Question

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

(Molecular mass of butan-1-ol $=74$)

• A. $14.8$
• B. $74$
• C. $37$
• D. $370$

Answer 1

Answer: (D)

$370$

Solution:- (D) $370$

Heat required to raise the temperature of $200g$ water by $5℃\left(q\right)=m.c.\Delta T$

whereas,

$m=$ mass of water $=200g$

$c=$ specific heat of water $=1Cal-g/℃$

$\Delta T=5℃$

$\therefore q=200\times 1\times 5=1\text{kcal}$

Heat liberated by $0.2g$ of $1$-butanol on burning $=1\text{kcal}$

$\therefore$ Heat liberated by $1$ mole of $1$-butanol $\left(74g\right)=\frac{1}{0.2}\times 74=370\text{kcal}$

Hence the heat liberated is $370\text{kcal}$.