Q.

Steam at 100 0C is passed into 22 g of water at 20 0C. The mass of water that will be present when the water acquires a temperature of 90 0C (Latent heat of steam is 540 cal g−1 ) is :

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a

24.8 gm

b

24 gm

c

36.6 gm

d

30 gm

answer is A.

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Detailed Solution

Heat loss = Heat gainmLV+mSW(100−90)=22Sw(90−20) m[540+10]=22×1×70m=22×70550=2.8gmMass o water = 22 + 2.8 = 24.8 gm
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Steam at 100 0C is passed into 22 g of water at 20 0C. The mass of water that will be present when the water acquires a temperature of 90 0C (Latent heat of steam is 540 cal g−1 ) is :