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

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answer is 160.

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Detailed Solution

Heat supplied to water is $Q_{1} = 1400 \times (80 - 16)$
Now this heat supplied to water will be released by condensation of vapor to liquid.
$\begin{array}{l} also, Q_{1} = m (L_{v a p} + 1 \times (100 - 80)) \\ so, m = 160 gm \end{array}$

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