Class 11 Physics: Thermal Properties of Matter | Physics Q&A

CBSE Class 11 › Unit VII: Properties of Bulk Matter

Thermal Properties of Matter

NCERT Chapter 10 • Heat, Expansion, Calorimetry & Heat Transfer

NCERT 2025–26 Unit VII • ~8 Marks JEE Main • 1-2 Questions

Blacksmith forging iron, illustrating heat transfer and high temperature physics. — Heat and Temperature: The science of hotness and energy transfer.

1. Temperature and Scales

Temperature is a relative measure of hotness or coldness. Heat is the form of energy transferred between two systems (or a system and its surroundings) by virtue of temperature difference.

1.1 Temperature Scales

Scale	Freezing Point	Boiling Point	Conversion
Celsius ( $^\circ$ C)	0	100	–
Fahrenheit ( $^\circ$ F)	32	212	$\dfrac{C}{5} = \dfrac{F-32}{9}$
Kelvin (K)	273.15	373.15	$T_K = T_C + 273.15$

1.2 Ideal Gas Equation

Combining Boyle’s Law ( $PV = \text{constant}$ at constant $T$ ) and Charles’ Law ( $V/T = \text{constant}$ at constant $P$ ), we get:

PV = \mu RT

where $\mu$ = number of moles, $R = 8.31$ J/mol·K (universal gas constant).

1.3 Absolute Zero

Experiments with low-density gases show that at constant volume, $P \propto T$ . Extrapolating to $P = 0$ gives $T = -273.15^\circ$ C, defined as absolute zero (0 K).

Pressure vs Temperature graph for gas thermometer showing extrapolation to absolute zero. — Absolute zero is the foundation of the Kelvin scale.

1.4 Triple Point of Water

The triple point is the unique temperature and pressure where water coexists as solid, liquid, and vapor in equilibrium:

Temperature = 273.16 K (0.01°C)
Pressure = 6.11 × 10⁻³ Pa

This is used to define the Kelvin scale because it’s a reproducible fixed point.

2. Thermal Expansion

Most substances expand on heating and contract on cooling.

Visualizing Linear, Superficial, and Cubical expansion in solids. — Solids expand in all dimensions when heated ( $\alpha, \beta, \gamma$ ).

2.1 Types of Expansion

Linear Expansion ( $\alpha$ ): $\Delta l = \alpha l \Delta T$ (1D: Length)
Area Expansion ( $\beta$ ): $\Delta A = \beta A \Delta T$ (2D: Area)
Volume Expansion ( $\gamma$ ): $\Delta V = \gamma V \Delta T$ (3D: Volume)

Relation: $\alpha : \beta : \gamma \approx 1 : 2 : 3$ . For solids, $\gamma = 3\alpha$ .

2.2 Thermal Stress

When a rod is rigidly fixed and heated, it cannot expand. This induces compressive stress:

\text{Thermal Stress} = Y \alpha \Delta T

where $Y$ = Young’s modulus.

Anomalous Expansion of Water:

Water contracts on heating from 0 $^\circ$ C to 4 $^\circ$ C. Its density is maximum at 4 $^\circ$ C. This property allows aquatic life to survive in frozen lakes.

Problem 1: A steel tape 1m long is correctly calibrated for a temperature of 27.0 $^\circ$ C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 $^\circ$ C. What is the actual length of the steel rod on that day? ( $\alpha_{steel} = 1.2 \times 10^{-5} \text{ K}^{-1}$ )

Show Answer

The tape expands on the hot day, so its markings are farther apart. It measures less than the actual length.
$\Delta T = 45 - 27 = 18^\circ$ C.
Expansion factor = $1 + \alpha \Delta T = 1 + 1.2 \times 10^{-5} \times 18 = 1.000216$ .
True Length = Measured Length $\times$ Expansion factor
$= 63.0 \times 1.000216 = 63.0136$ cm.

3. Calorimetry

Calorimetry implies measurement of heat. Principle: Heat Lost = Heat Gained.

3.1 Specific Heat Capacity ( $s$ or $c$ )

The amount of heat required to raise the temperature of unit mass of a substance by 1 $^\circ$ C.

Q = ms\Delta T

For water, $s_w = 4186$ J kg $^{-1}$ K $^{-1}$ (very high). This makes water an excellent coolant.

3.2 Change of State & Latent Heat ( $L$ )

During a phase change (Melting/Boiling), temperature remains constant. Heat supplied is used to break bonds.

Q = mL

Latent Heat of Fusion ( $L_f$ ): Solid $\leftrightarrow$ Liquid. (Ice: $3.33 \times 10^5$ J/kg)
Latent Heat of Vaporisation ( $L_v$ ): Liquid $\leftrightarrow$ Gas. (Water: $22.6 \times 10^5$ J/kg)

Temperature-Heat graph showing phase changes and latent heat plateaus. — Temperature remains constant during phase change as Latent Heat is absorbed.

3.3 Molar Specific Heat for Gases

For gases, heat capacity depends on process:

$C_v$ : At constant volume (no work done).
$C_p$ : At constant pressure (work done in expansion).

Always $C_p > C_v$ because extra heat is needed for work. For monatomic gases, $C_p - C_v = R$ .

3.4 Regelation

When pressure is applied to ice, its melting point lowers. A weighted wire can pass through an ice block because ice melts under the wire and refreezes above it. This is called regelation.

Problem 2: Calculate the heat required to convert 3 kg of ice at -12 $^\circ$ C to steam at 100 $^\circ$ C.
( $s_{ice}=2100, s_{water}=4186, L_f=3.35 \times 10^5, L_v=2.256 \times 10^6$ J/kg)

Show Answer

1. Ice (-12 to 0): $Q_1 = ms_i \Delta T = 3(2100)(12) = 75600$ J.
2. Melt Ice (0 to 0): $Q_2 = mL_f = 3(3.35 \times 10^5) = 1.005 \times 10^6$ J.
3. Water (0 to 100): $Q_3 = ms_w \Delta T = 3(4186)(100) = 1.2558 \times 10^6$ J.
4. Boil Water (100 to 100): $Q_4 = mL_v = 3(2.256 \times 10^6) = 6.768 \times 10^6$ J.
Total $Q = Q_1 + Q_2 + Q_3 + Q_4 \approx 9.1 \times 10^6$ J.

4. Heat Transfer

Heat flows from high temperature to low temperature via three modes:

Diagram illustrating Conduction, Convection, and Radiation in a single setup. — The three modes of heat transfer: Conduction, Convection, and Radiation.

4.1 Conduction

Heat transfer by molecular collision without bulk motion of matter (mostly solids). Rate of heat flow ( $H$ ):

H = \dfrac{dQ}{dt} = -KA \dfrac{dT}{dx} \approx KA \dfrac{T_1 - T_2}{L}

Where $K$ is Thermal Conductivity. $K_{metal} \gg K_{insulator}$ .

4.2 Thermal Conductivity Table

Material	K (W/m·K)
Silver	406
Copper	385
Water	0.8
Air	0.024

4.3 Series Combination of Rods

For two rods in series (same heat current $H$ ):

H = \dfrac{T_1 - T_2}{\dfrac{L_1}{K_1 A_1} + \dfrac{L_2}{K_2 A_2}}

4.4 Convection

Heat transfer by actual motion of fluid particles (liquids & gases). Examples: Sea breeze, trade winds.

4.5 Radiation

Heat transfer via electromagnetic waves (infrared). Needs no medium. Example: Sunlight.

4.6 Dewar Flask (Thermos)

A thermos minimizes heat transfer by:

Conduction: Vacuum between double walls.
Convection: No air in vacuum.
Radiation: Silvered walls reflect heat.

5. Newton’s Law of Cooling

The rate of loss of heat ( $dQ/dt$ ) of a body is directly proportional to the temperature difference between the body and its surroundings ( $\Delta T = T - T_s$ ), provided $\Delta T$ is small.

-\dfrac{dQ}{dt} = -ms\dfrac{dT}{dt} = k (T - T_s)

-\dfrac{dT}{dt} = \dfrac{k}{ms} (T - T_s)

-\dfrac{dT}{dt} = K (T - T_s)

Exponential cooling curve showing temperature drop over time. — The rate of cooling depends on the temperature difference with surroundings.

Problem 3: A body cools from 80 $^\circ$ C to 50 $^\circ$ C in 5 minutes. Calculate the time it takes to cool from 60 $^\circ$ C to 30 $^\circ$ C. Temperature of surroundings is 20 $^\circ$ C.

Show Answer

Using Avg Rate form: $\dfrac{T_1 - T_2}{t} = K \left( \dfrac{T_1+T_2}{2} - T_s \right)$
Case 1: $\dfrac{80-50}{5} = K(65 - 20) \Rightarrow 6 = 45K \Rightarrow K = \dfrac{6}{45} = \dfrac{2}{15}$ .
Case 2: $\dfrac{60-30}{t} = \dfrac{2}{15}(45 - 20) \Rightarrow \dfrac{30}{t} = \dfrac{2}{15}(25)$ .
$\dfrac{30}{t} = \dfrac{10}{3} \Rightarrow t = \dfrac{90}{10} = 9$ minutes.

6. Blackbody Radiation Laws

Stefan-Boltzmann Law

Energy radiated per second per unit area by a blackbody is proportional to the fourth power of its absolute temperature.

E = \sigma T^4

$\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{K}^{-4}$ (Stefan’s Constant).

6.1 Emissivity

A perfect blackbody has emissivity $e = 1$ . Real bodies have $e < 1$ . The net power radiated is:

H = e\sigma A(T^4 - T_s^4)

Wien’s Displacement Law

The wavelength ( $\lambda_m$ ) corresponding to maximum emission intensity is inversely proportional to absolute temperature.

\lambda_m T = b \quad (\text{Constant})

This explains why hotter stars appear blue (lower $\lambda$ ) and cooler stars appear red (higher $\lambda$ ).

Problem 4 (Thermal Stress): A steel rail (length 10 m, $\alpha = 1.2 \times 10^{-5}$ /K, $Y = 2 \times 10^{11}$ Pa) is prevented from expanding. Find thermal stress if temperature rises by 20°C.

Show Answer

Thermal Stress = $Y \alpha \Delta T = 2 \times 10^{11} \times 1.2 \times 10^{-5} \times 20 = 4.8 \times 10^7$ Pa.

← Previous Chapter Next Chapter: Thermodynamics →

1. Temperature and Scales

1.1 Temperature Scales

1.2 Ideal Gas Equation

1.3 Absolute Zero

1.4 Triple Point of Water

2. Thermal Expansion

2.1 Types of Expansion

2.2 Thermal Stress

3. Calorimetry

3.1 Specific Heat Capacity ( or )

3.2 Change of State & Latent Heat ()

3.3 Molar Specific Heat for Gases

3.4 Regelation

4. Heat Transfer

4.1 Conduction

4.2 Thermal Conductivity Table

4.3 Series Combination of Rods

4.4 Convection

4.5 Radiation

4.6 Dewar Flask (Thermos)

5. Newton’s Law of Cooling

6. Blackbody Radiation Laws

Stefan-Boltzmann Law

6.1 Emissivity

Wien’s Displacement Law

3.1 Specific Heat Capacity ( $s$ or $c$ )

3.2 Change of State & Latent Heat ( $L$ )