Chapter 10: Thermal Properties of Matter
Unit VII – Properties of Bulk Matter
10.1 Temperature and Heat
- Temperature: A measure of the average kinetic energy of molecules (degree of hotness/coldness).
- Heat: Energy transferred due to temperature difference.
Temperature Scales:
\[ \frac{T_C}{100} = \frac{T_F - 32}{180} = \frac{T_K - 273.15}{100} \]
\[ T_K = T_C + 273.15 \]
10.2 Thermal Expansion
Linear Expansion
\[ \Delta L = L_0 \alpha \Delta T \Rightarrow L = L_0(1 + \alpha \Delta T) \]
Area Expansion
\[ \Delta A = A_0 \beta \Delta T \Rightarrow \beta \approx 2\alpha \]
Volume Expansion
\[ \Delta V = V_0 \gamma \Delta T \Rightarrow \gamma \approx 3\alpha \]
| Material | \(\alpha\) (× 10⁻⁶ K⁻¹) |
|---|---|
| Steel | 11 |
| Aluminium | 23 |
| Copper | 17 |
| Glass | 9 |
Anomalous expansion of water: Water expands on cooling from 4°C to 0°C — density maximum at 4°C.
10.3 Specific Heat Capacity
The heat required to raise temperature of mass \(m\) by \(\Delta T\):
\[ Q = mc\Delta T \]
where \(c\) = specific heat capacity (J kg⁻¹ K⁻¹)
| Substance | \(c\) (J kg⁻¹ K⁻¹) |
|---|---|
| Water | 4186 |
| Ice | 2090 |
| Aluminium | 900 |
| Steel | 490 |
Molar specific heat capacity:
\[ Q = n C_m \Delta T \]
10.4 Calorimetry
The principle of calorimetry: Heat lost = Heat gained (in an isolated system)
\[ m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2) \]
10.5 Change of State and Latent Heat
During a phase change (melting, boiling), temperature does not change despite heat being added.
\[ Q = mL \]
where \(L\) = latent heat:
| Change of State | Latent Heat of Water |
|---|---|
| Fusion (melting) | \(L_f = 3.34 \times 10^5 \text{ J/kg}\) |
| Vaporisation (boiling) | \(L_v = 22.6 \times 10^5 \text{ J/kg}\) |
10.6 Heating Curve
10.7 Heat Transfer
Conduction
Heat flows from hot to cold through a material.
Rate of heat flow (Fourier’s law):
\[ \frac{dQ}{dt} = -kA\frac{dT}{dx} \]
where \(k\) = thermal conductivity (W m⁻¹ K⁻¹)
Convection
Heat transfer by bulk fluid motion (e.g., boiling water, sea breeze).
Radiation
Heat transfer via electromagnetic waves — does not need a medium.
Stefan-Boltzmann Law:
\[ P = \sigma A e T^4 \]
where \(\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{K}^{-4}\), \(e\) = emissivity.
Wien’s Displacement Law:
\[ \lambda_{max} T = b = 2.898 \times 10^{-3} \text{ m·K} \]
Key Formulas Summary
| Formula | Quantity |
|---|---|
| \(Q = mc\Delta T\) | Heat absorbed |
| \(Q = mL\) | Latent heat |
| \(L = L_0(1 + \alpha\Delta T)\) | Linear expansion |
| \(\frac{dQ}{dt} = kA\frac{\Delta T}{d}\) | Fourier’s law |
| \(P = \sigma e A T^4\) | Stefan’s law |
| \(\lambda_{max}T = 2.898 \times 10^{-3}\) | Wien’s law |
Practice Questions
Section A – MCQ (1 mark each)
Q1. Which of the following is a good absorber of radiation?
(a) White shiny surface (b) Black rough surface (c) Silver surface (d) Mirror
Answer
(b) Black rough surface — good absorbers are also good emitters (Kirchhoff’s law).
Q2. During the melting of ice, the temperature:
(a) Increases (b) Decreases (c) Remains constant (d) First decreases then increases
Answer
(c) Remains constant — heat absorbed goes into changing state, not changing temperature.
Section B – Short Answer (2–3 marks)
Q3. A 100 g piece of metal at 200°C is dropped into 200 g of water at 20°C. The final temperature is 30°C. Find the specific heat of the metal. (\(c_{water} = 4200 \text{ J/kg·K}\))
Answer
Heat lost by metal = Heat gained by water:
\[ m_{m}c_{m}(T_{m}-T_f) = m_w c_w (T_f - T_w) \]
\[ 0.1 \times c_m \times (200 - 30) = 0.2 \times 4200 \times (30 - 20) \]
\[ 0.1 \times c_m \times 170 = 0.2 \times 4200 \times 10 = 8400 \]
\[ c_m = \frac{8400}{17} \approx \mathbf{494 \text{ J/kg·K}} \]
Section D – Competency-Based Questions
Q4. (Case Study) The surface temperature of the Sun is 5800 K.
(i) Find the wavelength of maximum radiation emitted. (\(b = 2.898 \times 10^{-3} \text{ m·K}\))
(ii) In which part of the EM spectrum does this lie?
(iii) If a body has emissivity 0.8 and area 1 m², find the power radiated at 1000 K. (\(\sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\))
Answer
(i) \(\lambda_{max} = \dfrac{b}{T} = \dfrac{2.898 \times 10^{-3}}{5800} \approx \mathbf{5 \times 10^{-7} \text{ m} = 500 \text{ nm}}\)
(ii) 500 nm falls in the visible light spectrum (green-yellow region).
(iii) \(P = \sigma e A T^4 = 5.67 \times 10^{-8} \times 0.8 \times 1 \times (1000)^4\)
\(= 5.67 \times 10^{-8} \times 0.8 \times 10^{12} = \mathbf{45360 \text{ W} \approx 45.36 \text{ kW}}\)