Entropy change of a thermal reservoir and of the universe (Carnot heat engine)

Solution:

The Carnot cycle consists of these four reversible processes:

A-B Isothermal expansion
B-C Adiabatic expansion
C-D Isothermal compression
D-A Adiabatic compression

We will make use of the adiabatic equation for the B-C process to determine the temperature of the cold thermal reservoir temperature (T₂ in the figure):

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The Carnot heat engine efficiency is given by:

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The entropy change of the gas during the adiabatic processes of the Carnot cycle is zero (because Q_R = 0).

The entropy change of the gas during the isothermal expansion AB is given by:

We now calculate the volume in state D using the adiabatic equation applied to states D and A:

The entropy change of the gas during the isothermal compression CD is given by:

The total entropy change of the gas over the entire cycle is equal to the sum of the entropy changes for the different processes of the cycle:

And it is equal to zero because entropy is a state function and in a cycle, the initial and final states are the same.

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In a Carnot cycle, the ideal gas absorbs heat (Q₁) from the hot thermal reservoir during the isothermal expansion. So the hot thermal reservoir discharges the same amount of heat, -Q₁, at constant temperature T₁. Remember that a thermal reservoir is a system that can exchange heat indefinitely while keeping its temperature constant.

The following figure represents the heat flow of a Carnot heat engine:

The isothermal compression of the Carnot cycle takes place while in contact with the cold thermal reservoir at temperature T₂. During this process, the heat Q₂ is removed from gas and absorbed by the cold thermal reservoir (thus Q₂ is positive from the point of view of the cold bath).

The entropy change between any two states A and B is defined:

And, as the temperature of the thermal reservoir does not change, we can pull T out of the integral. As a consequence, the entropy changes of the hot and cold thermal reservoirs are respectively given by:

As you can see, the entropy change of the hot thermal reservoir is minus the entropy change of the gas during the isothermal expansion; and the entropy change of the cold thermal reservoir is minus the entropy change of the gas during the isothermal compression So:

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The entropy change of the universe is the sum of the entropy change of both the thermodynamic system (in this case the Carnot heat engine) and its surroundings (in this case the thermal reservoirs).

Adding all these quantities together we can see that the entropy change of the universe is zero because the Carnot heat engine operates reversibly.

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If a heat engine had 20% of the efficiency of a Carnot heat engine operating between the same two thermal reservoirs at temperatures T₁ y T₂, it would operate irreversible, because all reversible heat engines operating between the same two thermal reservoirs have the same thermal efficiency.

The efficiency of this engine would be:

Where W is the net work output of the engine and Q₁ the heat absorbed by its working fluid from the hot thermal reservoir.

The heat absorbed from the hot thermal reservoir is the same as in the case of the Carnot heat engine:

And, since the efficiency of this engine is 20% of that of Carnot’s engine we have:

We can now deduce the heat Q₂ discharged to the cold thermal reservoir:

And after substituting we have:

The entropy change of the working fluid of this heat engine is zero, because entropy is a state function and therefore, its change depends only on the initial and final states and in a cycle, they are the same.

The entropy change of the thermal reservoirs is given by:

The entropy change of the universe is the sum of these three:

The entropy change of the universe is positive because the heat engine operates irreversibly.

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