A bomb calorimeter with a heat capacity of 13.9 kJ °C^-1 has an initial temperature of 21.9 °C. If 5.00 g of propanal (C₃H₆O, molar mass = 58.0791 g mol^-1, Δ_rU = -1822.7 kJ mol^-1 for combustion) is combusted, calculate the final temperature of the calorimeter.

Question

Quizplus · Accepted Answer

First, we need to calculate the heat released from the combustion of propanal:
n = 5.00 g / 58.0791 g/mol = 0.0861 mol
ΔHcomb = -1822.7 kJ/mol
q = ΔHcomb * n = (-1822.7 kJ/mol) * (0.0861 mol) = -156.8 kJ

Since the bomb calorimeter is at constant volume, all the heat released by the combustion is absorbed by the calorimeter:
q = CΔT
ΔT = q / C = (-156.8 kJ) / (13.9 kJ/°C) = -11.27 °C

Therefore, the final temperature of the calorimeter is:
Tf = Ti + ΔT = 21.9 °C - 11.27 °C = 10.63 °C

However, we need to remember that the temperature change is negative because the heat released by the combustion is absorbed by the calorimeter. Therefore, the actual final temperature is:
Tf = Ti - ΔT = 21.9 °C + 11.27 °C = 33.17 °C

The closest answer choice is C) 33.2 °C.

A Bomb Calorimeter with a Heat Capacity of 13