22 Jun 2025

Heat Engines, Carnot Cycle, and Refrigerators

Second-law statements, reversible engines, Carnot efficiency, refrigeration, and coefficients of performance.

bsc semester-iv mj-6 heat-and-thermodynamics carnot-cycle heat-engines

A heat engine operates cyclically between a hot reservoir at $T_h$ and a cold reservoir at $T_c<T_h$. Let $Q_h>0$ be heat absorbed from the hot reservoir, $Q_c>0$ the magnitude rejected to the cold reservoir, and $W>0$ the net work output. Since the working substance returns to its initial state,

\[W=Q_h-Q_c.\]

Its thermal efficiency is

\[\boxed{\eta=\frac{W}{Q_h}=1-\frac{Q_c}{Q_h}}.\]

Statements of the second law

The Kelvin-Planck statement says that no cyclic device can take heat from a single reservoir and convert it completely into work. Thus $Q_c$ cannot vanish for an engine operating in a cycle.

The Clausius statement says that no cyclic device can have as its sole effect the transfer of heat from a colder body to a hotter body. Such a transfer requires work input.

The statements are equivalent. A perfect single-reservoir engine could drive a refrigerator without net work and violate Clausius; a refrigerator that required no work could combine with an ordinary engine to convert the heat taken from one reservoir completely into work and violate Kelvin-Planck.

Reversible Carnot cycle

For $n$ moles of ideal gas with constant heat capacities, the Carnot engine consists of four reversible processes:

1. $1\to2$: isothermal expansion at $T_h$, absorbing $Q_h$;
2. $2\to3$: adiabatic expansion from $T_h$ to $T_c$;
3. $3\to4$: isothermal compression at $T_c$, rejecting $Q_c$;
4. $4\to1$: adiabatic compression from $T_c$ to $T_h$.

The two isothermal steps give

\[Q_h=nRT_h\ln\!\frac{V_2}{V_1}, \qquad Q_c=nRT_c\ln\!\frac{V_3}{V_4}.\]

On each reversible adiabat, $TV^{\gamma-1}$ is constant. Hence

\[T_hV_2^{\gamma-1}=T_cV_3^{\gamma-1}, \qquad T_cV_4^{\gamma-1}=T_hV_1^{\gamma-1},\]

which imply

\[\frac{V_2}{V_1}=\frac{V_3}{V_4}.\]

Therefore

\[\frac{Q_c}{Q_h}=\frac{T_c}{T_h}, \qquad \boxed{\eta_{\mathrm C}=1-\frac{T_c}{T_h}}.\]

Temperatures must be absolute. Unit efficiency would require $T_c=0$, which is unattainable.

Carnot theorem

No engine operating between two fixed reservoirs can be more efficient than a reversible engine between them. Suppose an engine were more efficient than a reversible Carnot engine while absorbing the same $Q_h$. It would produce more work and reject less heat. Use part of its work to run the Carnot engine backward. The hot-reservoir exchanges cancel, leaving net work while an equal amount of heat is removed from the cold reservoir: a Kelvin-Planck violation. Thus the supposed engine cannot exist.

Running either of two reversible engines backward gives the same contradiction unless their efficiencies are equal. Therefore all reversible engines between $T_h$ and $T_c$ have the same Carnot efficiency, independent of working substance.

Refrigerator and coefficient of performance

A refrigerator is the reversed heat-engine cycle. It removes $Q_c$ from the cold reservoir, receives work $W$, and rejects

\[Q_h=Q_c+W\]

to the hot reservoir. Its coefficient of performance is

\[\boxed{\mathrm{COP}_{R}=\frac{Q_c}{W}}.\]

For a reversible refrigerator,

\[\boxed{\mathrm{COP}_{R,\mathrm C}=\frac{T_c}{T_h-T_c}}.\]

If the desired output is heating, the same reversed device is a heat pump with

\[\mathrm{COP}_{H}=\frac{Q_h}{W} =\frac{T_h}{T_h-T_c}=\mathrm{COP}_{R}+1.\]

The linked Unit II Maxima worksheet verifies

\[\eta_{\mathrm C}-\left(1-\frac{T_c}{T_h}\right)=0.\]

Solved Problems

1. Derive Carnot efficiency from reservoir entropy balance

Over one reversible cycle the working substance has zero net entropy change. The hot reservoir loses $Q_h/T_h$ and the cold reservoir gains $Q_c/T_c$. Reversibility requires zero entropy generation, hence

\[-\frac{Q_h}{T_h}+\frac{Q_c}{T_c}=0.\]

Therefore $Q_c/Q_h=T_c/T_h$. With $W=Q_h-Q_c$,

\[\boxed{\eta=\frac{W}{Q_h}=1-\frac{T_c}{T_h}}.\]

No ideal-gas equation has been used, demonstrating why Carnot efficiency is independent of working substance. If either heat transfer occurred through a finite temperature difference, entropy generation would be positive and the efficiency would be smaller.

2. Couple an engine directly to a refrigerator

Let an engine of efficiency $\eta_E$ absorb $Q_h^{(E)}$. Its work output is

\[W_E=\eta_EQ_h^{(E)}.\]

If all of this work drives a refrigerator of coefficient of performance $\mathrm{COP}_R$, the heat removed from the refrigerator’s cold space is

\[\boxed{Q_c^{(R)}=\mathrm{COP}_R\,\eta_EQ_h^{(E)}}.\]

The refrigerator rejects $Q_h^{(R)}=Q_c^{(R)}+W_E$. These relations separately conserve energy in both cyclic devices; they do not imply reversibility unless each device reaches its corresponding Carnot bound.

Descriptive Questions

1. Show logically why the Kelvin-Planck and Clausius statements of the second law are equivalent.
2. Why must all temperatures in Carnot formulae be thermodynamic temperatures rather than Celsius temperatures?
3. State Carnot’s theorem and explain which step would fail for an irreversible comparison engine.
4. Distinguish thermal efficiency, refrigerator COP, and heat-pump COP.

Numerical Problems

1. Find the maximum efficiency of an engine operating between $600\ \mathrm K$ and $300\ \mathrm K$.

   Final answer: $\eta_C=1-300/600=0.500$ or $50.0\%$.

2. Such a reversible engine absorbs $1.20\ \mathrm{kJ}$ per cycle. Find its work output and rejected heat.

   Final answer: $W=0.600\ \mathrm{kJ}$ and $Q_c=0.600\ \mathrm{kJ}$.

3. Find the maximum refrigerator COP between $270\ \mathrm K$ and a room at $300\ \mathrm K$.

   Final answer: $\mathrm{COP}_{R,C}=270/(300-270)=9.00$.

4. A real engine absorbs $2.00\ \mathrm{kJ}$ and rejects $1.30\ \mathrm{kJ}$ per cycle. Find its efficiency.

   Final answer: $\eta=(2.00-1.30)/2.00=0.350$ or $35.0\%$.

The Carnot and refrigerator Maxima worksheet verifies the heat-sign convention, energy balances, efficiency and COP identities, and all numerical answers.

References

1. Carnot cycle, Wikipedia.
2. H. B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed., Wiley, 1985, chapter 4.
3. M. W. Zemansky and R. H. Dittman, Heat and Thermodynamics, 7th ed., McGraw-Hill, 1997, chapters “The second law” and “Ideal engines.”