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MHT CET 2026 (16 Apr, Shift 2) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from MHT CET 2026 (16 Apr, Shift 2)

Q6MHT CET 2026 · 16 APR · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

A Carnot engine with efficiency 50% takes heat from a source at 600 K. To increase the efficiency by 20%, keeping temperature of the sink same, the new temperature of the source will be

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Answer: (c)

$0.5 = 1 - \frac{T_2}{600} \Rightarrow T_2 = 300$ K
Efficiency rises by 20 percentage points, to $70\%$ : $0.7 = 1 - \frac{300}{T_1'}$
$T_1' = \frac{300}{0.3} = 1000$ K
So the answer is (c) 1000 K.

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