**Problem:**

What is the heat in Joules required to convert 25 grams of water into steam? What is the heat in calories?

Useful information: heat of vaporization of water = 2257 J/g = 540 cal/g

**Solution:**

**Part I**

Use the formula

q = m·ΔH

_{v}

where

q = heat energy

m = mass

ΔH

_{v}= heat of vaporization

q = (25 g)x(2257 J/g)

q = 56425 J

**Part II**

q = m·ΔH

_{f}

q = (25 g)x(540 cal/g)

q = 13500 cal

**Answer:**

The amount of heat required to melt 25 grams of ice is 56425 Joules or 13500 calories.