Heat of Fusion Formula

The heat of fusion of any substance is the important calculation of the heat. It is the change in the value of the enthalpy by providing energy i.e. heat, for a specific quantity of the substance. It will change its state from a solid to a liquid keeping the pressure constant. The heat of fusion of any sample will measure the amount of heat that needs to be introduced to convert its crystalline fraction into the disordered state. The topic explains the heat of fusion formula with examples. Let us learn it!

Concept of the Heat of Fusion

‘Heat of fusion’ measures the amount of energy required to melt a given amount of a solid at its melting point temperature. In other words, it also represents the amount of energy given up when a given mass of liquid solidifies. For example, water has a heat of fusion of 80 calories per gram. It means that it takes 80 calories of energy to melt 1 gram of ice at the temperature of zero degrees C into the water at zero degrees C. Heat of fusion values will differ for the different materials.

For example, we may see that heat gained by ice is equal to the heat lost by the water. We denote the Heat of fusion by the symbol \(\Delta H_f.\)

When a solid material turns into the liquid, then it is what we know as melting. This melting process will need an increase in energy to allow the solid-state particles to break free from each other. This energy input is the heat of fusion. The heat of fusion is not the same for all substances, but it is a constant value for each individual kind of substance.

The Formula for the Heat of Fusion:

We compute it as: \(\Delta H_f = \frac{q}{m}\)

\(\Delta H_f\)	heat of fusion
q	Heat
m	mass

Solved Examples for Heat of Fusion Formula

Q.1: Calculate the heat in Joules which is required to melt 26 grams of the ice. It is given here that heat of fusion of water is 334 J/g i.e. equals to 80 cal per gram.

Solution: Given parameters are,

Mass, m = 26 g

We know that,

\(\Delta H_f = \frac{q}{m}\)

Rearranging the formula,

\(q = m \times \Delta H_f,\\\)

\(= 26 \times 334 \\\)

= 8684 Joules.

Thus heat required will be 8684 Joules.

Q.2: What will be the heat of fusion for the water, if it takes 668 Joules of the heat energy to melt 2 grams?

Solution: Known values are,

Q = 668 joules

M = 2 grams

Formula is:

\(\Delta H_f = \frac{q}{m}\)

\(= \frac{668}{2}\)

= 334 J per gram.

Thus heat of fusion will be 334 J per gram.

Q.3: What mass of water will be melted at zero degrees C, if 1500 J of heat energy is applied? Use Heat of Fusion Formula.

Solution: The heat of fusion for water is applicable here and the equation has to be rearranged to solve it for the mass.

Here, \(H_f =1500 J\)

Q = 334 C per gram

\(\Delta H_f = \frac{q}{m}\)

Then,

\(m = \frac {H_f}{q}\)

\(= \frac {1500}{ 334 }\)

= 4.49 gram

Mass of water will be 4.49 gram.