Question

1.6 kg of boiling water at 100°C is poured into 2 kg of crushed ice at 0°C, such that final temperature recorded is 0°C. Calculate the specific latent heat of ice. [sp. heat capacity of water = 4200 Jkg⁻¹ K⁻¹]

Answer 1

The heat given out by 1.6 kg of water cooling from 100°C to 0°C is 1.6 × 4200 × 100 = 672000 J.

Let L be the specific latent heat of fusion of ice.

Heat absorbed by 2 kg of ice to melt at 0°C is 2L.

Using heat lost = heat gained

2L = 672000

L = 336000 J kg⁻¹

= 336 × 10³ J kg⁻¹