Question

20 g of ice at 0°C is added to 200g of water at 20°C. Calculate the drop in temperature ignoring the heat capacity of the container. (Specific latent heat of ice = 80 cal/g)

Answer 1

Let x be the drop in temperature of water

Heat lost by water = 200 × x

Heat gained by ice in melting = 20 × 80 = 1600 calories

Heat gained by the water from ice in rising from 0°C to (20 - x)°C = 20 × 1 × (20 - x)

∵ Heat gained = Heat lost

∴ 200 × x = 20 × 80 + 20 (20 - x)

or 200x + 20x = 400 + 1600

x = `2000/220 = 9.1^circ "C"`