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JEE Main 2020 (7 Jan, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2020 (7 Jan, Shift 2)

Q35JEE Main 2020 · 7 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

M grams of steam at 100oC is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produced liquid water at 40oC [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g] the value of M is ____

Numerical answer type. Work it out, then open the answer.

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Answer: 40

Heat lost by steam (condense + cool to $40^{\circ}$ C) = heat gained by ice (melt + warm to $40^{\circ}$ C).
$M(540 + 1 \times 60) = 200(80 + 1 \times 40)$
$600M = 24000$
So $M = 40\ \text{g}$ .

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