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JEE Main 2020 (3 Sep, Shift 2) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2020 (3 Sep, Shift 2)

Q56JEE Main 2020 · 3 SEP · SHIFT 2Difficult

Heat Engines, Refrigerators & Second Law

If minimum possible work is done by a refrigerator in converting 100 grams of water at 0oC to ice, how much heat (in calories) is released to the surroundings at temperature 27oC (Latent heat of ice = 80 Cal/gram) to the nearest integer?

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

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Answer: 8791

Minimum work means a Carnot (reversible) refrigerator: $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$ .
Heat removed: $Q_2 = mL = 100 \times 80 = 8000$ cal.
$Q_1 = 8000 \times \frac{300}{273} \approx 8791$ cal
So the heat released is $8791$ cal.

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