JEE Main 2025 · Physics · Thermodynamics and Kinetic Theory of Gases · easy

A Carnot engine has an efficiency of 40%. When the temperature of its source is raised by 100 K keeping the sink temperature unchanged, the…

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A Carnot engine has an efficiency of

40\%

. When the temperature of its source is raised by

100\ \mathrm{K}

keeping the sink temperature unchanged, the efficiency becomes

50\%

. The temperature of the sink is

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Answer: (B) $300\ \mathrm{K}$

Let the source and sink temperatures be

T_1

and

T_2

. Efficiency

\eta = 1 - T_2/T_1

. First case:

T_2 = 0.6\,T_1

. Second case:

T_2 = 0.5\,(T_1+100)

. Equating,

0.6\,T_1 = 0.5\,T_1 + 50

, so

T_1 = 500\ \mathrm{K}

and

T_2 = 0.6\times500 = 300\ \mathrm{K}

. The value

500\ \mathrm{K}

is the original source temperature and

400\ \mathrm{K}

results from using the raised source with the wrong ratio.

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