K and solve.

Memory Aids

• Boyle's Law: \( P \propto \frac{1}{V} \) at constant T.
• Charles' Law: \( V \propto T \) at constant P.
• Ideal Gas Equation: \( PV = nRT \).
• Kelvin = Celsius + 273.15.
• Absolute zero = -273.15°C = 0 K.

Thermal Expansion

Concepts

• Thermal expansion: Increase in dimensions due to temperature rise.
• Occurs in solids (length, area, volume), liquids and gases (volume only).
• Gas state depends on volume, pressure, temperature; expansion can be at constant volume or pressure.
• Different substances and states have different expansion properties.

Solved Examples

• Ex 1: Calculate length after heating: Use \( L_T = L_0 (1 + \alpha \Delta T) \).
• Ex 2: Find volume increase: Use \( V_T = V_0 (1 + \gamma \Delta T) \).

Practice Set

• Q1: Thermal expansion occurs in which states? A) Solids only B) Liquids only C) Gases only D) All states
• Q2: Which expansion is possible only in solids? A) Linear B) Volume C) Area D) Both A and C
• Q3: If a gas is heated at constant volume, which property changes? A) Pressure B) Volume C) Temperature D) Mass

Answer Key

• Q1: D; All states expand on heating.
• Q2: D; Solids show linear and area expansion.
• Q3: A; Pressure changes at constant volume.

Memory Aids

• Thermal expansion = size increase with temperature.
• Solids: length, area, volume expand; Liquids/gases: volume only.
• Gas expansion depends on pressure/volume constraints.

Thermal Expansion in Solids

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: Calculate new length: \( L_T = L_0 (1 + \alpha \Delta T) \).
• Ex 2: Calculate new area: \( A_T = A_0 (1 + \beta \Delta T) \).
• Ex 3: Calculate new volume: \( V_T = V_0 (1 + \gamma \Delta T) \).

Practice Set

• Q1: Coefficient of linear expansion is increase in length per unit length per degree. True or False? A) True B) False C) Sometimes D) None
• Q2: Relation between coefficients: \( \alpha = \frac{\beta}{2} = \frac{\gamma}{3} \). What is \( \beta \) if \( \alpha = 2 \times 10^{-5} \)? A) \(4 \times 10^{-5}\) B) \(1 \times 10^{-5}\) C) \(6 \times 10^{-5}\) D) \(2 \times 10^{-5}\)
• Q3: Volume expansion coefficient \( \gamma \) unit is? A) m B) \( ^\circ C^{-1} \) C) m/s D) kg

Answer Key

• Q1: A; Definition of linear expansion coefficient.
• Q2: A; \( \beta = 2\alpha = 4 \times 10^{-5} \).
• Q3: B; Units are per degree Celsius or Kelvin.

Memory Aids

• \( \alpha \) for length, \( \beta = 2\alpha \) for area, \( \gamma = 3\alpha \) for volume.
• Units: inverse temperature (\( ^\circ C^{-1} \) or K⁻¹).
• Volume expansion > area expansion > linear expansion.

Anomalous Expansion of Water

Key Points

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

Solved Examples

• Ex 1: Explain why ice floats using density variation with temperature.
• Ex 2: Describe volume change of water from 0°C to 10°C.

Practice Set

• Q1: Water has maximum density at? A) 0°C B) 4°C C) 10°C D) 100°C
• Q2: Between 0°C and 4°C, water volume? A) Increases B) Decreases C) Constant D) None
• Q3: Anomalous expansion means? A) Normal expansion B) Negative expansion C) No expansion D) Random expansion

Answer Key

• Q1: B; Maximum density at 4°C.
• Q2: B; Volume decreases on heating from 0°C to 4°C.
• Q3: B; Negative volume expansion in 0–4°C range.

Memory Aids

• Water densest at 4°C.
• Volume decreases 0–4°C, increases above 4°C.
• Ice floats due to lower density than water.

Coefficient of Volume Expansion for Ideal Gas

Concepts

• Ideal gas law: \( PV = \mu RT \).
• At constant pressure: \( P \Delta V = \mu R \Delta T \).
• Volume change: \( \frac{\Delta V}{V} = \frac{\Delta T}{T} \).
• Coefficient of volume expansion \( \gamma \propto \frac{1}{T} \).

Solved Examples

• Ex 1: Calculate volume change for ideal gas heated from 300 K to 330 K at constant pressure.

Practice Set

• Q1: For ideal gas, volume expansion coefficient \( \gamma \) is proportional to? A) T B) 1/T C) P D) V
• Q2: If temperature doubles, volume change ratio is? A) 1 B) 2 C) 0.5 D) 0
• Q3: At constant pressure, volume change \( \Delta V \) is proportional to? A) \( \Delta T \) B) \( \Delta P \) C) \( \Delta V \) D) None

Answer Key

• Q1: B; \( \gamma \propto 1/T \).
• Q2: B; Volume doubles if temperature doubles.
• Q3: A; Volume change proportional to temperature change.

Memory Aids

• Ideal gas volume expansion ~ temperature increase.
• \( \gamma = \frac{1}{T} \) approx. for ideal gases.

Thermal Stress

Concepts

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

Solved Examples

• Ex 1: Calculate thermal stress if rod fixed at ends heated by \( \Delta T \).
• Ex 2: Find force developed in rod with given area and Young's modulus.

Practice Set

• Q1: Thermal stress occurs when expansion is? A) Allowed B) Prevented C) Ignored D) None
• Q2: Thermal strain formula is? A) \( \alpha \Delta T \) B) \( Y \alpha \Delta T \) C) \( \frac{\Delta L}{L} \) D) Both A and C
• Q3: Force due to thermal stress is? A) \( Y A \alpha \Delta T \) B) \( \alpha \Delta T \) C) \( Y \alpha \) D) \( A \Delta T \)

Answer Key

• Q1: B; Stress arises if expansion is prevented.
• Q2: D; Thermal strain = \( \frac{\Delta L}{L} = \alpha \Delta T \).
• Q3: A; Force = Young's modulus × area × expansion.

Memory Aids

• Stress = Young's modulus × strain.
• Strain from thermal expansion = \( \alpha \Delta T \).
• Force depends on cross-sectional area.

Density and Temperature

Concepts

• Volume increases with temperature; mass constant ⇒ density decreases.
• Density change: \( \rho' = \frac{\rho}{1 + \gamma \Delta \theta} \approx \rho (1 - \gamma \Delta \theta) \) if \( \gamma \Delta \theta << 1 \).
• For floating solid: immersed fraction \( f = \frac{\rho_s}{\rho_l} \).
• With temperature increase: \( f' = f \frac{1 + \gamma_l \Delta \theta}{1 + \gamma_s \Delta \theta} \).
• If \( \gamma_l > \gamma_s \), immersed fraction increases.
• Length of pendulum increases with temperature, increasing time period.
• Time period change: \( T' = T (1 + \alpha \Delta \theta)^{1/2} \approx T (1 + \frac{1}{2} \alpha \Delta \theta) \).
• Time lost in interval \( t \): \( \Delta t = \frac{\Delta T}{T} \times t \).

Solved Examples

• Ex 1: Calculate density change for given \( \gamma \) and temperature rise.
• Ex 2: Find time lost by pendulum clock heated by \( \Delta \theta \).

Practice Set

• Q1: Density decreases with temperature because? A) Mass decreases B) Volume increases C) Both D) None
• Q2: Immersed fraction increases if? A) \( \gamma_l > \gamma_s \) B) \( \gamma_s > \gamma_l \) C) Equal D) None
• Q3: Time period of pendulum increases with temperature by? A) \( \alpha \Delta \theta \) B) \( \frac{1}{2} \alpha \Delta \theta \) C) \( 2 \alpha \Delta \theta \) D) None

Answer Key

• Q1: B; Volume increase lowers density.
• Q2: A; Liquid expands more, increasing immersed fraction.
• Q3: B; Time period increases approx. by half the linear expansion factor.

Memory Aids

• Density inversely proportional to volume.
• Immersed fraction depends on relative expansion coefficients.
• Pendulum time period increases with length expansion.

Specific Heat Capacity

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 (kg)
• \( s \): specific heat capacity (J/kg·K)

Specific Heat Capacity

• Amount of heat per unit mass to raise temperature by 1 K
• 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: \( C = \frac{S}{\mu} = \frac{1}{\mu} \frac{\Delta Q}{\Delta T} \)
• \( \mu \): number of moles
• Two types:
  • At constant pressure: \( C_p \)
  • At constant volume: \( C_V \)

Solved Examples

• Ex 1: Calculate heat capacity if \( \Delta Q = 500\,J \), \( \Delta T = 10\,K \), and mass \( m = 2\,kg \).
  • Heat capacity \( S = \frac{\Delta Q}{\Delta T} = \frac{500}{10} = 50\,J/K \)
  • Specific heat \( s = \frac{S}{m} = \frac{50}{2} = 25\,J/kg\cdot K \)
• Ex 2: Find molar specific heat if \( S = 100\,J/K \) and \( \mu = 4\,mol \).
  • \( C = \frac{S}{\mu} = \frac{100}{4} = 25\,J/mol\cdot K \)

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) \( \frac{\Delta Q}{\Delta T} \) is the definition of heat capacity.
• Q2: B) It depends on the nature of substance 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 \( C = \frac{S}{\mu} \)
• Remember: \( C_p > C_V \) for gases

Calorimetry

Concept and Principle

• Calorimetry measures specific heat or latent heat of substances.
• A calorimeter is the device used for heat measurement.
• Heat lost by a hot body = Heat gained by a cold body (no heat loss to surroundings).
• Based on the law of conservation of energy.
• Formula: Heat Lost = Heat Gained

Solved Examples

• Ex 1: A 100 g hot metal at 100°C is dropped into 200 g water at 20°C. Final temperature is 25°C. Find specific heat of metal.
  • Heat lost by metal = Heat gained by water
  • m_metalc_metal(T_initial - T_final) = m_waterc_water(T_final - T_initial)
  • Calculate c_metal using known values.

Practice Set

• Q1: What is the principle behind calorimetry?
  • A) Conservation of momentum
  • B) Conservation of energy
  • C) Newton's third law
  • D) Pascal's law
• Q2: A 50 g metal at 150°C is placed in 100 g water at 25°C. Final temperature is 30°C. If specific heat of water is 4.2 J/g°C, what is the heat lost by metal?
  • A) 2100 J
  • B) 3150 J
  • C) 1050 J
  • D) 4200 J
• Q3: In a calorimeter, 200 g of water at 20°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) 10°C
  • C) 15°C
  • D) 20°C

Answer Key

• Q1: B) Conservation of energy — Calorimetry is based on heat lost equals heat gained.
• Q2: C) 1050 J — Heat gained by water = m c ΔT = 100 × 4.2 × (30-25) = 2100 J; heat lost by metal equals heat gained, so 1050 J is half, check carefully: Actually, heat gained by water = 100 × 4.2 × 5 = 2100 J, so heat lost by metal = 2100 J (Option A). Correction: Option A is correct.
• Q3: A) 0°C — Ice melts absorbing heat; water temperature drops to 0°C before any temperature rise.

Memory Aids

• Heat lost = Heat gained (no heat loss to surroundings).
• Calorimeter measures heat exchange.
• Use m c ΔT for sensible heat; m L for latent heat.

Change of State

Key Concepts

• Matter exists in three states: solid, liquid, gas.
• Change of state: transition between these states.
• Melting: solid to liquid.
• Fusion: liquid to solid.
• Melting point: temperature where solid and liquid coexist in equilibrium; depends on pressure; normal melting point at 1 atm.
• Vaporisation: liquid to vapor.
• Boiling point: temperature where liquid and vapor coexist; depends on pressure; normal boiling point at 1 atm.
• Sublimation: solid to vapor without liquid phase.
• Latent heat (L): heat required to change state of unit mass at constant temperature; formula: L = Q/m.
• SI unit of latent heat: J/kg; practical unit: cal/g.
• Latent heat of fusion: heat to convert solid to liquid; e.g., ice = 80 cal/g.
• Latent heat of vaporisation: heat to convert liquid to vapor; e.g., water = 540 cal/g.
• Triple point: temperature and pressure where solid, liquid, and vapor coexist; for water: 0.016C and 6.11d7 10² Pa.

Solved Examples

• Ex 1: Calculate heat to melt 2 kg ice at 06C. Given latent heat of fusion = 3.33 d7 10⁵ J/kg.
  • Q = mL = 2 d7 3.33 d7 10⁵ = 6.66 d7 10⁵ J
• Ex 2: Heat required to vaporise 1.5 kg water at 1006C. Latent heat of vaporisation = 22.6 d7 10⁵ J/kg.
  • Q = mL = 1.5 d7 22.6 d7 10⁵ = 3.39 d7 10⁶ J

Practice Set

• Q1: What is the process of solid changing directly to vapor called?
  • A) Melting
  • B) Sublimation
  • C) Vaporisation
  • D) Fusion
• Q2: The latent heat of fusion of ice is approximately:
  • A) 80 cal/g
  • B) 540 cal/g
  • C) 100 cal/g
  • D) 2260 cal/g
• Q3: How much heat is required to convert 500 g of ice at 06C to water at 06C? (Latent heat of fusion = 3.33 d7 10⁵ J/kg)
  • A) 1.665 d7 10⁵ J
  • B) 6.66 d7 10⁴ J
  • C) 3.33 d7 10⁵ J
  • D) 8.33 d7 10⁴ J

Answer Key

• Q1: B) Sublimation - solid changes directly to vapor without liquid phase.
• Q2: A) 80 cal/g - standard latent heat of fusion for ice.
• Q3: A) 1.665 d7 10⁵ J - Q = mL = 0.5 d7 3.33 d7 10⁵.

Memory Aids

• Melting = solid to liquid; Fusion = liquid to solid.
• Latent heat formula: L = Q/m.
• Ice fusion latent heat ~ 80 cal/g; water vaporisation ~ 540 cal/g.
• Triple point water: 0.016C, 6.11 d7 10² Pa.

Heat Transfer

Modes of Heat Transfer

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

Solved Examples

• Ex 1: Heat flows from 1006C to 06C through a rod of area 0.01 m8 and thickness 0.5 m with thermal conductivity 200 W/mK. Calculate heat flow in 10 s.
  • Use \( Q = \frac{K A (T_1 - T_2) t}{d} \)
  • \( Q = \frac{200 \times 0.01 \times (100 - 0) \times 10}{0.5} = 4000 \) J

Practice Set

• Q1: Heat transfer by conduction depends on
  • A) Temperature difference only
  • B) Area, temperature difference, time, and thickness
  • C) Thickness only
  • D) Time only
• Q2: Which mode of heat transfer involves actual motion of particles?
  • A) Conduction
  • B) Convection
  • C) Radiation
  • D) None
• Q3: Heat flow \( Q \) through a slab is directly proportional to
  • A) Thickness \( d \)
  • B) Cross-sectional area \( A \)
  • C) Time \( t \)
  • D) Both B and C

Answer Key

• Q1: B - Heat conduction depends on area, temperature difference, time, and inversely on thickness.
• Q2: B - Convection involves actual motion of heated particles.
• Q3: D - Heat flow is directly proportional to area and time.

Memory Aids

• Conduction: Heat flows through solids without particle movement.
• Convection: Heat transfer by fluid motion.
• Radiation: Heat transfer without medium.
• Formula: \( Q = \frac{K A (T_1 - T_2) t}{d} \)

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 the mode of heat transfer responsible for monsoons.
  • Answer: Convection

Practice Set

• Q1: Heat transfer in fluids due to particle movement is called
  • A) Conduction
  • B) Convection
  • C) Radiation
  • D) Evaporation
• Q2: Which of the following is an application of convection?
  • A) Heat conduction in metals
  • B) Formation of trade winds
  • C) Heat from the sun
  • D) Reflection of heat
• Q3: Convection occurs in
  • A) Solids only
  • B) Liquids and gases
  • C) Vacuum
  • D) All states of matter

Answer Key

• Q1: B - Convection involves heat transfer by fluid particle motion.
• Q2: B - Trade winds form due to convection currents.
• Q3: B - Convection occurs in liquids and gases.

Memory Aids

• Convection = fluid motion heat transfer.
• Think: Winds and ocean currents.

Radiation

Key Points

• Heat transfer without heating the intervening medium.
• Thermal radiation \( Q \) is partly absorbed (\( Q_a \)), reflected (\( Q_r \)), and transmitted (\( Q_t \)).
• Relation: \( Q = Q_a + Q_r + Q_t \) and \( a + r + t = 1 \) where \( a, r, t \) are absorptance, reflectance, and transmittance.
• Coefficients \( a, r, t \) are dimensionless ratios.

Solved Examples

• Ex 1: If a body absorbs 70% of radiation and reflects 20%, find transmittance.
  • \( a + r + t = 1 \Rightarrow t = 1 - (a + r) = 1 - (0.7 + 0.2) = 0.1 \)

Practice Set

• Q1: For a body, if absorptance \( a = 0.5 \) and reflectance \( r = 0.3 \), transmittance \( t \) is
  • A) 0.2
  • B) 0.8
  • C) 0.5
  • D) 0.3
• Q2: A perfectly black body has
  • A) \( a = 1, r = 0, t = 0 \)
  • B) \( a = 0, r = 1, t = 0 \)
  • C) \( a = 0, r = 0, t = 1 \)
  • D) \( a = 0.5, r = 0.5, t = 0 \)
• Q3: Which of the following is true for a good reflector?
  • A) High absorptance
  • B) High transmittance
  • C) Low absorptance
  • D) None

Answer Key

• Q1: A - \( t = 1 - (0.5 + 0.3) = 0.2 \)
• Q2: A - Perfect black body absorbs all radiation.
• Q3: C - Good reflector has low absorptance.

Memory Aids

• \( a + r + t = 1 \)
• Black body: \( a=1, r=0, t=0 \)
• Good reflector = bad absorber.

Wien's Displacement Law

Key Points

• Wavelength of maximum radiation intensity \( \lambda_m \) and temperature \( T \) satisfy \( \lambda_m T = b \).
• Constant \( b = 2.89 \times 10^{-3} \) mK.
• As temperature increases, \( \lambda_m \) shifts to shorter wavelengths.
• Used to find temperature of stars from their radiation.

Solved Examples

• Ex 1: Find \( \lambda_m \) for a body at 5800 K.
  • \( \lambda_m = \frac{b}{T} = \frac{2.89 \times 10^{-3}}{5800} = 4.98 \times 10^{-7} \) m (498 nm)

Practice Set

• Q1: Wien's constant \( b \) has value
  • A) \( 2.89 \times 10^{-3} \) mK
  • B) \( 5.67 \times 10^{-8} \) W/m8K8
  • C) \( 3.0 \times 10^{8} \) m/s
  • D) \( 1.38 \times 10^{-23} \) J/K
• Q2: If \( \lambda_m = 500 \) nm, temperature \( T \) is
  • A) 5780 K
  • B) 2890 K
  • C) 1000 K
  • D) 2000 K
• Q3: As temperature increases, \( \lambda_m \) shifts
  • A) To longer wavelengths
  • B) To shorter wavelengths
  • C) Remains constant
  • D) Randomly

Answer Key

• Q1: A - Wien's constant is \( 2.89 \times 10^{-3} \) mK.
• Q2: A - \( T = \frac{b}{\lambda_m} = \frac{2.89 \times 10^{-3}}{500 \times 10^{-9}} = 5780 \) K.
• Q3: B - Peak wavelength shifts to shorter wavelengths with higher temperature.

Memory Aids

• \( \lambda_m T = b \)
• Hotter body emits shorter wavelength radiation.

Stefan-Boltzmann Law

Key Points

• Emissive power \( E \) of a perfect black body: \( E = \sigma T^4 \).
• Stefan's constant \( \sigma = 5.67 \times 10^{-8} \) W/m8K8.
• For ordinary body: \( E = e \sigma T^4 \), where \( e \) is emissivity.
• Net energy radiated when surrounded by temperature \( T_0 \): \( E = e A \sigma (T^4 - T_0^4) t \).

Solved Examples

• Ex 1: Calculate radiant energy emitted by a black body at 500 K with area 2 m8 in 10 s.
  • \( E = \sigma T^4 = 5.67 \times 10^{-8} \times 500^4 = 354.4 \) W/m8
  • Energy \( Q = E A t = 354.4 \times 2 \times 10 = 7088 \) J

Practice Set

• Q1: Stefan's constant \( \sigma \) has value
  • A) \( 5.67 \times 10^{-8} \) W/m8K8
  • B) \( 2.89 \times 10^{-3} \) mK
  • C) \( 3.0 \times 10^{8} \) m/s
  • D) \( 1.38 \times 10^{-23} \) J/K
• Q2: Radiant energy emitted by a body is proportional to
  • A) \( T \)
  • B) \( T^2 \)
  • C) \( T^3 \)
  • D) \( T^4 \)
• Q3: If emissivity \( e = 0.8 \), area \( A = 1 \) m8, temperature \( T = 400 \) K, time \( t = 5 \) s, calculate radiant energy.
  • A) 1.16 J
  • B) 2.91 J
  • C) 3.72 J
  • D) 4.56 J

Answer Key

• Q1: A - Stefan's constant is \( 5.67 \times 10^{-8} \) W/m8K8.
• Q2: D - Radiant energy is proportional to \( T^4 \).
• Q3: B - \( Q = e A \sigma T^4 t = 0.8 \times 1 \times 5.67 \times 10^{-8} \times 400^4 \times 5 = 2.91 \) J.

Memory Aids

• \( E = \sigma T^4 \)
• Emissivity \( e \) scales radiant energy.

Thermal Conduction: Ohm's Law Analogy

Key Points

• Heat flow rate: \( H = \frac{dQ}{dt} = \frac{K A (T_1 - T_2)}{L} \).
• Thermal resistance: \( R_{th} = \frac{L}{K A} \).
• Analogy: \( H = \frac{\Delta T}{R_{th}} \) similar to \( I = \frac{V}{R} \) in electricity.
• Results from Ohm's law apply to thermal conduction.

Solved Examples

• Ex 1: Calculate thermal resistance of a rod of length 0.5 m, area 0.01 m8, and thermal conductivity 200 W/mK.
  • \( R_{th} = \frac{L}{K A} = \frac{0.5}{200 \times 0.01} = 0.25 \) K/W

Practice Set

• Q1: Thermal resistance \( R_{th} \) is
  • A) \( \frac{K A}{L} \)
  • B) \( \frac{L}{K A} \)
  • C) \( K A L \)
  • D) \( \frac{L K}{A} \)
• Q2: Heat flow rate \( H \) is
  • A) \( \frac{\Delta T}{R_{th}} \)
  • B) \( R_{th} \Delta T \)
  • C) \( K A L \)
  • D) \( \frac{L}{K A} \)
• Q3: For a rod with \( R_{th} = 0.5 \) K/W and temperature difference 100 K, heat flow rate is
  • A) 50 W
  • B) 200 W
  • C) 100 W
  • D) 0.5 W

Answer Key

• Q1: B - Thermal resistance \( R_{th} = \frac{L}{K A} \).
• Q2: A - Heat flow rate \( H = \frac{\Delta T}{R_{th}} \).
• Q3: C - \( H = \frac{100}{0.5} = 200 \) W.

Memory Aids

• Thermal resistance analogous to electrical resistance.
• Heat flow \( H = \frac{\Delta T}{R_{th}} \).

Thermal Conduction in Composite Slabs

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 equivalent conductivity for series slabs with \( x_1 = x_2 = x_3 = 1 \) m, \( K_1 = 2 \), \( K_2 = 4 \), \( K_3 = 6 \) W/mK.
  • \( \frac{1}{K} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} = 0.5 + 0.25 + 0.1667 = 0.9167 \)
  • \( K = 1.09 \) W/mK

Practice Set

• Q1: In series slabs, total thermal resistance is
  • A) Sum of individual resistances
  • B) Product of individual resistances
  • C) Average of individual resistances
  • D) None
• Q2: Equivalent conductivity for parallel slabs is
  • A) Weighted average by thickness
  • B) Weighted average by area
  • C) Harmonic mean
  • D) Geometric mean
• Q3: Heat flow through parallel slabs depends on
  • A) Total area and conductivity
  • B) Thickness only
  • C) Temperature difference only
  • D) None

Answer Key

• Q1: A - Thermal resistances add in series.
• Q2: B - Equivalent conductivity is area-weighted average.
• Q3: A - Heat flow depends on total area and conductivity.

Memory Aids

• Series: Resistances add.
• Parallel: Conductivities weighted by area.

Ice Layer Growth Time

Key Points

• Time to grow ice layer from thickness \( x_1 \) to \( x_2 \):
  • \( t = \frac{\rho L}{2 K T} (x_2^2 - x_1^2) \)
  • \( \rho \): density, \( L \): latent heat, \( K \): thermal conductivity, \( T \): atmospheric temperature.
• Time intervals for thickness changes 0 to \( x \), \( x \) to \( 2x \), etc. are in ratio 1:3:5.

Solved Examples

• Ex 1: If time to grow ice from 0 to \( x \) is 1 hour, time from \( x \) to \( 2x \) is
  • 3 hours (ratio 1:3)

Practice Set

• Q1: Time to grow ice layer is proportional to
  • A) Thickness \( x \)
  • B) Square of thickness \( x^2 \)
  • C) Cube of thickness \( x^3 \)
  • D) None
• Q2: Ratio of time intervals for ice growth from 0 to \( x \), \( x \) to \( 2x \), \( 2x \) to \( 3x \) is
  • A) 1:2:3
  • B) 1:3:5
  • C) 1:4:9
  • D) 1:1:1
• Q3: In the formula \( t = \frac{\rho L}{2 K T} (x_2^2 - x_1^2) \), \( T \) is
  • A) Temperature of ice
  • B) Temperature of water
  • C) Temperature of atmosphere
  • D) None

Answer Key

• Q1: B - Time is proportional to square of thickness difference.
• Q2: B - Time intervals ratio is 1:3:5.
• Q3: C - \( T \) is atmospheric temperature.

Memory Aids

• Ice growth time \( \propto x^2 \).
• Time intervals ratio odd numbers: 1, 3, 5...

Newton's Law of Cooling

Key Concepts

• Rate of cooling \( \left(-\frac{dQ}{dt}\right) \) 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: \( T = T_0 + (T_i - T_0) e^{-kt} \), \( T_i \) = initial temp, \( T_0 \) = surrounding temp.
• Plotting \( \ln(T - T_0) \) vs time \( t \) gives 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: Initial temp \( T_i = 80^\circ C \), surrounding temp \( T_0 = 30^\circ C \), \( k = 0.1 \ \text{s}^{-1} \). Find temp after 10 s.
  • Use \( T = T_0 + (T_i - T_0) e^{-kt} \)
  • \( T = 30 + (80 - 30) e^{-0.1 \times 10} = 30 + 50 e^{-1} \approx 30 + 50 \times 0.3679 = 48.4^\circ C \)
• Ex 2: Given \( T_1 = 70^\circ C \), \( T_2 = 50^\circ C \), \( T_0 = 30^\circ C \), time \( t = 5 \) min. Find \( k \).
  • Use shortcut: \( \frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_0 \right) \)
  • \( \frac{70 - 50}{5} = k (60 - 30) \Rightarrow 4 = 30k \Rightarrow k = \frac{4}{30} = 0.1333 \ \text{min}^{-1} \)

Practice Set

• Q1: If the initial temperature of a body is 90°C and surrounding temperature is 30°C, which formula gives temperature at time \( t \)?
  • A) \( T = T_i e^{-kt} \)
  • B) \( T = T_0 + (T_i - T_0) e^{-kt} \)
  • C) \( T = T_0 e^{-kt} \)
  • D) \( T = (T_i - T_0) e^{kt} \)
• Q2: For a cooling body, the plot of \( \ln(T - T_0) \) vs time \( t \) is:
  • A) A horizontal line
  • B) A straight line with positive slope
  • C) A straight line with negative slope
  • D) A parabola
• Q3: A body cools from 80°C to 60°C in 5 minutes in surroundings of 30°C. Using shortcut formula, find \( k \).
  • A) 0.1 min⁻¹
  • B) 0.133 min⁻¹
  • C) 0.2 min⁻¹
  • D) 0.05 min⁻¹

Answer Key

• Q1: B - Formula includes surrounding temperature and exponential decay.
• Q2: C - \( \ln(T - T_0) \) decreases linearly with time, slope = \( -k \).
• Q3: B - Using \( \frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_0 \right) \), \( k = 0.133 \) min⁻¹.

Memory Aids

• Cooling rate ∝ temperature difference (small difference < 40°C).
• Temperature decays exponentially: \( T = T_0 + (T_i - T_0) e^{-kt} \).
• Plot \( \ln(T - T_0) \) vs \( t \) → straight line with slope \( -k \).
• Shortcut: \( \frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_0 \right) \) avoids exponentials.