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
(C^O)^-1 and latent heat of steam = 540 cal g^-1)

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22.5 g

31.5 g

24 g

42.5 g

answer is D.

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Detailed Solution

Heat lost = Heat gained
$\Rightarrow {mL}_{v} + {ms}_{w} ∆ θ = m_{w} S_{w} ∆ θ \Rightarrow m \times 540 + m \times 1 \times (100 - 80) = 20 \times 1 \times (80 - 10) \Rightarrow m = 2.5 g$
Total mass of water = (20 + 2.5) g = 22.5 g

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