The combustion of 1.00 mol of glucose, \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \), releases 2820 kJ of heat. If 2.00 g of glucose is burned in a calorimeter containing 1.00 kg of water, and the temperature increases by \( 3.50^{\circ} \mathrm{C} \), what is the heat capacity of the calorimeter? \( 2.38 \mathrm{~J} /{ }^{\circ} \mathrm{C} \) \( 2.38 \mathrm{~kJ} /{ }^{\circ} \mathrm{C} \) \( 4.76 \mathrm{~kJ} /{ }^{\circ} \mathrm{C} \) \( 4.76 \mathrm{~J} /{ }^{\circ} \mathrm{C} \)

To find the heat capacity of the calorimeter, we need to calculate the heat released by the combustion of glucose and then use the temperature change of the water to determine the heat capacity of the calorimeter. Given: - 1.00 mol of glucose (\( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \)) releases 2820 kJ of heat. - 2.00 g of glucose is burned in a calorimeter containing 1.00 kg of water. - The temperature increases by \( 3.50^{\circ} \mathrm{C} \). First, let's calculate the heat released by the combustion of 2.00 g of glucose: \[ \text{Heat released by 2.00 g of glucose} = \frac{2820 \text{ kJ}}{180 \text{ g}} \times 2.00 \text{ g} \] Next, we can use the heat released by the combustion of glucose and the temperature change of the water to find the heat capacity of the calorimeter: \[ \text{Heat capacity of the calorimeter} = \frac{\text{Heat released by glucose}}{\text{Temperature change of water}} \] Let's calculate the heat released by the combustion of 2.00 g of glucose and then find the heat capacity of the calorimeter. Simplify the expression by following steps: - step0: Solution: \(\frac{2820kJ}{180g}\times 2g\) - step1: Reduce the fraction: \(\frac{47Jk}{3g}\times 2g\) - step2: Multiply the terms: \(\frac{94Jk}{3g}\times g\) - step3: Reduce the fraction: \(\frac{94Jk}{3}\) Calculate or simplify the expression \( 94Jk/3 \). Simplify the expression by following steps: - step0: Solution: \(\frac{94Jk}{3}\) The heat released by the combustion of 2.00 g of glucose is \( \frac{94 \text{ J}}{3} \) kJ. Now, we can calculate the heat capacity of the calorimeter using the heat released by the combustion of glucose and the temperature change of the water: \[ \text{Heat capacity of the calorimeter} = \frac{\text{Heat released by glucose}}{\text{Temperature change of water}} \] Let's calculate the heat capacity of the calorimeter. Simplify the expression by following steps: - step0: Solution: \(\frac{\frac{94Jk}{3}}{3.5}\) - step1: Convert the expressions: \(\frac{\frac{94Jk}{3}}{\frac{7}{2}}\) - step2: Multiply by the reciprocal: \(\frac{94Jk}{3}\times \frac{2}{7}\) - step3: Multiply the terms: \(\frac{94Jk\times 2}{3\times 7}\) - step4: Multiply the terms: \(\frac{188Jk}{3\times 7}\) - step5: Multiply the terms: \(\frac{188Jk}{21}\) The heat capacity of the calorimeter is \( \frac{188 \text{ J}}{21} \) kJ/°C. Therefore, the heat capacity of the calorimeter is \( \frac{188 \text{ J}}{21} \) kJ/°C, which is approximately 8.95 kJ/°C. So, the correct answer is \( 4.76 \text{ kJ/}^{\circ} \text{C} \).