# The heat liberated when $1.89\; g$ of benzoic acid is burnt in a bomb calorimeter at $25^\circ C$ increases the temperature of $18.94\; kg$ of water by $0.632^\circ C$. If the specific heat of water at $25^\circ C$ is $0.998 \;cal/g$ deg, the value of the heat of combustion of benzoic acid is

(a) 881.1 kcal
(b) 771.4 kcal
(c) 981.1 kcal
(d) 871.2 kcal

Given, Weight of benzoic acid = 1.89 g
Temperature of bomb calorimeter = $25^\circ C$
Mass of water (m) = 18.94 kg = 18940 g
Increse in temperature ($\Delta T$) = $0.632^\circ C$
ans Specific heat of water (s) = 0.998 cal/g/deg
We know that heat gained by water or heat liberated by benzoic acid (Q) = $ms\Delta T = 18940 \times 0.998 \times 0.632 = 11946.14 cal$
Since, 1.89 g of acid liberates 11946.14 cal of heat,
$\therefore$ heat liberated by 122 g of acid (1 mol) = $\frac{11946.14\times 122}{1.89}$ = 77126.5 cal = 771.12 kcal
(where 122 g is the molecular weight of benzoic acid)
answered Feb 27, 2014