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JEE Main 2023 (25 Jan, Shift 1) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2023 (25 Jan, Shift 1)

Q22JEE Main 2023 · 25 JAN · SHIFT 1Easy

Heat Engines, Refrigerators & Second Law

A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be:

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Answer: (b)

$0.5 = 1 - \frac{T_2}{600} \Rightarrow T_2 = 300$ K
$0.7 = 1 - \frac{300}{T_1'} \Rightarrow T_1' = \frac{300}{0.3} = 1000$ K
So the answer is (b) 1000 K.

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