Question

Steam at ${100}^{\circ }C$ is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at ${15}^{o}C$ till the temperature of the calorimeter and its contents rises to ${80}^{\circ }C.$

What will be the mass of the steam condensed (in kg)?

Take:

Specific heat of water$=1calg{m}^{-1}$

Latent heat of vaporization$=540calg{m}^{-1}$

• A. 0.130
• B. 0.065
• C. 0.260
• D. 0.135

Answer 1

The correct option is D

0.135

$Given:$

$\text{Mass of water}=1.1kg$

$\text{water equivalent of calorimeter}=0.02kg$

Specific heat of water$=1calg{m}^{-1}$

Latent heat of vaporization$=540calg{m}^{-1}$

$\text{Heat required}$
$Q=(1.1+0.02)\times {10}^3\times 1\times (80-15)=72800cal.$
$\therefore$ mass of steam condensed (in kg), $m=\frac{Q}{L}=\frac{72800}{540}\times {10}^{-3}=0.135kg.$