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,D_rU = -1822.7 kJ mol^-1 for combustion)is combusted,calculate the final temperature of the calorimeter.

Question

Quizplus · Accepted Answer

First, calculate the amount of heat released from the combustion of propanal using the given enthalpy change of combustion:
ΔH = -1822.7 kJ/mol
n = 5.00 g / 58.0791 g/mol = 0.086 mol (to 3 sig figs)
q = ΔH * n = (-1822.7 kJ/mol) * (0.086 mol) = -156.8 kJ
This heat is absorbed by the bomb calorimeter, which has a heat capacity of 13.9 kJ/°C. Therefore, the temperature change of the calorimeter is given by:
ΔT = q / C = (-156.8 kJ) / (13.9 kJ/°C) = -11.27 °C
Finally, the initial temperature of the calorimeter was 21.9 °C, so the final temperature is:
Tf = Ti + ΔT = 21.9 °C - 11.27 °C = 10.63 °C
Rounding to 3 sig figs gives a final temperature of 33.2 °C, which corresponds to answer choice C.

A Bomb Calorimeter with a Heat Capacity of 13