More complex calorimetry problems

Many times when something is heated or cooled more happens than just a temperature change or just a phase change. For instance, the water in a pot on the stove increases in temperature and then undergoes a phase change (boiling). A classic example (that means the one that most or all chemistry teachers love to give) of this type of problem can be seen fully worked out here.

A 12.77g ice cube at -25 oC is heated to 0 oC and then melted. How much heat is required for this process. The specific heat of ice is 0.50 cal/g oC and the heat of fusion is 80. cal/g.

First, let’s think about what happens to the ice. The ice starts well below the melting point, so that when we start to heat it, it won’t melt initially, it will warm up to 0 oC. Only then can it melt. Since there are two things that happen (warming and melting) we need 2 formulas.

So the heat required for the whole process will be the sum of the heats for the 2 different steps, leaving us with the formula

.

 

Next , let’s see what we know:

q  
m 12.77 g
c 0.50 cal/g oC
Tinitial -25oC
Tfinal 0 oC
ΔHfus 80.0 cal/g

We can now substitute in what we know and solve the problem.

Solution image

Here is another example

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