Samacheer Kalvi 10th Science Chapter 3 Numerical Problems — Thermal Physics with Solutions

Numerical Problems — Chapter 3: Thermal Physics

Important Formulas:

Useful values:

• Specific heat of water: c = 4200 J kg⁻¹ K⁻¹
• Specific heat of ice: c = 2100 J kg⁻¹ K⁻¹
• Latent heat of fusion of ice: L_f = 3.36 × 10⁵ J/kg
• Latent heat of vaporisation of water: L_v = 22.6 × 10⁵ J/kg

Problem 1. How much heat is required to raise the temperature of 2 kg of water from 25°C to 75°C? (Specific heat of water = 4200 J kg⁻¹ K⁻¹)

• Mass, m = 2 kg
• Initial temperature, T₁ = 25°C
• Final temperature, T₂ = 75°C
• ΔT = 75 − 25 = 50°C
• c = 4200 J kg⁻¹ K⁻¹

Q = mcΔT = 2 × 4200 × 50

Q = 420,000 J = 4.2 × 10⁵ J

Problem 2. A piece of iron of mass 0.5 kg is heated from 30°C to 130°C. Find the heat absorbed. (Specific heat of iron = 470 J kg⁻¹ K⁻¹)

• Mass, m = 0.5 kg
• ΔT = 130 − 30 = 100°C
• c = 470 J kg⁻¹ K⁻¹

Q = mcΔT = 0.5 × 470 × 100

Q = 23,500 J = 2.35 × 10⁴ J

Problem 3. How much heat is required to melt 5 kg of ice at 0°C? (Latent heat of fusion of ice = 3.36 × 10⁵ J/kg)

• Mass, m = 5 kg
• Latent heat of fusion, L_f = 3.36 × 10⁵ J/kg
• The melting happens at constant temperature (0°C)

Q = mL_f = 5 × 3.36 × 10⁵

Q = 16.8 × 10⁵ J = 1.68 × 10⁶ J

Problem 4. How much heat is needed to convert 2 kg of water at 100°C to steam at 100°C? (Latent heat of vaporisation = 22.6 × 10⁵ J/kg)

• Mass, m = 2 kg
• Temperature = constant at 100°C
• L_v = 22.6 × 10⁵ J/kg

Q = mL_v = 2 × 22.6 × 10⁵

Q = 45.2 × 10⁵ J = 4.52 × 10⁶ J

Note: This is more than 5× the heat needed to heat the same mass of water from 0°C to 100°C (Q = 2 × 4200 × 100 = 840,000 J), which is why steam burns are so severe.

Problem 5. A 200 g block of copper (specific heat = 385 J kg⁻¹ K⁻¹) at 80°C is dropped into 500 g of water at 20°C. Find the final temperature of the mixture. (Specific heat of water = 4200 J kg⁻¹ K⁻¹)

• m_copper = 200 g = 0.2 kg; c_copper = 385 J kg⁻¹ K⁻¹; T_copper = 80°C
• m_water = 500 g = 0.5 kg; c_water = 4200 J kg⁻¹ K⁻¹; T_water = 20°C
• Final temperature = T (same for both at equilibrium)

By principle of calorimetry: Heat lost by copper = Heat gained by water

m_c × c_c × (80 − T) = m_w × c_w × (T − 20)

0.2 × 385 × (80 − T) = 0.5 × 4200 × (T − 20)

77(80 − T) = 2100(T − 20)

6160 − 77T = 2100T − 42000

6160 + 42000 = 2100T + 77T

48160 = 2177T

T = 48160 / 2177 ≈ 22.1°C

Problem 6. Convert: (a) 100°C to Kelvin (b) 373 K to Celsius (c) 0 K to Celsius

(a) K = °C + 273 = 100 + 273 = 373 K

(b) °C = K − 273 = 373 − 273 = 100°C

(c) °C = K − 273 = 0 − 273 = −273°C (absolute zero)

Problem 7. A gas has a volume of 4 L at 27°C. If it is heated at constant pressure to 127°C, find the new volume. (Use Charles' Law)

• V₁ = 4 L
• T₁ = 27°C = 27 + 273 = 300 K
• T₂ = 127°C = 127 + 273 = 400 K
• V₂ = ?

By Charles' Law: V₁/T₁ = V₂/T₂

V₂ = V₁ × T₂/T₁ = 4 × 400/300 = 4 × 4/3 = 16/3 ≈ 5.33 L

Problem 8. A gas occupies 500 mL at a pressure of 2 atm. Find the volume when the pressure is reduced to 1 atm at constant temperature. (Use Boyle's Law)

• P₁ = 2 atm; V₁ = 500 mL
• P₂ = 1 atm; V₂ = ?

By Boyle's Law: P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂ = (2 × 500) / 1 = 1000 mL = 1 L

Summary Table

📚 Samacheer Kalvi 10th Science Guide — Buy Online