Q.

Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be : (Take specific heat of water =1 cal g-1(CO)-1 and latent heat of steam = 540 cal g-1)

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a

24 g

b

31.5 g

c

42.5 g

d

22.5 g

answer is D.

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Detailed Solution

Heat lost = Heat gained⇒      mLv+msw∆θ=mwSw∆θ ⇒    m×540+m×1×(100-80)                              =20×1×(80-10) ⇒                     m=2.5 g Total mass of water = (20 + 2.5) g = 22.5 g
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Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be : (Take specific heat of water =1 cal g-1(CO)-1 and latent heat of steam = 540 cal g-1)