When a substance changes its state of matter, it takes a specific amount of energy to complete the change. When the phase change is between solid and liquid, the amount of energy per unit mass is called the heat of fusion. These heat of fusion example problems will show how to apply heat of fusion to heat equations.

The equation to find this energy is rather simple.

Q = m · ΔH_{f}

where

Q = Energy (heat)

m = mass

ΔH_{f} = heat of fusion

Heat of Fusion Example Problem 1

**Question:** The heat of fusion of water is 334 J/g. How much energy is required to melt 50 grams of ice into liquid water?

**Solution:** Apply this information to the formula.

m = 50 grams

ΔH_{f} = 334 J/g

Q = m · ΔH_{f}

Q = (50 g) · (334 J/g)

Q = 16700 J = 16.7 kJ

**Answer:** It takes 16700 joules or 16.7 kilojoules of to melt 50 grams of ice into water.

### Heat of Fusion Example Problem 2

**Question:** If it takes 41000 joules of heat to melt 200 grams solid copper to liquid copper, what is the heat of fusion of copper?

**Solution:** First, list what we know.

m = 200 g

Q = 41000 J

Apply this to the heat of fusion equation.

Q = m·ΔH_{f}

41000 J = (200 g) · ΔH_{f}

Solve for ΔH_{f}.

ΔH_{f} = 41000 J/200 g

ΔH_{f} = 205 J/g**Answer:** The heat of fusion of copper is 205 J/g.

### Heat of Fusion Example Tip: Check Your Units!

The main source of errors working problems involving heat of fusion is the units. Be sure to match the units of

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