Steam is passed into water taken in a vessel (total thermal capacity $110 c a l /^{0} C$ ) at $30^{0} C$ till the resultant temperature is $90^{0} C$ . If latent heat of steam is $540 c a l / g$ , the increase in mass of the water is:

60 g

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110 g

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12 g

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550 g

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Solution

The correct option is A $12 g$
Heat absorbed by the water plus the container in raising the temp by $90 - 30 = 60^{0} C$ is $Q = s Δ T$ $= 110 \times 60 = 6600 c a l$
Latent heat of condensation for steam is $540 c a l / g$ .
Hence, the increase in the mass of water is giveb by $m = \frac{Q}{L}$ = $\frac{6600}{540} = 12.22 g$

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