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Steam at $100 ° C$ is added slowly to $1400 g m$ of water at $16 ° C$ until the temperature of water is raised to $80 ° C$ . The mass of steam required to do this is ( $L_{v}$ $= 540 c a l / g m)$ :

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$160 g m$

$125 m g$

$250 g m$

$320 g m$

answer is A.

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Detailed Solution

Heat lost from steam = Heat gained by water

$m \times 540 + m \times 1 \times 20 = 1400 \times 1 (80 - 16)$

$\Rightarrow m = \frac{1400 \times 64}{560} = 160 gm$

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