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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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31.5 g

42.5 g

22.5 g

24 g

answer is C.

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Detailed Solution

$\begin{array}{l} (ML + MSΔθ) = (MSΔθ) W \\ x (540) + x (1) (100 - 80) = 20 (80 - 10) \\ x 560 = 1400 \\ x = 2 .5 gm \\ mass of water = 22 .5 gm \end{array}$

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