Question

1. a chemist wishes to calibrate a bomb calorimeter, so she combusts 7.529 g of d - galactose (molar mass = 180.156 g mol⁻¹; $delta_{combustion}u=-15.48$ kj g⁻¹) which causes a temperature change from 26.38 °c to 34.60 °c. what should the chemist report for the value of $c_{cal}$?

a) 0.05917 kj °c⁻¹
b) 0.2501 kj °c⁻¹
c) 45.06 kj °c⁻¹
d) 14.18 kj °c⁻¹
e) 0.07870 kj °c⁻¹

Explanation:

Step1: Calculate the heat released by the combustion

The heat released $q$ is calculated by multiplying the mass of D - galactose by the heat of combustion per gram.
$q = m\times\Delta_{combustion}U$
$m = 7.529\ g$ and $\Delta_{combustion}U=- 15.48\ kJ\ g^{-1}$
$q=7.529\ g\times(-15.48\ kJ\ g^{-1})=- 116.55992\ kJ$

Step2: Calculate the temperature change

The temperature change $\Delta T=T_{final}-T_{initial}$
$T_{initial}=26.38^{\circ}C$ and $T_{final}=34.60^{\circ}C$
$\Delta T = 34.60^{\circ}C - 26.38^{\circ}C=8.22^{\circ}C$

Step3: Calculate the heat capacity of the calorimeter

The heat absorbed by the calorimeter $q_{cal}=-q$ (since the heat released by the reaction is absorbed by the calorimeter). And $q_{cal}=C_{cal}\times\Delta T$
$C_{cal}=\frac{q_{cal}}{\Delta T}$
Since $q_{cal}=116.55992\ kJ$ and $\Delta T = 8.22^{\circ}C$
$C_{cal}=\frac{116.55992\ kJ}{8.22^{\circ}C}=14.18\ kJ\ ^{\circ}C^{-1}$

Answer:

D. $14.18\ kJ\ ^{\circ}C^{-1}$