Question

Water at 80°C is poured into a bucket containing 1.5 kg of crushed ice at 0°C, such that all the ice melts and the final temperature recorded is 0°C. Calculate the amount of hot water added to the ice.

[Take the specific heat capacity of water as 4200 Jkg⁻¹ K⁻¹ and sp. latent heat of ice = 336 × 10³ Jkg⁻¹]

Answer 1

Let the mass of hot water be m kg.

Hot water cools from 80°C to 0°C, so its fall in temperature is 80°C.

Heat given out by hot water = m × 4200 × 80 J

The ice at 0°C melts completely to water at 0°C, so heat absorbed by ice is 1.5 × 336000 J.

By the principle of calorimetry

Heat lost = Heat gained

m × 4200 × 80 = 1.5 × 336000

$$m = \frac{1.5 \times 336000}{4200 \times 80}$$

= 1.5 kg