The bomb calorimeter in Exercise 108 is filled with \(987 \mathrm{~g}\) water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C}\). A \(1.056-\mathrm{g}\) sample of benzoic acid \(\left(\Delta E_{\text {comb }}=-26.42 \mathrm{~kJ} / \mathrm{g}\right)\) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?

To calculate the total heat released, we will multiply the mass of the benzoic acid sample and the heat of combustion per gram: \(q = m_{BA} \Delta E_{comb}\) where \(m_{BA}\) is the mass of benzoic acid sample and \(\Delta E_{comb}\) is the heat of combustion per gram. \(q = (1.056 g)\left(-26.42 \frac{kJ}{g}\right)\) First, convert kJ to J: \(q = (1.056 g)\left(-26.42 \frac{1000 J}{g}\right)\) Now, calculate q: \(q = -27,932.32 J\) The negative sign indicates that the heat is released during the combustion process.

Now, we will use the heat transfer formula to find the final temperature of the calorimeter contents. The heat released by the combustion of benzoic acid will be absorbed by the water in the calorimeter: \(q = m_{H2O}c\Delta T\) where \(m_{H2O}\) is the mass of water in the calorimeter and c is specific heat capacity of water. Substitute the values: \(-27,932.32 J = (987g)\left(4.184 \frac{J}{g^{\circ}C}\right)(T_f - 23.32^{\circ}C)\) Now, divide by the product of the mass of water and the specific heat capacity of water: \(\frac{-27,932.32 J}{(987g)(4.184 \frac{J}{g^{\circ}C})} = T_f - 23.32^{\circ}C\) Calculate the value on the left side: \(-6.41^{\circ}C = T_f - 23.32^{\circ}C\) Now, add the initial temperature to both sides: \(-6.41^{\circ}C + 23.32^{\circ}C = T_f\) Finally, calculate the final temperature: \(T_f = 16.91^{\circ}C\) The final temperature of the calorimeter contents is approximately \(16.91^{\circ}C\).