Q51 / 180 · AIIMS 2016 · Chemistry
The heat 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 $0.998 \mathrm{cal} / \mathrm{g}$ - $\mathrm{deg}$, the value of the heat of combustion of benzoic acid is
- A$881.1 \mathrm{kcal}$
- Correct answerB$771.12 \mathrm{kcal}$
- C$981.1 \mathrm{kcal}$
- D$871.2 \mathrm{kcal}$
Solution
Given : Weight of benzoic acid $=1.89 \mathrm{~g}$;
Temperature of bomb calorimeter $=25^{\circ} \mathrm{C}=298 \mathrm{~K}$;
Mass of water $(m)=18.94 \mathrm{~kg}=18940 \mathrm{~g}$;
Increase in temperature $(\Delta t)=0.632^{\circ} \mathrm{C}$ and specific heat of water $(s)=0.998 \mathrm{cal} / \mathrm{g}$-deg.
We know that heat gained by water or heat liberated by benzoic acid $(Q)=m s \Delta t$
$=\frac{11946.14 \times 122}{1.89}=771126.5 \mathrm{cal}=771.12 \mathrm{kcal}$
Hence, the correct answer is option: 'B'.