Question

during the reaction, the water temperature rises from 25.00 to 28.54 °c. the heat capacity of the calorimeter, including water, is 5860 j/°c. calculate the heat of reaction in joules.
q_rxn = ? j
hint: let the units guide your calculation set up!
enter either a + or - sign and the magnitude. do not round until the end.

Explanation:

Step1: Find the temperature change

The temperature change $\Delta T$ is the final temperature minus the initial temperature. So $\Delta T = 28.54^{\circ}\text{C} - 25.00^{\circ}\text{C} = 3.54^{\circ}\text{C}$.

Step2: Calculate the heat absorbed by the calorimeter

The formula for heat absorbed by the calorimeter (including water) is $q = C \times \Delta T$, where $C$ is the heat capacity and $\Delta T$ is the temperature change. Here, $C = 5860 \text{ J/}^{\circ}\text{C}$ and $\Delta T = 3.54^{\circ}\text{C}$. So $q_{\text{cal}} = 5860 \text{ J/}^{\circ}\text{C} \times 3.54^{\circ}\text{C}$.
Calculating that: $5860\times3.54 = 5860\times(3 + 0.54)=5860\times3 + 5860\times0.54 = 17580+3164.4 = 20744.4 \text{ J}$.

Step3: Determine the heat of reaction

The heat of the reaction $q_{\text{rxn}}$ is equal to the negative of the heat absorbed by the calorimeter (because if the calorimeter absorbs heat, the reaction is releasing heat, so $q_{\text{rxn}}$ is negative). So $q_{\text{rxn}} = -q_{\text{cal}} = - 20744.4 \text{ J}$.

Answer:

-20744.4