Chemistry - Thermodynamics - I Question with Solution | TestHub

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 $\mathrm{acid}(Q)=ms\Delta t$ $=18940\times 0.998\times 0.632=11946.14\mathrm{cal}$ Since, $1.89\mathrm{g}$ of acid, liberates $11946.14\mathrm{cal}$ of heat, therefore heat liberated by $122\mathrm{g}$ (mol. $\begin{array}{rl}& \text{ wt. of benzoic acid) of acid }=\frac{11946.14\times 122}{1.89}\\ & =771126.5\mathrm{cal}\approx 771.1\mathrm{kcal}\end{array}$