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  • A Carnot engine operates between two reservoirs at temperatures <img alt="\mathrm{T_{\text {hot }}=500 \mathrm{~K} \: \text{and}\: T_{\text {cold }}=300 \mathrm{~K}}" src="https://entrancecorner.on

A Carnot engine operates between two reservoirs at temperatures \mathrm{T_{\text {hot }}=500 \mathrm{~K} \: \text{and}\: T_{\text {cold }}=300 \mathrm{~K}}. Calculate the efficiency of the Carnot engine.
 

Option: 1

0.6


Option: 2

0.8


Option: 3

0.4


Option: 4

0.55


Answers (1)

best_answer

Step 1: The efficiency \mathrm{(\eta)} of a Carnot engine is given by the formula:

\mathrm{ \eta=1-\frac{T_{\text {cold }}}{T_{\text {hot }}} }

Given: Hot reservoir temperature \mathrm{\left(T_{\text {hot }}\right)=500 \mathrm{~K}} Cold reservoir temperature \mathrm{\left(T_{\text {cold }}\right)=300 \mathrm{~K}}

Step 2: Substitute the given values and calculate the efficiency of the Carnot engine.

\mathrm{ \eta=1-\frac{300}{500} }

Calculating the numerical value of the efficiency:

\mathrm{ \eta=1-0.6=0.4 }

So, the efficiency of the Carnot engine operating between reservoirs at temperatures

\mathrm{ T_{\text {hot }}=500 \mathrm{~K}\: and \: T_{\text {cold }}=300 \mathrm{~K}\: is\: 0.4 \: or\: 40 }

Hence option 3 is correct.
 

Posted by

Ritika Harsh

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