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 ${15}^{\circ }C$ till the temperature of the calorimeter and its contents rises to ${80}^{\circ }C$.The mass of the steam condensed in kg is

• A. 0.13
• B. 0.065
• C. 0.26
• D. 0.135

Answer 1

The correct option is A 0.13

Let $M$ be the mass of steam condensed.

Amount of heat lost by the steam due to latent heat of condensation and due to cooling to ${80}^{\circ }C$

$=M\times 540+m(100-80)$

Amount of heat absorbed by the water and calorimeter system=$1100\times 1\times (80-15)+20\times 1\times (80-15)$

From conservation of heat energy,

$=M\times 540+M(100-80)$ $=1100\times 1\times (80-15)+20\times 1\times (80-15)$

$⟹M=130g=0.130kg$