Question

Find the amount of heat required to convert 100 g of water at 100°C into steam at 100°C. Latent heat of vapourisation of water is 540 cal g⁻¹.

Answer 1

Given: Mass of water, m = 100 g

Latent heat of vapourisation, L= 540 cal g⁻¹

Formula: Q = mL

= 100 × 540

= 54000 cal

= 54 kcal

∴ The amount of heat required is $$54,000\text{ cal}$$ (or $$54\text{ kcal}$$).