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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 cal g–1 c–1 latent heat of steam = 540 cal/g-1]

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a

31.5 g

b

42.5 g

c

22.5 g

d

24 g

answer is C.

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Detailed Solution

Let x gm of steam be converted into water.x(540) + x(1)(100-80) = 20(80 - 10) ⇒ x (560) = 1400x = 2.5 gm, mass of water = 22.5gm
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