Questions: The combustion of 4.00 grams of milk in a bomb calorimeter resulted in a temperature increase of 3.4°C. qcomb = -840 J What is the heat capacity of the calorimeter? Ccal = [?] J / °C Enter the magnitude. Do not round until the end. Use significant figures.

The combustion of 4.00 grams of milk in a bomb calorimeter resulted in a temperature increase of 3.4°C. qcomb = -840 J What is the heat capacity of the calorimeter? Ccal = [?] J / °C Enter the magnitude. Do not round until the end. Use significant figures.
Transcript text: The combustion of 4.00 grams of milk in a bomb calorimeter resulted in a temperature increase of $3.4^{\circ} \mathrm{C}$. \[ \mathrm{q}_{\text {comb }}=-840 \mathrm{~J} \] What is the heat capacity of the calorimeter? \[ \mathrm{C}_{\mathrm{cal}}=[?] \mathrm{J} /{ }^{\circ} \mathrm{C} \] Enter the magnitude. Do not round until the end. Use significant figures.
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Solution

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Solution Steps

Step 1: Understand the Problem

We need to find the heat capacity of the calorimeter, denoted as \( C_{\text{cal}} \), given the heat of combustion (\( q_{\text{comb}} \)) and the temperature change (\( \Delta T \)).

Step 2: Use the Formula for Heat Capacity

The formula to calculate the heat capacity of the calorimeter is:

\[ C_{\text{cal}} = \frac{q_{\text{comb}}}{\Delta T} \]

where:

  • \( q_{\text{comb}} = -840 \, \text{J} \) (the negative sign indicates exothermic reaction)
  • \( \Delta T = 3.4^{\circ} \text{C} \)
Step 3: Calculate the Heat Capacity

Substitute the given values into the formula:

\[ C_{\text{cal}} = \frac{-840 \, \text{J}}{3.4^{\circ} \text{C}} \]

Calculate the magnitude:

\[ C_{\text{cal}} = \frac{840}{3.4} \approx 247.0588 \, \text{J/}^{\circ} \text{C} \]

Step 4: Round to Significant Figures

Since the temperature change is given to two significant figures, we should round the final answer to two significant figures:

\[ C_{\text{cal}} \approx 250 \, \text{J/}^{\circ} \text{C} \]

Final Answer

\[ \boxed{C_{\text{cal}} = 250 \, \text{J/}^{\circ} \text{C}} \]

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