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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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Vakul

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