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Steam is passed into $22\, g$ of water at $20\,^oC$ . The mass of water that will be present when the water acquires a temperature of $90\,^oC$ is ............ $\mathrm{gm}$ (Latent heat of steam is $540\, cal/gm$)
$24.8\, gm$
$24\, gm$
$36.6\, gm$
$30\, gm$
Solution
Let $m\, g$ of steam get condensed into water (By heat loss). This happens in following two steps.
Heat gained by water $\left(20^{\circ} \mathrm{C}\right)$ to raise it's temperature upto $90^{\circ} \mathrm{C}=22 \times 1{\times}(90-20)$
Hence, in equilibrium, heat lost = Heat gain $\Rightarrow \mathrm{m} \times 540+\mathrm{m} \times 1 \times(100-90)$
$=22 \times 1 \times(90-20)$
$\Rightarrow \mathrm{m}=2.8 \mathrm{\,gm}$
The net mass of the water present in the mixture $=22+2.8=24.8 \mathrm{\,gm}$
