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Standard XII

Chemistry

Enthalpy of Combustion

The heat line...

Question

The heat linerated when $1.89 g$ of benzoic acid is burnt in a bomb calorimeter at $25^{o} C$ increases the temperature of $18.94 k g$ of water by ${0.632}^{o} C$ . If the specific heat of water at $25^{o} C$ is $0.998 c a l g^{- 1}$ ${d e g}^{- 1}$ , the value of heat of combustion of benzoic acid is:

881.1 k c a l

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771.4 k c a l

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981.1 k c a l

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871.2 k c a l

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Solution

The correct option is B $771.4 k c a l$
$Given, weight of benzoic acid =1.89g$
$Temperature of bomb calorimeter=25 degree celsius$
$Mass of water (m)= 18.94 kg=18940g$
$Increse in temperature(T)=0.632$
$and specific heat of water(s)=0.998 cal/g/deg$
$We know that heat gained by water or heat liberated by benzoic acid(Q)=$
$m s δ T = 18940 * 0.998 - 0.632 = 11946.14 c a l$
$t e x t S i n c e, 1.89 g o f a c i d l i b e r a t e s 11946.14 c a l o f h e a t,$
$hence, heat liberated by 112g of acid(1 mol)=$ $\frac{11946.14 X 112}{1.89} = 771.12 k c a l$

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The heat linerated when 1.89g of benzoic acid is burnt in a bomb calorimeter at 25oC increases the temperature of 18.94kg of water by 0.632oC. If the specific heat of water at 25oC is 0.998calg−1 deg−1, the value of heat of combustion of benzoic acid is: