Question

Steam at $100{}^{\circ }C$ is passed into $1.1kg$ of water contained in a calorimeter of water equivalent $0.02kg$ at $15{}^{\circ }C$ till the temperature of the calorimeter and its contents reaches $80{}^{\circ }C$. Considering the latent heat of steam $=540cal{g}^{-1}$ $\mathrm{and}{s}^{}$pecific heat capacity of water $=4183Jk{g}^{-1}{K}^{-1}$ ${\mathrm{or}}^{}$ $=1calk{g}^{-1}{K}^{-1}$,
calculate the mass of the steam condensed.

• A. $0.139$
• B. $0.065$
• C. $0.263$
• D. $0.144$

Answer 1

The correct option is A

$0.139$

Given:
Mass of water and calorimeter $=1.10+0.02=1.12kg$
Specific heat capacity of water $=4.184\times {10}^{3}Jk{g}^{-1}{K}^{-1}=1calk{g}^{-1}{K}^{-1}$
Let the mass of steam$=mkg$
Latent heat of vaporization of water at ${100}^{\circ }C=540cal{g}^{-1}$

We know that,
$\text{Heat lost by steam}=\text{Heat gained by water}+\text{calorimeter}$
or, $mL+mc(100-80)=1.2\times c\times (80-15)$
where, $m$ is the mass of the steam, $L$ is the latent heat of vaporization and $c$ is the specific heat of water.
or, $m[540+(1\times 20\left)\right]=1.2\times 1\times 65$
or, $m=\frac{1.2\times 1\times 65}{560}=\frac{72.8}{560}kg$
or, $m=0.139kg$