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Q.

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

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a

80 g

b

60 g

c

50 g

d

24 g

answer is B.

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Detailed Solution

Let mass of steam be m thenm×536+m×1×(100-90)=54×1×(90-30) ⇒ m=5.93 g≈6 g Total mass of mixture =54+6=60 g.

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