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Interconversion of State

Steam at 10...

Question

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

160 g m

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125 g m

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250 g m

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320 g m

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Solution

The correct option is A $160 g m$
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.
also, $Q_{1} = m (L_{v a p} + 1 \times (100 - 80))$
so, $m = 160 g m$

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