Carnot`s cycle is said to have $25 %$ efficiency when it operates between T(source) and $300 K$ (sink). Temperaure T is

200 K

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300 K

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400 K

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None of these

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Solution

The correct option is B 400 K
Efficiency of carnot cycle,

$η = \frac{T_{2} - T_{1}}{T_{2}}$

$T_{1} = t e m p e r a t u r e o f s i n k = 300 K$

$T_{2} = t e m p e r a t u r e o f t h e s o u r c e$

$0.25 = \frac{T_{2} - T_{1}}{T_{2}}$

$T_{2} = 400 K$

Hence, option C is correct.

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