Heat

Introduction

Heat is a form of energy that flows from a body at a higher temperature to a body at a lower temperature. In ICSE Class 10 Physics, you study calorimetry, specific heat capacity, latent heat, and the changes of state.

Specific Heat Capacity

The amount of heat required to raise the temperature of a unit mass of a substance by 1 K (or 1°C).

Q = m × c × ΔT

Where Q = heat energy (J), m = mass (kg), c = specific heat capacity (J kg⁻¹ K⁻¹), ΔT = temperature change (K or °C).

Specific Heat Capacities

Water has the highest specific heat capacity among common substances.

Calorimetry

Calorimetry is the measurement of heat transfer. The principle of calorimetry states:

Heat lost by the hot body = Heat gained by the cold body (assuming no heat loss to the surroundings)

Principle of Heat Exchange

m₁c₁(T₁ − T) = m₂c₂(T − T₂)

Where m₁, m₂ = masses, c₁, c₂ = specific heat capacities, T₁ = initial temperature of hot body, T₂ = initial temperature of cold body, T = final temperature.

Latent Heat

The heat energy absorbed or released when a substance changes its state without any change in temperature.

Q = m × L

Where L = latent heat (J kg⁻¹).

Specific Latent Heat of Fusion (Lf)

Heat required to convert 1 kg of solid to liquid at its melting point.

• Ice: Lf = 3.36 × 10⁵ J kg⁻¹ (or 336 J/g)
• The heat absorbed is used to overcome the force of attraction between molecules.

Specific Latent Heat of Vaporisation (Lv)

Heat required to convert 1 kg of liquid to vapour at its boiling point.

• Water: Lv = 2.26 × 10⁶ J kg⁻¹ (or 2260 J/g)

Melting and Boiling

Worked Numericals

Example 1: Heat Required to Raise Temperature

How much heat is required to raise the temperature of 2 kg of water from 20°C to 80°C? (c_water = 4200 J/kg°C)

Solution: Q = mcΔT = 2 × 4200 × (80 − 20) = 2 × 4200 × 60 = 5,04,000 J = 504 kJ

Example 2: Calorimetry — Mixing

2 kg of water at 90°C is mixed with 3 kg of water at 30°C. Find the final temperature.

Solution: Heat lost by hot water = Heat gained by cold water 2 × 4200 × (90 − T) = 3 × 4200 × (T − 30) 2(90 − T) = 3(T − 30) 180 − 2T = 3T − 90 270 = 5T T = 54°C

Example 3: Latent Heat

How much heat is required to convert 500 g of ice at 0°C into water at 20°C? (Lf = 336 J/g, c_water = 4.2 J/g°C)

Solution: Step 1: Melting ice at 0°C: Q₁ = mLf = 500 × 336 = 1,68,000 J Step 2: Heating water from 0°C to 20°C: Q₂ = mcΔT = 500 × 4.2 × 20 = 42,000 J Total heat = 168,000 + 42,000 = 2,10,000 J = 210 kJ

Common Mistakes and Fixes

ICSE Exam Focus

This chapter carries 6–8 marks. Key topics: Q = mcΔT numericals, calorimetry mixture problems, latent heat problems, principle of heat exchange.

Marks Blueprint: Specific heat capacity — 2 marks, Calorimetry mixtures — 3 marks, Latent heat problems — 3 marks.

Self-Test Questions

1. How much heat is needed to raise the temperature of 500 g of iron from 25°C to 75°C? (c_iron = 450 J/kg°C)

2. 500 g of water at 100°C is mixed with 300 g of water at 20°C. Find the final temperature.

3. Calculate the heat required to convert 200 g of ice at −10°C to steam at 100°C. (c_ice = 2100 J/kg°C, Lf = 3.36 × 10⁵ J/kg, c_water = 4200 J/kg°C, Lv = 2.26 × 10⁶ J/kg)

4. Define specific latent heat of fusion and vaporisation.

5. Explain why water is used as a coolant in car radiators.

6. State the principle of calorimetry and write its equation.