Heat Capacity Problem

Q. How much heat must be removed from 50 grams of water vapor at 100°C to freeze it into a solid at 0°C?

A. There are three steps to this process:

Step Formula Heat Removed
Condense the vapor to liquid water at 100°C 50 g times 540 cal/g (heat of vaporization) 27,000 cal
Cool the liquid from 100°C to 0°C 50 g times 100°C (temperature change) times 1 cal/g°C 5,000 cal
Freeze the liquid to ice at 0°C 50 g times 80 cal/g (heat of fusion) 4,000 cal
The total heat that must be removed to freeze this water vapor is:
27,000 + 5,000 + 4000 = 36,000 calories