madey21
Answered

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What temperature increase would be expected if the same [tex]$1.200 \, \text{g}$[/tex] sample of benzoic acid were combusted when the calorimeter contained [tex]$1.00 \, \text{kg}$[/tex] of water?

[tex]\[
\begin{aligned}
q_{\text{rxn}} & = -31.66 \, \text{kJ} \\
C_{\text{cal (dry)}} & = 2.39 \, \text{kJ/}^{\circ}\text{C} \\
-q_{\text{rxn}} & = q_{\text{cal}} = \left[ C_{\text{cal, dry}} (\Delta T) + C_{\text{cal, } H_2O} (\Delta T) \right] \\
q_{\text{cal}} & = \left[ C_{\text{cal, dry}} (\Delta T) + (m_{H_2O}) (c_{H_2O}) (\Delta T) \right]
\end{aligned}
\][/tex]

Combine like terms using [tex] \Delta T [/tex]:

[tex]\[
31.66 \, \text{kJ} = \left[ 2.39 \, \text{kJ/}^{\circ}\text{C} + (1.00 \, \text{kg} \times 4.184 \, \text{kJ/kg}^{\circ}\text{C}) \right] (\Delta T)
\][/tex]

[tex]\[
\Delta T = [?] \, ^{\circ}\text{C}
\][/tex]

Sagot :

GinnyAnswer
Sure, let's solve this step-by-step.

Given:
[tex]\[ q_{\text{rxn}} = -31.66 \, \text{kJ} \][/tex]
[tex]\[ C_{\text{cal, dry}} = 2.39 \, \text{kJ}/{^\circ}\text{C} \][/tex]
[tex]\[ m_{\text{H}_2\text{O}} = 1.00 \, \text{kg} \][/tex]
[tex]\[ c_{\text{H}_2\text{O}} = 4.184 \, \text{kJ}/(\text{kg} \cdot {^\circ}\text{C}) \][/tex]

We need to calculate the temperature increase [tex]\( \Delta T \)[/tex] expected when the sample of benzoic acid is combusted in the calorimeter that contains 1.00 kg of water.

First, let's express the combined heat capacity term:

[tex]\[ C_{\text{total}} = C_{\text{cal, dry}} + (m_{\text{H}_2\text{O}} \cdot c_{\text{H}_2\text{O}}) \][/tex]

Substituting the given values:

[tex]\[ C_{\text{total}} = 2.39 \, \text{kJ}/{^\circ}\text{C} + (1.00 \, \text{kg} \cdot 4.184 \, \text{kJ}/(\text{kg} \cdot {^\circ}\text{C})) \][/tex]

[tex]\[ C_{\text{total}} = 2.39 \, \text{kJ}/{^\circ}\text{C} + 4.184 \, \text{kJ}/{^\circ}\text{C} \][/tex]

[tex]\[ C_{\text{total}} = 6.574 \, \text{kJ}/{^\circ}\text{C} \][/tex]

Now, using the energy conservation equation, we know that the heat released by the reaction [tex]\( q_{\text{rxn}} \)[/tex] will be absorbed by the calorimeter and the water, causing a temperature change [tex]\(\Delta T\)[/tex]:

[tex]\[ q_{\text{cal}} = C_{\text{total}} \cdot \Delta T \][/tex]

Since [tex]\( q_{\text{cal}} = -q_{\text{rxn}} \)[/tex]:

[tex]\[ 31.66 \, \text{kJ} = 6.574 \, \text{kJ}/{^\circ}\text{C} \cdot \Delta T \][/tex]

Solving for [tex]\(\Delta T\)[/tex]:

[tex]\[ \Delta T = \frac{31.66 \, \text{kJ}}{6.574 \, \text{kJ}/{^\circ}\text{C}} \][/tex]

[tex]\[ \Delta T \approx 4.816 \, {^\circ}\text{C} \][/tex]

Thus, the expected temperature increase [tex]\(\Delta T\)[/tex] would be approximately 4.816°C.
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