Heat of Fusion Example Problem: Melting Ice

Heat of fusion is the amount of heat energy required to change the state of matter of a substance from a solid to a liquid. It's also known as enthalpy of fusion. Its units are usually Joules per gram (J/g) or calories per gram (cal/g). This example problem demonstrates how to calculate the amount of energy required to melt a sample of water ice.

Example 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

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

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·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 8,350 Joules or 2,000 calories.

Note: Heat of fusion should be a positive value. (The exception is helium.) If you get a negative number, check your math.