How to determine the specific heat of aluminum?

Question

I have a question from my AP preparation book that I got wrong and do not understand the explanation:

A student designs an experiment to determine the specific heat of aluminum. The student heats a piece of aluminum with a mass of $\pu{5.86 g}$ to various temperatures, then drops it into a calorimeter containing $\pu{25 mL}$ of water. The following data is gathered during one of the trials:

Initial Temperature of $\ce{Al}$: 109.1
Initial Temperature of $\ce{H2O}$: 23.2
Final Temperature of $\ce{Al + H2O}$: 26.8

I correctly calculated the heat gained by the water ($\pu{376.2 J}$), and the specific heat of aluminum ($\pu{0.78 J/g*C}$). However, how do I calculate the enthalpy change for the cooling of aluminum in water? The book's procedure is as follows:

$$\frac{\pu{5.86g}~\ce{Al}}{\pu{26.98g}~\ce{Al}} = \pu{0.217 mol}~\ce{Al}$$

$$\frac{\pu{376 J}}{\pu{0.217 mol}} = \pu{1730 J//mol} = \pu{1.73 kJ//mol}$$

I see that one may calculate specific heat by dividing the heat gained by water by moles of aluminum and converting to kilo-joules per mole in this scenario. However, why is that? What is the reasoning behind this procedure? How would you phrase this procedure to apply to this general type of question?

Answer 1

If the water and aluminum are regarded as incompressible, then the internal energy is a function only of temperature. In addition, the heat capacity at constant pressure is equal to the heat capacity at constant volume. If the pressure is constant, then the change in enthalpy will be equal to the change in internal energy. So the change in enthalpy of the aluminum is equal to its heat capacity times its temperature change. The same goes for the water. So the total change in internal energy and/or enthalpy for the combination of water and aluminum is zero.

Answer 2

This is simply conservation of energy. Assume you controlled for loses of heat to, say, the container and atmosphere; using a styrofoam cup and cover is a good way to do this is a student-lab situation. Then, the total heat lost by the metal is equal to the total heat gained by the water.

The last line of calculation in your post is using the total heat lost by the metal, same as that gained by the water, and finding how much to cool/heat 1 kg. This amount applies to the entire temperature difference through which the metal cooled. Convert to J/g, then calculate how many J/g per each degree.