A Carnot engine has same efficiency between (i) $100 K$ and $500 K$ and (ii) $T_{k}$ and $900 K$ . the value of T is

360 K

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270 K

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180 K

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90 K

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Solution

The correct option is C $180 K$
The efficiency of the cycle is given as $η = 1 - \frac{T_{2}}{T_{1}}$ . Here $T_{1}$ is the temperature of the source and $T_{2}$ is the temperature of the sink. . Thus we get $η = 1 - \frac{100}{500} = 1 - \frac{T_{k}}{900}$
or, $T_{k} = \frac{100}{500} \times 900 = 180 K$

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100 K

500 K

T

900 K

T

T_{1} = 500 K

T_{2} = 300 K

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