Heat of Fusion Example Problem - Melting Ice

How Calculate Energy Needed to Change a Solid Into a Liquid

It takes a lot of energy to melt ice into water. The heat of fusion equation can tell you exactly how much energy you need.
It takes a lot of energy to melt ice into water. The heat of fusion equation can tell you exactly how much energy you need. Atomic Imagery / Getty Images

Heat of fusion is the amount of heat energy required to change the state of matter 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.

Heat of Fusion Problem - Melting Ice

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:
In the problem, the heat of fusion is given.

This isn't a number you're expected to know off the top of your head. There are chemistry tables that state common heat of fusion values. To solve this problem, you'll need the formula that relates heat energy to mass and heat of fusion:

q = m·ΔHf

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

Keep in mind, temperature is not anywhere in the equation because it doesn't change when matter changes state. The equation is straightforward, so the key is to make sure you're using the right units for the answer. To get heat in Joules:

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

It's just as easy to express the heat in terms of calories:

q = m·ΔHf
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.