An insulating container contains 250 g of water at 20°C. Steam at 100°C is passed through it, heating it up to 25°C. During this process, part of the steam also gets condensed, raising the mass of water to 252 g. If the specific heat of water is 1 Cal g⁻¹ °C⁻¹, then what is the value of the latent heat of steam?

The correct answer is C, 550 Cal/g. The heat given out by 2 g of steam equals the heat taken in by 250 g of water, and that balance gives L = 550 Cal/g.

The water warms from 20°C to 25°C, so it absorbs 250 × 1 × 5 = 1250 Cal. The mass rose from 250 g to 252 g, so 2 g of steam condensed. That steam first gives up its latent heat, 2L, on turning into water at 100°C, and then cools from 100°C to 25°C, giving 2 × 1 × 75 = 150 Cal. Setting 2L + 150 = 1250 gives 2L = 1100, so L = 550 Cal/g. The container is insulating, so no heat is lost to it or to the air.

A, 530 Cal/g, would make the heat given out 1210 Cal, less than the 1250 Cal absorbed.

B, 540 Cal/g, is close to the textbook value but ignores the 150 Cal released as the condensed water cools.

D, 560 Cal/g, would release 1270 Cal, more than the water took in.

Exam tip: in mixture problems, heat lost by the hot body (latent heat plus cooling) equals heat gained by the cold body.