Question

when 17.6 g hcl dissociates in 55.0 g of water in a coffee cup calorimeter, the temperature rises from 28.1 °c to 66.5 °c.

$c_{sol} = 4.04 \\, \text{j/g °c}$ $c_{cal} = 4.90 \\, \text{j/°c}$

what is the heat of the calorimeter, $q_{cal}$?

$q_{cal} = ? \\, \text{j}$

enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for heat absorbed by calorimeter

The formula for the heat absorbed (or released) by a calorimeter is \( q_{cal} = c_{cal} \times \Delta T \), where \( c_{cal} \) is the heat capacity of the calorimeter and \( \Delta T \) is the change in temperature.

Step2: Calculate the change in temperature (\(\Delta T\))

\(\Delta T = T_{final} - T_{initial}\). Given \( T_{initial} = 28.1^\circ C \) and \( T_{final} = 66.5^\circ C \), so \(\Delta T = 66.5 - 28.1 = 38.4^\circ C\).

Step3: Calculate \( q_{cal} \)

We know \( c_{cal} = 4.90 \, J/^\circ C \) and \( \Delta T = 38.4^\circ C \). Substitute into the formula: \( q_{cal} = 4.90 \, J/^\circ C \times 38.4^\circ C \).
Calculating the product: \( 4.90\times38.4 = 188.16 \). Since the temperature of the calorimeter rises (heat is absorbed by the calorimeter), the sign is positive.

Answer:

\( +188.16 \)