The second type of phase change graph you might see on the SAT II Chemistry exam is called a heating curve. This is a graph of the change in temperature of a substance as energy is added in the form of heat. The pressure of the system is assumed to be held constant, at normal pressure (1 atm). As you can see from the graph below, at normal pressure water freezes at 0ºC and boils at 100ºC.
The plateaus on this diagram represent the points where water is being converted from one phase to another; at these stages the temperature remains constant since all the heat energy added is being used to break the attractions between the water molecules.
Specific Heat
On the SAT II Chemistry test, you might see a diagram that looks something like this one, and you might come across a question that asks you to calculate the amount of energy needed to take a particular substance through a phase change. This would be one of the most difficult questions on the exam, but you might see something like it, or at least part of it. If you were asked to do this, you would need to use the following equation:
energy (in calories) = mCp DT
where m = the mass of the substance (in grams)
Cp = the specific heat of the substance (in cal/g ºC)
DT = the change in temperature of the substance (in either Kelvins or ºC, but make sure all your units are compatible!)
As you can see, this requires that you know the specific heat of the substance. A substance’s specific heat refers to the heat required to raise the temperature of 1 g of a substance by 1ºC. You will not be required to remember any specific heat values for the exam.
Work through the example below to get a feel for how to use this equation.
Example
If you had a 10.0 g piece of ice at -10ºC, under constant pressure of 1 atm, how much energy would be needed to melt this ice and raise the temperature to 25.0ºC?
Explanation
First, the temperature of the ice would need to be raised from -10ºC to 0ºC. This would require the following calculation. The specific heat for ice is 0.485 cal/g ºC. Substituting in the formula
energy = mCp DT; energy = (10.0 g) (0.485 cal/g ºC) (10.0ºC) = 48.5 cal
So 48.5 calories are needed to raise temperature.
Next, we must calculate the heat of fusion of this ice: we must determine how much energy is needed to completely melt the 10 g of it.
energy = mHfus
energy = (10.0 g) (80 cal/g) = 800 cal
So 800 cal of energy are needed to completely melt this sample of ice.
