Under constant-volume conditions, the heat of combustion of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) is \(15.57 \mathrm{~kJ} / \mathrm{g}\). A 3.500 -g sample of glucose is burned in a bomb calorimeter. The temperature of the calorimeter increased from \(20.94^{\circ} \mathrm{C}\) to \(24.72^{\circ} \mathrm{C}\). (a) What is the total heat capacity of the calorimeter? (b) If the size of the glucose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

To find the total heat released, multiply the heat of combustion of glucose by the mass of the glucose sample: \[q = m(\text{glucose}) \cdot \Delta H_c(\text{glucose})\] Plug in the given values for the mass of glucose and heat of combustion: \[q = (3.500 \; \text{g})\left(15.57 \; \frac{\text{kJ}}{\text{g}}\right)\] Now, convert the heat released to J by multiplying by 1,000 J/kJ: \[q = (3.500 \cdot 15.57 \cdot 1,000) \; \text{J}\] Calculate the value of q: \[q = 54,495 \; \text{J}\]

Using the formula: \[q = C\Delta T\] We can solve for the heat capacity of the calorimeter (C), where q is the heat released, and ΔT is the change in temperature: \[C = \frac{q}{\Delta T}\] Calculate the change in temperature using the given initial and final temperatures: \[\Delta T = T_{final} - T_{initial} = 24.72 - 20.94 = 3.78 \; \text{°C}\] Now plug in the values of q and ΔT: \[C = \frac{54,495}{3.78}\] Calculate the heat capacity C: \[C = 14,403.17 \; \frac{\text{J}}{\text{°C}}\]

Now, let's determine the temperature change for twice the mass of the glucose sample. First, find the heat released by twice the mass of glucose: \[q' = 2 \cdot (3.500 \; \text{g}) \cdot (15.57\; \frac{\text{kJ}}{\text{g}}) \cdot 1,000\; J\] Calculate q': \[q' = 108,990 \; \text{J}\] With q' and the heat capacity of the calorimeter known, we can find the new temperature change for the calorimeter: \[\Delta T' = \frac{q'}{C}\] Plug in the values of q' and C: \[\Delta T' = \frac{108,990}{14,403.17}\] Calculate the new temperature change ΔT': \[\Delta T' = 7.56 \; \text{°C}\] So, if the size of the glucose sample had been exactly twice as large, the temperature change of the calorimeter would have been 7.56°C.