A Carnot engine works between $600 K$ and $300 K$ . In each cycle of operation, the engine draws $1000 J$ of energy from the source at $600 K$ . The efficiency of the engine is

90 %

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70 %

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50 %

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20 %

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Solution

The correct option is C $50 %$
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 $η = 1 - \frac{300}{600} = \frac{1}{2} = 0.5 = 50 %$

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A Carnot engine works between 600K and 300K. In each cycle of operation, the engine draws 1000J of energy from the source at 600K. The efficiency of the engine is

The correct option is C 50%The efficiency of the cycle is given as η=1−T2T1. Here T1 is the temperature of the source and T2 is the temperature of the sink. Thus, we get η=1−300600=12=0.5=50%

A Carnot engine works between $600 K$ and $300 K$ . In each cycle of operation, the engine draws $1000 J$ of energy from the source at $600 K$ . The efficiency of the engine is

The correct option is C $50 %$
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 $η = 1 - \frac{300}{600} = \frac{1}{2} = 0.5 = 50 %$