### Molar Heat of Fusion

Here is the definition of the molar heat of fusion:

the amount of heat necessary to melt (or freeze) 1.00 mole of a substance at its melting point
Note the two important factors:
1) It's 1.00 mole of a substance
2) there is no temperature change

Keep in mind the fact that this is a very specific value. It is only for one mole of substance melting. The molar heat of fusion is an important part of energy calculations since it tells you how much energy is needed to melt each mole of substance on hand. (Or, if you are cooling off a substance, how much energy per mole to remove from a substance as it solidifies.

Every substance has its own molar heat of fusion.

The units for the molar heat of fusion are kilojoules per mole (kJ/mol). Sometimes, the unit J/g is used. In that case, the term heat of fusion is used, with the word 'molar' being eliminated. See Example #3 below.

The molar heat of fusion for water is 6.02 kJ/mol. As you go around the Internet, you will see other values used. For example, 6.01 is a popular value and you sometimes see 6.008. I grew up with 6.02, so I'll stick to it.

Molar heat values can be looked up in reference books.

The molar heat of fusion equation looks like this:

q = ΔHfus (mass/molar mass)

The meanings are as follows:

1) q is the total amount of heat involved
2) ΔHfus is the symbol for the molar heat of fusion. This value is a constant for a given substance.
3) (mass/molar mass) is the division to get the number of moles of substance

Example #1: 31.5 g of H2O is being melted at its melting point of 0 °C. How many kJ is required?

Solution:

plug the appropriate values into the molar heat equation shown above

q = (6.02 kJ/mol) (31.5 g / 18.0 g/mol)

Example #2: 53.1 g of H2O exists as a liquid at 0 °C. How many kJ must be removed to turn the water into a solid at 0 °C

Solution:

note that the water is being frozen and that there is NO temperature change. The molar heat of fusion value is used at the solid-liquid phase change, REGARDLESS of the direction (melting or freezing).

q = (6.02 kJ / mol) (53.1 g / 18.0 g/mol)

Example #3: Calculate the heat of fusion for water in J/g

Solution:

divide the molar heat of fusion (expressed in Joules) by the mass of one mole of water.

(6020 J / mol) / (18.015 g/mol)

This value, 334.166 J/g, is called the heat of fusion, it is not called the molar heat of fusion. When this value is used in problems, the 334 J/g value is what is most-often used.

Example #4: Using the heat of fusion for water in J/g, calculate the energy needed to melt 50.0 g of water at its melting point of 0 °C.

Solution:

multiply the heat of fusion (expressed in J/g) by the mass of the water involved.

(334.166 J/g) (50.0 g) = 16708.3 J = 16.7 kJ (to three sig figs)

Example #5: By what factor is the energy requirement to evaporate 75 g of water at 100 °C greater than the energy required to melt 75 g of ice at 0 °C?

Solution:

Notice how the amounts of water are the same. This is deliberate. The equality is important, not the amount.

Change the 75 g to one mole and solve:

40.7 kJ / 6.02 kJ = 6.76

Change the amount to 1 gram of water and solve:

2259.23 J / 334.166 J = 6.76

If you insisted that you must do it for 75 g, then we have this:

(75 g * 2259.23 J/g) / (75 g * 334.166 J/g) = ???

You can see that the 75 cancels out, leaving 6.76 for the answer.