Question

The efficiency of carnot cycle may be increased by (A) Increasing the highest temperature (B) Decreasing the highest temperature (C) increasing the lowest temperature (D) Decreasing the lowest temperature

Answer 1

Answer:

Increasing the highest temperature

Explanation:

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Answer 2

Answer:

The Carnot cycle's efficiency increases by decreasing the lowest temperature.

Therefore, option D) Decreasing the lowest temperature, is the correct option.

Explanation:

Carnot cycle:

• It describes the most efficient heat engine.
• It involves four processes, i.e., the two isothermal processes and the two adiabatic processes.

Isothermal process: The process in which temperature remains the same.

Adiabatic process: The process in which there is no heat transfer.

The Carnot cycle has shown in the graph.

• $T_{H}$ = Highest temperature or hot reservoir temperature.
• $T_{L}$ = Lowest temperature or cold reservoir temperature.

As we know,

Carnot efficiency ( η ) is given by:

• $\eta = \frac{T_{H}-T_{L} }{T_{H} }$

As we can see in the formula;

• The Carnot efficiency increases by decreasing the lowest temperature because the value of $T_{H} -T_{L}$ will increase.

Hence, the Carnot efficiency increases by decreasing the lowest temperature.