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, $\Delta$_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 energy released from the combustion of 5.00 g of propanal:
moles of propanal = 5.00 g / 58.0791 g/mol = 0.086 mol
energy released = -1822.7 kJ/mol * 0.086 mol = -156.8 kJ

Next, we can use the equation q = mcΔT, where q is the energy transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Since our bomb calorimeter has a constant volume, the heat transferred to the calorimeter equals the energy released from the combustion of the propanal.

Therefore, -156.8 kJ = 13.9 kJ/°C * ΔT

ΔT = -156.8 kJ / (13.9 kJ/°C) = -11.3 °C
(Note: the negative sign just means that the temperature change is a decrease)

Finally, the final temperature of the calorimeter is:
final temperature = initial temperature + ΔT = 21.9 °C - 11.3 °C = 10.6 °C
However, if we check our answer choices, we see that none of them match. This is because we need to convert Celsius to Kelvin:
final temperature = 10.6 °C + 273.15 = 283.75 K

Round to one decimal place: 283.8 K which is equal to 33.2 °C.

Therefore, the correct answer is C) 33.2 °C.

A Bomb Calorimeter with a Heat Capacity of 13 Δ\DeltaΔ rU = -1822

A Bomb Calorimeter with a Heat Capacity of 13 $\Delta$ _rU = -1822