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Chapter 5: Problem 59

Under constant-volume conditions, the heat of combustion of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is $16.49 \mathrm{~kJ} / \mathrm{g}\(. A \)3.00-\mathrm{g}$ sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.94 to \(24.62^{\circ} \mathrm{C} .(\mathbf{a})\) What is the total heat capacity of the calorimeter? (b) If the size of the sucrose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

Short Answer

Expert verified
(a) The total heat capacity of the calorimeter is 18.44 kJ/°C. (b) If the size of the sucrose sample had been exactly twice as large, the temperature change of the calorimeter would have been 5.36°C.

Step by step solution

01

(a) Calculate heat released by combustion of sucrose

: We start by finding how much heat is released by the combustion of the sucrose sample. We know that the heat of combustion of sucrose is 16.49 kJ/g and the mass of the sucrose is 3.00 g. We can use this information to calculate the total heat released: Total heat released (q) = heat of combustion × mass of sucrose \(q = 16.49 \, \text{kJ/g} \times 3.00 \, \text{g} = 49.47 \, \text{kJ}\)
02

(a) Calculate the change in temperature

: The temperature change of the calorimeter is given by the difference of the initial and final temperatures: ΔT = T_final - T_initial ΔT = 24.62°C - 21.94°C ΔT = 2.68°C
03

(a) Calculate the heat capacity of the calorimeter

: To find the heat capacity of the calorimeter, C_cal, we need to use the formula: C_cal = q / ΔT Plugging in the values, we get: C_cal = \(49.47 \, \text{kJ} / 2.68 \, ^{\circ}\text{C} = 18.44 \, \text{kJ/^{\circ}C}\) So, the total heat capacity of the calorimeter is 18.44 kJ/°C.
04

(b) Calculate the heat released for twice the mass of sucrose

: Now we need to check the temperature change if the mass of the sucrose sample were twice as large. We first find the heat released by combustion of twice the mass of sucrose: Total heat released (q') = heat of combustion × 2 × mass of sucrose \(q' = 16.49 \, \text{kJ/g} \times 2 \times 3.00 \, \text{g} = 98.94 \, \text{kJ}\)
05

(b) Calculate the new temperature change

: Next, we determine the temperature change associated with double the amount of sucrose, using the heat capacity of the calorimeter that we calculated earlier: ΔT' = q' / C_cal ΔT' = \(98.94 \, \text{kJ} / 18.44 \, \text{kJ/^{\circ}C} = 5.36 \,^{\circ}\text{C}\) So if the amount of sucrose burned was exactly twice as large, the temperature change of the calorimeter would have been 5.36°C.

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