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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24 g

31.5 g

42.5 g

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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