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The efficiency of an ideal heat engine working

between the freezing point and boiling point of water, is

  • Option 1)

    6.25%

  • Option 2)

    20%

  • Option 3)

    26.8%

  • Option 4)

    12.5%

 

Answers (1)

best_answer

As we have learned

Efficiency of a carnot cycle -

eta =frac{W}{Q_{1}-Q_{2}}=1-frac{T_{2}}{T_{1}}

T_{1}, and, T_{2}  are in kelvin
 

- wherein

T_{1}= Source temperature

T_{2}= Sink Temperature

left ( T_{1} > T_{2}
ight )

 

n=1-\frac{T_{2}}{T_{1}}= (1-\frac{273}{373})\times 100\%

=\frac{100}{373}*100\%=26.8\% 

 

 

 

 

 


Option 1)

6.25%

This is incorrect

Option 2)

20%

This is incorrect

Option 3)

26.8%

This is correct

Option 4)

12.5%

This is incorrect

Posted by

Avinash

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