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Entropy-Based Efficiency of a Heat Engine

Created by @nathanedwards
 at November 1st 2023, 6:32:07 pm.
AP_Physics_2-Questions
#efficiency
#entropy
#heat-engine

Question:

A heat engine operates between a hot reservoir at a temperature Th and a cold reservoir at a temperature Tc. The engine absorbs Qh amount of heat from the hot reservoir and rejects Qc amount of heat to the cold reservoir. The efficiency of the engine is given by the formula:

Efficiency=Useful work outputHeat input \text{{Efficiency}} = \frac{{\text{{Useful work output}}}}{{\text{{Heat input}}}}

Where the useful work output is given by:

Useful work output=Qh−Qc \text{{Useful work output}} = Qh - Qc

Using the concept of entropy, derive an expression for the efficiency of the heat engine in terms of the temperatures Th and Tc.

Answer:

Entropy change is given by the formula:

ΔS=QT \Delta S = \frac{{Q}}{{T}}

Where Q is the amount of heat absorbed or rejected by the system and T is the temperature in Kelvin at which the heat transfer occurs.

To derive an expression for the efficiency of the heat engine, we can consider the entropy change of the hot reservoir and the cold reservoir.

The entropy change of the hot reservoir is given by:

ΔSh=QhTh \Delta S_h = \frac{{Q_h}}{{T_h}}

Similarly, the entropy change of the cold reservoir is given by:

ΔSc=QcTc \Delta S_c = \frac{{Q_c}}{{T_c}}

According to the second law of thermodynamics, the total entropy change of a system and its surroundings must be greater than or equal to zero:

ΔStotal≥0 \Delta S_{\text{{total}}} \geq 0

In this case, the total entropy change is the sum of the entropy changes of the hot and cold reservoirs:

ΔStotal=ΔSh+ΔSc \Delta S_{\text{{total}}} = \Delta S_h + \Delta S_c

Substituting the entropy change formulas and simplifying:

ΔStotal=QhTh+QcTc \Delta S_{\text{{total}}} = \frac{{Q_h}}{{T_h}} + \frac{{Q_c}}{{T_c}}

Rearranging the equation:

Qh(1Th)=−Qc(1Tc) Q_h \left( \frac{{1}}{{T_h}} \right) = - Q_c \left( \frac{{1}}{{T_c}} \right)

Dividing by Q_h:

QcQh=−TcTh \frac{{Q_c}}{{Q_h}} = - \frac{{T_c}}{{T_h}}

Now, let's consider the efficiency of the heat engine:

Efficiency=Useful work outputHeat input \text{{Efficiency}} = \frac{{\text{{Useful work output}}}}{{\text{{Heat input}}}}

From the given information, the useful work output is Q_h - Q_c and the heat input is Q_h. Substituting these values:

Efficiency=Qh−QcQh \text{{Efficiency}} = \frac{{Q_h - Q_c}}{{Q_h}}

Dividing numerator and denominator by Q_h:

Efficiency=1−QcQh \text{{Efficiency}} = 1 - \frac{{Q_c}}{{Q_h}}

Using the equation we derived earlier:

Efficiency=1+TcTh \text{{Efficiency}} = 1 + \frac{{T_c}}{{T_h}}

Therefore, the efficiency of the heat engine can be expressed in terms of the temperatures Th and Tc as:

Efficiency=1+TcTh \boxed{{\text{{Efficiency}} = 1 + \frac{{T_c}}{{T_h}}}}

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