The heat liberated when $1.89 \mathrm{~g}$ of benzoic acid is burnt in a bomb calorimeter at $25^{\circ}…
The heat liberated when $1.89 \mathrm{~g}$ of benzoic acid is burnt in a bomb calorimeter at $25^{\circ} \mathrm{C}$ increases the temperature of $18.94 \mathrm{~kg}$ of water by $0.632^{\circ} \mathrm{C}$. If the specific heat of water at $25^{\circ} \mathrm{C}$ is $0.998 \mathrm{cal} /\left(\mathrm{g}^{\circ} \mathrm{C}\right)$, then find the heat of combustion of benzoic acid.
$2540 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$1975 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$3240 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$2825 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
Mass of benzoic acid $=1.89 \mathrm{~g}$
Temperature $=25^{\circ} \mathrm{C}=298 \mathrm{~K}$
Mass of water $=18.94 \mathrm{~kg}$
Increase in temperature $(\Delta t)=0.632^{\circ} \mathrm{C}$
Specific heat of $\mathrm{H}_2 \mathrm{O}=0.998 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}$
Heat gained by water or heat liberated by benzoic acid $(Q)$
$Q=m s \Delta t$
$\begin{aligned} & =18.94 \times 0.998 \times 0.632 \\ & =11.946 \mathrm{kcal}=49.982 \mathrm{~kJ}\end{aligned}$
$\because 1.89 \mathrm{~g}$ of acid, liberates $49.98 \mathrm{~kJ}$ of heat
$\therefore$ Heat liberated by $122 \mathrm{~g}$ of acid $=\frac{49.98 \times 122}{1.89}$
$=3226.2 \mathrm{~kJ} / \mathrm{mol}$