Measurement of Water's Heat of Fusion

Measuring WATER'S HEAT OF FUSION

In the measurement of the heat capacity of an ice cube, you can see that the ice lowers the temperature of the cup of water much more than you would expect of a volume of liquid water at 0°C. Thus we see two things:

That the heat capacity equation only works for items to do not exhibit a change of state such a changing from a solid into a liquid, and
We must amend the equation used to calculate heat capacity.

Let us now derive that new equation.

It might be good at this point in your student career to learn about the conservation of matter and energy equation. Because this can be written in many forms, this will suit our needs here (note how it is almost idiotically simple!):

energy lost = energy gained

Fitting the heat capacity equation into that:

heat lost by the water in the cup = heat gained by the object added
ΔT_water x mass_water = ΔT_object x mass_object

If that 'object' is an ice cube or other solid that will melt in the cup of water, then:

heat lost by the water in the cup = heat gained by the object added
ΔT_water x mass_water = ΔT_object x mass_object + the heat of fusion/gm x mass_object
or
ΔT_water x mass_water = ΔT_object x mass_object + ΔH_{f_ice} x mass_object

Thus:

ΔH_{f_ice} = ((ΔT_water x mass_water) - (ΔT_object x mass_object)) / mass_object = Heat of fusion per gram

So let's see how the data from the ice cube experiment works in this equation:

ΔH_{f_ice} = (200 g x 8.5° - 20 g x 13.5°) / 20 g
ΔH_{f_ice} = (1700 - 270) / 20 = 71.5 calories

This is surprisingly close to the real answer 79.7 calories considering the very crude temperature readings that were made - to the nearest half degree. (Note: the unit degree/gram for water is called a calorie.)*

* A food calorie in your diet equals 1,000 of these "small" water calories.