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The [tex]$1.200 \, g$[/tex] benzoic acid released [tex]$31.66 \, kJ$[/tex] of energy and the calorimeter absorbed the same [tex]$31.66 \, kJ$[/tex]. The temperature of the calorimeter increased from [tex]$22.45^{\circ} C$[/tex] to [tex]$26.10^{\circ} C$[/tex]. What is the total heat capacity of the calorimeter, in [tex]$kJ /{ }^{\circ} C$[/tex]?

[tex]\[
\begin{array}{l}
31.66 \, kJ = C_{cal} \times \Delta T \\
C_{cal} = \frac{31.66 \, kJ}{(26.10^{\circ} C - 22.45^{\circ} C)}
\end{array}
\][/tex]


Sagot :

To solve the problem of finding the total heat capacity of the calorimeter, let's go through the step-by-step calculations:

1. Understanding the Given Information:
- The benzoic acid sample released [tex]\( 31.66 \, \text{kJ} \)[/tex] of energy.
- The calorimeter absorbed the same amount of energy, [tex]\( 31.66 \, \text{kJ} \)[/tex].
- The initial temperature of the calorimeter was [tex]\( 22.45^\circ \text{C} \)[/tex].
- The final temperature of the calorimeter was [tex]\( 26.10^\circ \text{C} \)[/tex].

2. Finding the Temperature Change ([tex]\(\Delta T\)[/tex]):
- [tex]\(\Delta T\)[/tex] is the change in temperature of the calorimeter.
- [tex]\(\Delta T = \text{final temperature} - \text{initial temperature} = 26.10^\circ \text{C} - 22.45^\circ \text{C}\)[/tex].

3. Calculate the Temperature Change:
[tex]\[ \Delta T = 26.10^\circ \text{C} - 22.45^\circ \text{C} = 3.65^\circ \text{C} \][/tex]

4. Using the Heat Capacity Formula:
- The formula for the total heat capacity ([tex]\(C_{\text{cal}}\)[/tex]) of the calorimeter is:
[tex]\[ C_{\text{cal}} = \frac{\text{Energy released}}{\Delta T} \][/tex]
- Substituting the values we have:
[tex]\[ C_{\text{cal}} = \frac{31.66 \, \text{kJ}}{3.65^\circ \text{C}} \][/tex]

5. Calculating the Total Heat Capacity:
[tex]\[ C_{\text{cal}} = \frac{31.66 \, \text{kJ}}{3.65^\circ \text{C}} \approx 8.674 \, \text{kJ} / ^\circ \text{C} \][/tex]

6. Final Answer:
The total heat capacity of the calorimeter is approximately [tex]\( 8.674 \, \text{kJ} / ^\circ \text{C} \)[/tex].

So, in conclusion, the total heat capacity of the calorimeter, based on the given information and calculations, is [tex]\( 8.674 \, \text{kJ} / ^\circ \text{C} \)[/tex].
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