Sure, let's solve this step-by-step.
Given:
[tex]\[ q_{\text{rxn}} = -31.66 \, \text{kJ} \][/tex]
[tex]\[ C_{\text{cal, dry}} = 2.39 \, \text{kJ}/{^\circ}\text{C} \][/tex]
[tex]\[ m_{\text{H}_2\text{O}} = 1.00 \, \text{kg} \][/tex]
[tex]\[ c_{\text{H}_2\text{O}} = 4.184 \, \text{kJ}/(\text{kg} \cdot {^\circ}\text{C}) \][/tex]
We need to calculate the temperature increase [tex]\( \Delta T \)[/tex] expected when the sample of benzoic acid is combusted in the calorimeter that contains 1.00 kg of water.
First, let's express the combined heat capacity term:
[tex]\[ C_{\text{total}} = C_{\text{cal, dry}} + (m_{\text{H}_2\text{O}} \cdot c_{\text{H}_2\text{O}}) \][/tex]
Substituting the given values:
[tex]\[ C_{\text{total}} = 2.39 \, \text{kJ}/{^\circ}\text{C} + (1.00 \, \text{kg} \cdot 4.184 \, \text{kJ}/(\text{kg} \cdot {^\circ}\text{C})) \][/tex]
[tex]\[ C_{\text{total}} = 2.39 \, \text{kJ}/{^\circ}\text{C} + 4.184 \, \text{kJ}/{^\circ}\text{C} \][/tex]
[tex]\[ C_{\text{total}} = 6.574 \, \text{kJ}/{^\circ}\text{C} \][/tex]
Now, using the energy conservation equation, we know that the heat released by the reaction [tex]\( q_{\text{rxn}} \)[/tex] will be absorbed by the calorimeter and the water, causing a temperature change [tex]\(\Delta T\)[/tex]:
[tex]\[ q_{\text{cal}} = C_{\text{total}} \cdot \Delta T \][/tex]
Since [tex]\( q_{\text{cal}} = -q_{\text{rxn}} \)[/tex]:
[tex]\[ 31.66 \, \text{kJ} = 6.574 \, \text{kJ}/{^\circ}\text{C} \cdot \Delta T \][/tex]
Solving for [tex]\(\Delta T\)[/tex]:
[tex]\[ \Delta T = \frac{31.66 \, \text{kJ}}{6.574 \, \text{kJ}/{^\circ}\text{C}} \][/tex]
[tex]\[ \Delta T \approx 4.816 \, {^\circ}\text{C} \][/tex]
Thus, the expected temperature increase [tex]\(\Delta T\)[/tex] would be approximately 4.816°C.
