I have a question from my AP preparation book that I got wrong and do not understand the explanation. The question is as follows:

A student designs an experiment to determine the specific heat of aluminum. The student heats a piece of aluminum with a mass of 5.86 g to various temperatures, then drops it into a calorimeter containing 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 (376.2 J), and the specific heat of aluminum (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?

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?

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

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

Initial Temperature of Al Initial Temperature of H2O 109.1 23.2

Final Temperature of Al + H2O 26.8

I correctly calculated the heat gained by the water (376.2 J), and the specific heat of aluminum (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:

5.86g Al/26.98g Al = 0.217 mol Al

376 J / 0.217 mol = 1730 J/mol = 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?

Thank you!

I have a question from my AP preparation book that I got wrong and do not understand the explanation. The question is as follows:

A student designs an experiment to determine the specific heat of aluminum. The student heats a piece of aluminum with a mass of 5.86 g to various temperatures, then drops it into a calorimeter containing 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 (376.2 J), and the specific heat of aluminum (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?