A Carnot engine takes 3\times 10^{6}  cal  of heat from a reservoir at 627^{\circ}C,  and gives it to a sink at 27^{\circ}C.  The work done by the engine is

  • Option 1)

    4.2\times 10^{6}\, J

  • Option 2)

    8.4\times 10^{6}\, J

  • Option 3)

    16.8\times 10^{6}\, J

  • Option 4)

    zero

 

Answers (1)

As we learnt in 

Efficiency of a cyclic process -

\eta = \frac{work\, done \, per \, cyclic}{gross \, heat\, supplied \, per\, cyclic}

 

- wherein

Gross heat implied only part of heat absorbed.

 

 

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}\right )

 

 Given,

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

 =1- \frac{300}{900}=\frac{W}{3\times 10^{6}cal}

or \frac{W}{3\times 10^{6} cal}=\frac{2}{3}

or W= 2\times 10^{6} \ kcal

= 8.4\times 10^{6} Joule

 


Option 1)

4.2\times 10^{6}\, J

This option is incorrect.

Option 2)

8.4\times 10^{6}\, J

This option is correct.

Option 3)

16.8\times 10^{6}\, J

This option is incorrect.

Option 4)

zero

This option is incorrect.

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