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A Carnot engine, having an efficiency of \eta = 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

  • Option 1)

    100 J

  • Option 2)

    99 J

  • Option 3)

    90 J

  • Option 4)

    1 J

 

Answers (1)

As we learnt in 

Coefficient of performance (β) -

\beta = \frac{Q_{2}}{Q_{1}-Q_{2}}= \frac{T_{2}}{T_{1}-T_{2}}
 

- wherein

T_{1}> T_{2}

For a perfect refrigerator \beta \rightarrow \infty

 

 \eta \:=\frac{1}{10}

Coefficient of performance of refrigerator \beta \:=\frac{1}{\eta }\:-1\:=9

\beta \:=\frac{Q_{2}}{W}\:=\frac{Q_{2}}{Q_{1}\:-Q_{2}}

9\:=\frac{Q_{2}}{10{J}}

\therefore \:Q_{2}\:=90J


Option 1)

100 J

This option is incorrect.

Option 2)

99 J

This option is incorrect.

Option 3)

90 J

This option is correct.

Option 4)

1 J

This option is incorrect.

Posted by

Sabhrant Ambastha

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