Thermal Physics — Class 10 Science (Samacheer Kalvi)
TN State Board (Samacheer Kalvi) Class 10 Science, Physics — Chapter 3. Heat, temperature, expansion and the behaviour of gases.
1. About this chapter
This chapter covers heat and temperature, how much heat a substance absorbs (specific heat capacity and latent heat), how materials expand on heating, and the gas laws that relate pressure, volume and temperature.
2. Heat, temperature and specific heat
- Heat is energy transferred due to a temperature difference (unit: joule, J).
- Temperature measures the degree of hotness (unit: kelvin, K, or °C).
- Specific heat capacity (c): heat needed to raise 1 kg of a substance by 1 K. Q = m c ΔT (unit of c: J kg⁻¹ K⁻¹).
- Latent heat (L): heat absorbed/released during a change of state at constant temperature. Q = m L (latent heat of fusion, of vaporisation).
3. Thermal expansion of solids
| Type | Coefficient | Relation |
|---|---|---|
| Linear (length) | α | ΔL = L α ΔT |
| Areal / superficial (area) | β | β = 2α |
| Cubical / volume | γ | γ = 3α |
So α : β : γ = 1 : 2 : 3. Expansion is why gaps are left in railway tracks and bridges.
4. Gas laws
- Boyle's law (constant T): P V = constant → P₁V₁ = P₂V₂.
- Charles's law (constant P): V / T = constant → V₁/T₁ = V₂/T₂ (T in kelvin).
- Avogadro's law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
- (Pressure/Gay-Lussac's law, constant V): P / T = constant.
Always use absolute temperature (K) in gas-law calculations: T(K) = t(°C) + 273.
5. Worked examples
Example 1. How much heat raises 2 kg of water by 10 °C? (c = 4200 J kg⁻¹ K⁻¹) Q = mcΔT = 2 × 4200 × 10 = 84 000 J.
Example 2. α of a metal is 12×10⁻⁶ K⁻¹. Find β and γ. β = 2α = 24×10⁻⁶ K⁻¹; γ = 3α = 36×10⁻⁶ K⁻¹.
Example 3. A gas at 2 atm, 3 L is compressed to 1 L at constant temperature. Find the new pressure. P₁V₁ = P₂V₂ → 2×3 = P₂×1 → P₂ = 6 atm.
6. Common mistakes
- Mistake: Using °C in gas-law formulas. Fix: Convert to kelvin: T(K) = t(°C) + 273.
- Mistake: Confusing specific heat with latent heat. Fix: Specific heat changes temperature (Q = mcΔT); latent heat changes state at constant T (Q = mL).
- Mistake: Mixing the expansion coefficients. Fix: Remember β = 2α and γ = 3α (ratio 1 : 2 : 3).
7. Practice (book-back style)
- Define specific heat capacity and give its unit.
- State Boyle's law and Charles's law.
- Write the relations between α, β and γ.
- Find the heat needed to melt 0.5 kg of ice (L = 3.34×10⁵ J kg⁻¹).
- A gas at 300 K, 2 L is heated to 600 K at constant pressure. Find the new volume.
8. Answer key
- Heat to raise 1 kg by 1 K; unit J kg⁻¹ K⁻¹.
- Boyle: PV = constant at constant T. Charles: V/T = constant at constant P.
- β = 2α, γ = 3α (α : β : γ = 1 : 2 : 3).
- Q = mL = 0.5 × 3.34×10⁵ = 1.67×10⁵ J.
- V₁/T₁ = V₂/T₂ → 2/300 = V₂/600 → V₂ = 4 L.
9. Quick revision
- Physics Ch 3 · heat, specific & latent heat, expansion, gas laws.
- Q = mcΔT (temperature change); Q = mL (state change).
- α : β : γ = 1 : 2 : 3 (β = 2α, γ = 3α).
- Boyle: PV = const; Charles: V/T = const; Avogadro's law.
- Always use kelvin in gas laws.
