A Carnot engine, having an efficiency of 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)

    90 J

  • Option 2)

    1 J

  • Option 3)

    100 J

  • Option 4)

    99 J

 

Answers (1)
V Vakul

As we learnt in 

Coefficient of performance (beta) -

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

- wherein

T_{1}> T_{2}

For a perfect refrigerator eta 
ightarrow infty

 

 

AND

 

 

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 )

 

\beta=\frac{1}{\eta}-1=\frac{1}{\frac{1}{10}}-1=10-1

\beta =9=\frac{Q_{2}}{W}

\because W=10J

\therefore Q_{2}=10\times 9J=90J

Correct option is 1.


Option 1)

90 J

This option is correct

Option 2)

1 J

This option is incorrect

Option 3)

100 J

This option is incorrect

Option 4)

99 J

This option is incorrect

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