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?D
T;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.
Next, we need to see how much energy would be needed to raise the temperature of
water from 0ºC to 25ºC. The specific heat for liquid water is 1.00 cal/g ºC. So
again use
energy = mCp?DT to get energy = (10.0 g) (1.00 cal/g ºC) (25.0ºC) = 250 cal
Finally, add together all of the energies to get the total: 48.5 + 800 + 250 =
about 1100 calories are needed to convert the ice to water at these given
temperatures.