Heat of Fusion Example Problem

Heat of fusion is the amount of heat energy required to change the state of a substance from solid to liquid. This example problem demonstrates how to calculate the amount of energy required to melt a sample of water ice.

Problem:

What is the heat in Joules required to melt 25 grams of ice? What is the heat in calories?

Useful information: heat of fusion of water = 334 J/g = 80 cal/g

Solution:

Part I

Use the formula

q = m·ΔH_f

where
q = heat energy
m = mass
ΔH_f = heat of fusion

q = (25 g)x(334 J/g)
q = 8350 J

Part II

q = m·ΔH_f
q = (25 g)x(80 cal/g)
q = 2000 cal

Answer:

The amount of heat required to melt 25 grams of ice is 8350 Joules or 2000 calories.