Thermal Properties of Matter

CBSE Class 11 Physics â€¢ Chapter 10

Chapter Overview

This chapter covers the fundamental concepts of Thermal Properties of Matter. Ensure you understand the definitions and derivations thoroughly.

10.1 Introduction

Thermal Properties: Properties related to heat and temperature. Matter expands on heating and changes state (Solid $\rightarrow$ Liquid $\rightarrow$ Gas).

10.2 Temperature and Heat

Temperature: A relative measure or indication of hotness or coldness of a body. It determines the direction of flow of heat.

Heat: A form of energy transferred between two systems or a system and its surroundings by virtue of temperature difference.

SI Unit of Heat: Joule (J). Common unit: Calorie (cal). $1 \text{ cal} = 4.186 \text{ J}$.

10.3 Measurement of Temperature

Scales: Celsius ($^\circ C$), Fahrenheit ($^\circ F$), Kelvin ($K$).

$$ \frac{C}{100} = \frac{F - 32}{180} = \frac{K - 273.15}{100} $$

Absolute Zero: $-273.15^\circ C$ or $0 K$. Molecular motion theoretically ceases.

Practice Problem 1

Q: At what temperature do the Celsius and Fahrenheit scales give the same reading?

Ans: Let $C = F = x$.
$\frac{x}{5} = \frac{x - 32}{9} \Rightarrow 9x = 5x - 160 \Rightarrow 4x = -160 \Rightarrow x = -40$.
So, $-40^\circ C = -40^\circ F$.

10.4 Ideal Gas Equation

Boyle's Law: $PV = \text{const}$ (at constant T).

Charles' Law: $V/T = \text{const}$ (at constant P).

Combining these gives the Ideal Gas Equation:

$$ PV = nRT $$

where $n = \text{moles}$, $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$ (Universal Gas Constant).

Absolute Temperature: $T = 273.15 + t^\circ C$.

10.5 Thermal Expansion

Figure 10.1: Types of Thermal Expansion

Linear Expansion ($\alpha$): $\Delta L = \alpha L \Delta T$.
Area Expansion ($\beta$): $\Delta A = \beta A \Delta T$. ($\beta = 2\alpha$)
Volume Expansion ($\gamma$): $\Delta V = \gamma V \Delta T$. ($\gamma = 3\alpha$)

Practice Problem 2

Q: A steel tape 1m long is valid at $10^\circ C$. On a hot day at $35^\circ C$, what will be the percentage error in measurement? ($\alpha = 1.2 \times 10^{-5} K^{-1}$)

Ans: $\Delta L = \alpha L \Delta T = 1.2 \times 10^{-5} \times 1 \times (35 - 10)$
$\Delta L = 1.2 \times 10^{-5} \times 25 = 30 \times 10^{-5} = 0.0003 \text{ m}$.
% Error = $(\Delta L / L) \times 100 = 0.03 \%$. Tape expands, reads less.

10.6 Specific Heat Capacity

Specific Heat Capacity ($s$): Amount of heat required to raise the temperature of unit mass of substance by $1^\circ C$.

$$ s = \frac{\Delta Q}{m \Delta T} $$

SI Unit: $J kg^{-1} K^{-1}$. For Water: $s_w \approx 4186 \text{ J kg}^{-1} K^{-1}$.

Molar Specific Heat ($C$): Heat required for 1 mole. $C = \frac{\Delta Q}{n \Delta T}$.

$C_p$: At constant pressure.
$C_v$: At constant volume.
$C_p - C_v = R$ (Mayer's Relation for Ideal Gas).

10.7 Calorimetry

Figure 10.2: Calorimeter for measuring Specific Heat

Principle of Calorimetry: Heat Lost by Hot Body = Heat Gained by Cold Body (in an isolated system).

$$ m_1 s_1 (T_1 - T_{mix}) = m_2 s_2 (T_{mix} - T_2) $$

Practice Problem 3 (Mixture)

Q: 0.1 kg of ice at $0^\circ C$ is mixed with 0.3 kg of water at $50^\circ C$. Find final temperature. ($L_{fusion} = 3.35 \times 10^5$, $s_w = 4186$)

Ans: Heat to melt ice $Q_1 = mL = 0.1 \times 3.35 \times 10^5 = 33500 \text{ J}$.
Max heat from water cooling to $0^\circ C$: $Q_2 = m s \Delta T = 0.3 \times 4186 \times 50 \approx 62790 \text{ J}$.
Since $Q_2 > Q_1$, all ice melts. Let final temp be $T$.
Heat gained (Ice melt + warming) = Heat lost (Water cooling)
$33500 + 0.1 \times 4186 \times (T - 0) = 0.3 \times 4186 \times (50 - T)$
$33500 + 418.6 T = 62790 - 1255.8 T \Rightarrow 1674.4 T = 29290 \Rightarrow T \approx \mathbf{17.5^\circ C}$.

10.8 Change of State

Latent Heat ($L$): Heat required to change the state of unit mass substance at constant temperature.

$$ Q = m L $$

Heat of Fusion ($L_f$): Solid $\leftrightarrow$ Liquid. ($3.35 \times 10^5 \text{ J/kg}$ for water).
Heat of Vaporization ($L_v$): Liquid $\leftrightarrow$ Gas. ($2.26 \times 10^6 \text{ J/kg}$ for water).

Triple Point: Temperature and pressure at which solid, liquid, and vapour phases coexist in equilibrium. (Water: $273.16 \text{ K}, 611 \text{ Pa}$).

10.9 Heat Transfer

Modes of Heat Transfer:

Conduction: Heat transfer through direct contact without actual movement of matter (Solids).
Convection: Heat transfer by actual movement of matter (Fluids).
Radiation: Heat transfer via electromagnetic waves (No medium required).

Thermal Conductivity (Conduction):

Rate of heat flow $H = \frac{Q}{t}$ across a rod of Area A and length L:

$$ H = K A \frac{T_1 - T_2}{L} $$

where $K$ is the coefficient of thermal conductivity.

Figure 10.3: Conduction of Heat

Blackbody Radiation:

Stefan-Boltzmann Law: Energy emitted per unit time $E = \sigma A T^4$. ($\sigma = 5.67 \times 10^{-8} W m^{-2} K^{-4}$).

Wien's Displacement Law: $\lambda_m T = \text{constant} (b)$. ($b = 2.9 \times 10^{-3} m K$).

10.1 Introduction

Thermal Properties: Properties related to heat and temperature. Matter expands on heating and changes state (Solid $\rightarrow$ Liquid $\rightarrow$ Gas).

10.2 Temperature and Heat

Temperature: A relative measure or indication of hotness or coldness of a body. It determines the direction of flow of heat.

Heat: A form of energy transferred between two systems or a system and its surroundings by virtue of temperature difference.

SI Unit of Heat: Joule (J). Common unit: Calorie (cal). $1 \text{ cal} = 4.186 \text{ J}$.

10.3 Measurement of Temperature

Scales: Celsius ($^\circ C$), Fahrenheit ($^\circ F$), Kelvin ($K$).

$$ \frac{C}{100} = \frac{F - 32}{180} = \frac{K - 273.15}{100} $$

Absolute Zero: $-273.15^\circ C$ or $0 K$. Molecular motion theoretically ceases.

Practice Problem 1

Q: At what temperature do the Celsius and Fahrenheit scales give the same reading?

Ans: Let $C = F = x$.
$\frac{x}{5} = \frac{x - 32}{9} \Rightarrow 9x = 5x - 160 \Rightarrow 4x = -160 \Rightarrow x = -40$.
So, $-40^\circ C = -40^\circ F$.

10.4 Ideal Gas Equation

Boyle's Law: $PV = \text{const}$ (at constant T).

Charles' Law: $V/T = \text{const}$ (at constant P).

Combining these gives the Ideal Gas Equation:

$$ PV = nRT $$

where $n = \text{moles}$, $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$ (Universal Gas Constant).

Absolute Temperature: $T = 273.15 + t^\circ C$.

10.5 Thermal Expansion

Figure 10.1: Types of Thermal Expansion

Linear Expansion ($\alpha$): $\Delta L = \alpha L \Delta T$.
Area Expansion ($\beta$): $\Delta A = \beta A \Delta T$. ($\beta = 2\alpha$)
Volume Expansion ($\gamma$): $\Delta V = \gamma V \Delta T$. ($\gamma = 3\alpha$)

Practice Problem 2

Q: A steel tape 1m long is valid at $10^\circ C$. On a hot day at $35^\circ C$, what will be the percentage error in measurement? ($\alpha = 1.2 \times 10^{-5} K^{-1}$)

10.6 Specific Heat Capacity

Specific Heat Capacity ($s$): Amount of heat required to raise the temperature of unit mass of substance by $1^\circ C$.

$$ s = \frac{\Delta Q}{m \Delta T} $$

SI Unit: $J kg^{-1} K^{-1}$. For Water: $s_w \approx 4186 \text{ J kg}^{-1} K^{-1}$.

Molar Specific Heat ($C$): Heat required for 1 mole. $C = \frac{\Delta Q}{n \Delta T}$.

$C_p$: At constant pressure.
$C_v$: At constant volume.
$C_p - C_v = R$ (Mayer's Relation for Ideal Gas).

10.7 Calorimetry

Figure 10.2: Calorimeter for measuring Specific Heat

Principle of Calorimetry: Heat Lost by Hot Body = Heat Gained by Cold Body (in an isolated system).

$$ m_1 s_1 (T_1 - T_{mix}) = m_2 s_2 (T_{mix} - T_2) $$

Practice Problem 3 (Mixture)

Q: 0.1 kg of ice at $0^\circ C$ is mixed with 0.3 kg of water at $50^\circ C$. Find final temperature. ($L_{fusion} = 3.35 \times 10^5$, $s_w = 4186$)

10.8 Change of State

Latent Heat ($L$): Heat required to change the state of unit mass substance at constant temperature.

$$ Q = m L $$

Heat of Fusion ($L_f$): Solid $\leftrightarrow$ Liquid. ($3.35 \times 10^5 \text{ J/kg}$ for water).
Heat of Vaporization ($L_v$): Liquid $\leftrightarrow$ Gas. ($2.26 \times 10^6 \text{ J/kg}$ for water).

Triple Point: Temperature and pressure at which solid, liquid, and vapour phases coexist in equilibrium. (Water: $273.16 \text{ K}, 611 \text{ Pa}$).

10.9 Heat Transfer

Modes of Heat Transfer:

Conduction: Heat transfer through direct contact without actual movement of matter (Solids).
Convection: Heat transfer by actual movement of matter (Fluids).
Radiation: Heat transfer via electromagnetic waves (No medium required).

Thermal Conductivity (Conduction):

Rate of heat flow $H = \frac{Q}{t}$ across a rod of Area A and length L:

$$ H = K A \frac{T_1 - T_2}{L} $$

where $K$ is the coefficient of thermal conductivity.

Figure 10.3: Conduction of Heat

Blackbody Radiation:

Stefan-Boltzmann Law: Energy emitted per unit time $E = \sigma A T^4$. ($\sigma = 5.67 \times 10^{-8} W m^{-2} K^{-4}$).

Wien's Displacement Law: $\lambda_m T = \text{constant} (b)$. ($b = 2.9 \times 10^{-3} m K$).

10.10 Newton's Law of Cooling

Statement: The rate of loss of heat by a body is directly proportional to the temperature difference between the body and the surroundings (for small differences).

$$ -\frac{dQ}{dt} = k (T - T_s) $$

$$ \frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right) $$

(Approximate form for numericals).

Practice Problem 4 (Cooling)

Q: 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 $30^\circ C$. The temperature of the surroundings is $20^\circ C$.

Ans: Case 1: $\frac{80-50}{5} = K (\frac{80+50}{2} - 20) \Rightarrow 6 = K(65-20) \Rightarrow 6 = 45K \Rightarrow K = 2/15$.
Case 2: $\frac{60-30}{t} = K (\frac{60+30}{2} - 20) \Rightarrow \frac{30}{t} = \frac{2}{15} (45-20) \Rightarrow \frac{30}{t} = \frac{2}{15}(25)$.
$\frac{30}{t} = \frac{10}{3} \Rightarrow 10t = 90 \Rightarrow t = \mathbf{9 \text{ minutes}}$.

Important Question Bank

Q5. Conceptual

Question: Why are cooking utensils provided with copper bottoms?

Answer: Copper is a very good conductor of heat ($K_{copper} \gg K_{steel}$). It distributes heat uniformly over the bottom, preventing food from getting burnt at hot spots and cooking faster.

Study Tip: Master the derivations in this chapter as they are frequently asked in exams. Practice numericals based on the formulas.