Question

When $0.2\, g$ of $1$-butanol was burnt in a suitable apparatus, the heat evolved was sufficient to raise the temperature of $200 \,g$ water by $ 5{}^\circ C $ . The enthalpy of combustion of $1$-butanol in kcal $ mo{{l}^{-1}} $ will be

• A. $37$
• B. $370$
• C. $ -370 $
• D. $ -740 $
• E. $ -14.8 $

Answer 1

Answer: (C)

Heat required to raise the temperature of 200 g water by
$ 5{}^\circ C $ $ =m.\text{ }c.\text{ }\Delta t $
$ =200\times 1\times 5=1000\text{ }cal=1\text{ }kcal $
$ \therefore $ 0.2 g 1-butanol on burning liberates = 1 kcal heat
$ \therefore $ 74 g (1 mol) 1-butanol on burning will liberate
$ =\frac{1}{0.2}\times 74 $ $ =370 $
Since, the heat is liberated,
$ \Delta H=-370\text{ }kcal $ .