Q47 / 180 · AIIMS 2017 · Chemistry
The heat is liberated when $1.89 \mathrm{~g}$ of benzoic acid is burnt in a bomb calorimeter at $25^{\circ} \mathrm{C}$ and it 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 $0.998 \mathrm{cal} / \mathrm{g}$-deg, then value of the heat of combustion of benzoic acid is
- A$881.1 \mathrm{kcal}$
- B$981.1 \mathrm{kcal}$
- Correct answerC$771.1 \mathrm{kcal}$
- D$871.2 \mathrm{kcal}$
Solution
Mass of benzoic acid $=1.89 \mathrm{~g}$
Temperature of bomb calorimeter
$=25^{\circ} \mathrm{C}=298 \mathrm{~K}$
Mass of water $(m)=18.94 \mathrm{~kg}=18940 \mathrm{~g}$
[ncrease in temperature $(\Delta t)=0.632^{\circ} \mathrm{C}$
Specific heat of water $(s)=0.998 \mathrm{cal} / \mathrm{g}-\mathrm{deg}$
Heat gained by water or heat liberated by benzoic $\operatorname{acid}(Q)=m s \Delta t$
$=18940 \times 0.998 \times 0.632=11946.14 \mathrm{cal}$
Since, $1.89 \mathrm{~g}$ of acid, liberates $11946.14 \mathrm{cal}$ of heat, therefore heat liberated by $122 \mathrm{~g}$ (mol.
$\begin{aligned} & \text { wt. of benzoic acid) of acid }=\frac{11946.14 \times 122}{1.89} \\ & =771126.5 \mathrm{cal} \approx 771.1 \mathrm{kcal}\end{aligned}$
Hence, the correct answer is option: 'C'.