Answer
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Hint
From Carnot's theorem we know that no heat engine operating between two reservoirs can be more efficient than a Carnot's engine operating between the same reservoirs. So from here, we can say that the efficiency $ \eta $ will be maximum when the cyclic process is reversible.
Complete step by step answer
A heat engine works by taking heat from a higher temperature region and then by utilizing this heat to do some work. The remaining heat is then rejected in a lower temperature sink. A Carnot’s heat engine is a theoretical engine that always operates on the Carnot’s cycle. This means that it is a reversible engine that always follows the same path and does not leave any effect on the system as well as the surrounding.
So, therefore, in the case of a reversible engine, since there is no loss of energy so it can be reversed.
But irreversible engines are those that undergo loss of energy in their process.
According to Carnot’s cycle,
There can be no heat engine that operates between the same two reservoirs and can be more efficient than a Carnot’s engine. Where the efficiency of the Carnot’s engine is given by,
$ {\eta _{\max }} = \dfrac{W}{{{Q_H}}} = 1 - \dfrac{{{T_C}}}{{{T_H}}} $
where $ W $ is the work done by the system
$ {Q_H} $ is the input heat,
$ {T_C} $ and $ {T_H} $ are the absolute temperature of the cold and hot reservoirs.
Therefore we can say that from Carnot’s theorem, the efficiency of a reversible heat engine $ {\eta _r} $ will always be more than the irreversible heat engine $ {\eta _i} $ .
Hence, $ {\eta _r} > {\eta _i} $
So the option (A). $ {\eta _r} > {\eta _i} $ is correct.
Note
The amount of heat that is converted to work in a reversible heat engine is always more than the amount of heat converted to work in an irreversible heat engine because unlike irreversible heat engines, there is no loss of energy in a reversible heat engine.
From Carnot's theorem we know that no heat engine operating between two reservoirs can be more efficient than a Carnot's engine operating between the same reservoirs. So from here, we can say that the efficiency $ \eta $ will be maximum when the cyclic process is reversible.
Complete step by step answer
A heat engine works by taking heat from a higher temperature region and then by utilizing this heat to do some work. The remaining heat is then rejected in a lower temperature sink. A Carnot’s heat engine is a theoretical engine that always operates on the Carnot’s cycle. This means that it is a reversible engine that always follows the same path and does not leave any effect on the system as well as the surrounding.
So, therefore, in the case of a reversible engine, since there is no loss of energy so it can be reversed.
But irreversible engines are those that undergo loss of energy in their process.
According to Carnot’s cycle,
There can be no heat engine that operates between the same two reservoirs and can be more efficient than a Carnot’s engine. Where the efficiency of the Carnot’s engine is given by,
$ {\eta _{\max }} = \dfrac{W}{{{Q_H}}} = 1 - \dfrac{{{T_C}}}{{{T_H}}} $
where $ W $ is the work done by the system
$ {Q_H} $ is the input heat,
$ {T_C} $ and $ {T_H} $ are the absolute temperature of the cold and hot reservoirs.
Therefore we can say that from Carnot’s theorem, the efficiency of a reversible heat engine $ {\eta _r} $ will always be more than the irreversible heat engine $ {\eta _i} $ .
Hence, $ {\eta _r} > {\eta _i} $
So the option (A). $ {\eta _r} > {\eta _i} $ is correct.
Note
The amount of heat that is converted to work in a reversible heat engine is always more than the amount of heat converted to work in an irreversible heat engine because unlike irreversible heat engines, there is no loss of energy in a reversible heat engine.
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