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Can someone help me with this, - Thermodynamics - NEET

A Carnot engine having an efficiency of \frac{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)

    1 J

  • Option 2)

    90 J

  • Option 3)

    99 J

  • Option 4)

    100 J

 
Answers (1)
100 Views

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

 

 

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 )

 

 For a carnot cycle

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

\beta = Coefficient of performance

\eta = Efficiency of carnot engine

\eta=\frac{1}{10}\:\:\:\:\:\:\:\:\:\:\:\:=> \beta =9

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

Q_{2}=\beta.W=9\times 10 J=90J

 


Option 1)

1 J

This option is incorrect.

Option 2)

90 J

This option is correct.

Option 3)

99 J

This option is incorrect.

Option 4)

100 J

This option is incorrect.

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