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Q.

Steam is passed into 54 gm of water at 30°C till the temperature of mixture becomes 90°C. If the latent heat of steam is 536 cal/gm, the mass of the mixture will be

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a

80 gm

b

59.5 gm

c

50 gm

d

24.5 gm

answer is B.

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Detailed Solution

Let mass of steam condensed = M gmM x 536 + M×1×(100-90)= 54×1×(90-30)Mx536 = 54x60M = 54×60536 = 5.5 gmHence, mass of mixture = 54 + 5.5 = 59.5 gm.
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