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Q.

Steam at 1000C is passed into 20 g of water at 100C. When water acquires a temperature of 800C, the mass of water present will be (Take, specific heat of water = 1 calg-1 0C-1 and latent heat of steam = 540 cal g-1) 

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a

22.5 g

b

31.5 g

c

24 g

d

42.5 g

answer is D.

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Detailed Solution

Heat lost by system = Heat gained by water 

Let  m' be amount of heat converted into water

m'×L=mcΔt

m'×540=20×1×(80−10)

⇒ m'=20×70540=2.5g

Now, net mass of water = 20 + 2.5 = 22.5 g

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