A calorimter of water equivalent $20 \mathrm{~g}$ contains $180 \mathrm{~g}$ of water at $25^{\circ} \mathrm{C} . ~ ' m$ ' grams of steam at $100^{\circ} \mathrm{C}$ is mixed in it till the temperature of the mixture is $31^{\circ} \mathrm{C}$. The value of $' m^{\prime}$ is close to (Latent heat of water $=540 \mathrm{cal} \mathrm{g}^{-1}$, specific heat of water $=1 \mathrm{cal} \mathrm{g}^{-1^{\circ}}{ }^{\circ}{ }^{-1}$ )
Correct Option: 1
(1) Heat given by water $=m_{w} C_{w}\left(T_{\operatorname{mix}}-T_{w}\right)$
$=200 \times 1 \times(31-25)$
Heat taken by steam $=m L_{\text {stem }}+m C_{w}\left(T_{s}-T_{\text {mix }}\right)$
$=\mathrm{m} \times 540+\mathrm{m}(1) \times(100-31)$
$=\mathrm{m} \times 540+\mathrm{m}(1) \times(69)$
From the principal of calorimeter,
Heat lost = Heat gained
$\therefore(200)(31-25)=m \times 540+m(1)(69)$
$\Rightarrow 1200=m(609) \Rightarrow m \approx 2$