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A carnote engine absorbs `8 KJ` of energy at `400K`. If sink is maintained at `300 K`, Calculate useful work done per cyclic (in joule) by the engine.

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Correct Answer - B
Here, `T_(1)= 400K, Q_(1)= 8 KJ, W=?, T_(2)= 300 K`
As `Q_(2)/(Q_(1))=(T_(2))/(T_(1))`
`:. Q_(2)=Q_(1)xx(T_(2))/(T_(1))= 8xx(300)/(400)= 6 J`
`W= Q_(1)-Q_(2)= 8-6= 2KJ`

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