Question

a 25.5 g piece of metal with a temperature of 105.00 °c is dropped into a calorimeter containing 145 g of water. the water temperature increases from 20.00 °c to 23.50 °c. the heat capacity of the calorimeter is 6.2 j/°c and the specific heat capacity of water is 4.18 j/g °c. what is the energy change for the calorimeter? $q_{cal} = ? j$ enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for heat change of a calorimeter

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

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

The initial temperature of the calorimeter (same as water initially) is \(T_{initial} = 20.00\ ^{\circ}\text{C}\) and the final temperature is \(T_{final}=23.50\ ^{\circ}\text{C}\). So, \(\Delta T=T_{final}-T_{initial}=23.50 - 20.00=3.50\ ^{\circ}\text{C}\).

Step3: Substitute values into the formula

We know that \(C_{cal} = 6.2\ \text{J/}^{\circ}\text{C}\) and \(\Delta T = 3.50\ ^{\circ}\text{C}\). Substituting these into the formula \(q_{cal}=C_{cal}\times\Delta T\), we get \(q_{cal}=6.2\ \text{J/}^{\circ}\text{C}\times3.50\ ^{\circ}\text{C}\).
Calculating the product: \(6.2\times3.50 = 21.7\). Since the calorimeter is absorbing heat (temperature increases), the sign is positive.

Answer:

\(+21.7\)