madey21
Answered

A [tex]$50.0 \, \text{g}$[/tex] sample of unknown metal at [tex]$230.0^{\circ} \text{C}$[/tex] is placed in a calorimeter containing [tex]$110 \, \text{g}$[/tex] of water at [tex]$15.0^{\circ} \text{C}$[/tex]. The final temperature of the combination is [tex]$19.3^{\circ} \text{C}$[/tex]. The heat capacity of the calorimeter is [tex]$8.2 \, \text{J}/{}^{\circ} \text{C}$[/tex] and the specific heat capacity of the water is [tex]$4.18 \, \text{J}/\text{g} \, {}^{\circ} \text{C}$[/tex].

What is the energy change for the calorimeter?

[tex]\[
q_{\text{cal}} = [?] \, \text{J}
\][/tex]

Enter either a [tex]$+$[/tex] or [tex]$-$[/tex] sign and the magnitude.



Answer :

To find the energy change for the calorimeter, we need to use the formula for heat transfer, which is given by \( q = C \Delta T \), where \( q \) is the heat transferred, \( C \) is the heat capacity of the calorimeter, and \( \Delta T \) is the change in temperature.

Given values:
- Heat capacity of the calorimeter, \( C = 8.2 \, \text{J/°C} \)
- Initial temperature of the water/calorimeter, \( T_i = 15.0 \, \text{°C} \)
- Final temperature, \( T_f = 19.3 \, \text{°C} \)

First, calculate the change in temperature \( \Delta T \):
[tex]\[ \Delta T = T_f - T_i = 19.3 \, \text{°C} - 15.0 \, \text{°C} = 4.3 \, \text{°C} \][/tex]

Next, calculate the energy change for the calorimeter using the formula:
[tex]\[ q_{\text{cal}} = C \Delta T = 8.2\, \text{J/°C} \times 4.3 \, \text{°C} = 35.26 \, \text{J} \][/tex]

Therefore, the energy change for the calorimeter is:
[tex]\[ q_{\text{cal}} = +35.26 \, \text{J} \][/tex]