Question

143 kj of energy was released during the reaction, raising the temperature of the calorimeter by 51.2 °c. what is the heat capacity of the calorimeter? $c_{cal} = ? kj/°c$ hints: use the units to guide your calculation. did the calorimeter gain or lose energy?

Explanation:

Step1: Recall the formula for heat capacity

The heat capacity \( C \) is related to the heat \( q \) and the temperature change \( \Delta T \) by the formula \( q = C \times \Delta T \). We need to solve for \( C \), so rearranging the formula gives \( C=\frac{q}{\Delta T} \).

Step2: Identify the values of \( q \) and \( \Delta T \)

The energy released during the reaction is \( 143 \, \text{kJ} \), which is the heat gained by the calorimeter (\( q = 143 \, \text{kJ} \)). The temperature change \( \Delta T=51.2 \, ^\circ\text{C} \).

Step3: Calculate the heat capacity

Substitute the values into the formula: \( C_{\text{cal}}=\frac{143 \, \text{kJ}}{51.2 \, ^\circ\text{C}} \approx 2.79 \, \text{kJ/}^\circ\text{C} \) (rounded to three significant figures).

Answer:

\( \approx 2.79 \)