Question

the coffee cup calorimeter was determined to have a heat capacity of 8.5 j/°c. the initial and final temperatures match the solution, at 21.50 °c and 28.25 °c, respectively. what is the heat change for the calorimeter during the dissociation process?
q_cal = (c_cal)(t_f - t_i)
q_cal = (8.5)(28.25 - 21.50)
q_cal = ? j
enter either a + or - sign and the magnitude.

Explanation:

Step1: Calculate the temperature difference

First, find the difference between the final temperature ($T_f$) and the initial temperature ($T_i$).
$T_f - T_i = 28.25 - 21.50 = 6.75$ $^\circ$C

Step2: Calculate the heat change for the calorimeter

Use the formula $q_{cal} = (c_{cal})(T_f - T_i)$. Substitute $c_{cal} = 8.5$ J/$^\circ$C and the temperature difference we just calculated.
$q_{cal} = 8.5 \times 6.75$

Calculate the product: $8.5 \times 6.75 = 57.375$

Since the temperature of the calorimeter increased (final temperature is higher than initial), the calorimeter absorbed heat, so the heat change $q_{cal}$ is positive.

Answer:

+57.375