Question

25 g of ice at −25°C is heated by a burner till it forms boiling water at 100°C. If the specific heat capacity of ice is 2 J g⁻¹°C⁻¹ and the specific latent heat of fusion of ice is 336 J g⁻¹, calculate:

1. Amount of heat energy supplied by the burner to bring the ice to 0°C.
2. Amount of heat energy supplied by burner to melt ice.
3. Amount of heat energy supplied by burner to bring water to boil.

Answer 1

Given: m = 25 g, c_ice = 2 J g⁻¹°C⁻¹, L = 336 J g⁻¹

Assume the specific heat capacity of water = 4.2 J g⁻¹°C⁻¹

(i) Heat to bring ice from −25°C to 0°C:

Q₁ = m × c_ice × ΔT

= 25 × 2 × 25

= 1250 J

(ii) Heat to melt ice at 0°C:

Q₂ = m × L

= 25 × 336

= 8400 J

(iii) Heat to bring water from 0°C to 100°C:

Q₃ = m × c_water × ΔT

= 25 × 4.2 × 100

= 10500 J

Total heat supplied = 1250 + 8400 + 10500 = 20150 J