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 calg^-1 ⁰C^-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 by system = Heat gained by water

Let $m'$ be amount of heat converted into water

$m' \times L = m c Δ t$

$m' \times 540 = 20 \times 1 \times (80 - 10)$

$\Rightarrow m' = \frac{20 \times 70}{540} = 2.5 g$

Now, net mass of water = 20 + 2.5 = 22.5 g

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