The heat liberated when 1.89g of benzoic acid is burnt in a bomob calorimeter at `25^(@)C` and it increases the temperature of 18.94 kg of water by `0.632^(@)C`. If the specific heat of water at `25^(@)C` is `0.998cal*g^(-1)*""^(@)C^(-1)`, the value of the heat of combustion of benzoic acid is-

Options

A. 881.1 kcal
B. 771.124 kcal
C. 981.1 kcal
D. 871.2 kcal

Answer

B

Explanation

Given: Weight of benzoic acid=1.89 g, Temperature of bomb calorimeter`=25^(@)C=298K`, Mass of water
`(m)=18.94kg=18940g,` Increase in temperature `(DeltaT)=0.632^(@)C` and specific heat of water `(s)=0.998cal*g*""^(@)C^(-1)`
We know that, heat gained by water or heat liberated by benzoic acid `(q)=msDeltaT`
`=18940xx0.998xx0.632=11946.14cal`
Since 1.89g of acid liberates 11946.14 cal of heat, therefore heat liberated by 122g of acid
`=(11946.14xx122)/(1.89)`
`=771126.5cal=771.12kcal`
(where 122 g is the molecular weight of benzoic acid)