Efficiency and coefficients of performance: the Carnot bounds from the Clausius inequality

Question 1. The three modes of operation differ in the actual direction of the transfers.

• As an engine, the fluid receives heat from the hot reservoir, gives up heat to the cold reservoir and supplies work: $Q_{\mathrm{H}}>0$, $Q_{\mathrm{C}}<0$ and $W<0$.
• As a refrigerator or as a heat pump, the fluid receives work, extracts heat from the cold reservoir and gives up heat to the hot reservoir: $W>0$, $Q_{\mathrm{C}}>0$ and $Q_{\mathrm{H}}<0$. The two machines are identical; only the desired service changes, cooling the cold reservoir or heating the hot reservoir.

Over one cycle, the fluid returns to its initial state, so its internal energy does not change: $\Delta U=0$. The first law $\Delta U=Q+W$ gives, in all three cases,

$$Q_{\mathrm{H}}+Q_{\mathrm{C}}+W=0.$$

Question 2. As an engine, $W<0$, so $|W|=-W$. The first law gives $-W=Q_{\mathrm{H}}+Q_{\mathrm{C}}$, hence

$$\eta =\frac{-W}{Q_{\mathrm{H}}}=\frac{Q_{\mathrm{H}}+Q_{\mathrm{C}}}{Q_{\mathrm{H}}}=1+\frac{Q_{\mathrm{C}}}{Q_{\mathrm{H}}}.$$

As a refrigerator or heat pump, $W>0$ and the first law gives $W=-Q_{\mathrm{H}}-Q_{\mathrm{C}}$. Since $Q_{\mathrm{H}}<0$, we have $|Q_{\mathrm{H}}|=-Q_{\mathrm{H}}$. Thus

$${\mathrm{COP}}_{\mathrm{R}}=\frac{Q_{\mathrm{C}}}{-Q_{\mathrm{H}}-Q_{\mathrm{C}}},{\mathrm{COP}}_{\mathrm{HP}}=\frac{-Q_{\mathrm{H}}}{-Q_{\mathrm{H}}-Q_{\mathrm{C}}}.$$

Since $-Q_{\mathrm{H}}=W+Q_{\mathrm{C}}$,

$${\mathrm{COP}}_{\mathrm{HP}}=\frac{W+Q_{\mathrm{C}}}{W}=1+\frac{Q_{\mathrm{C}}}{W}=1+{\mathrm{COP}}_{\mathrm{R}}.$$

As an engine, $Q_{\mathrm{C}}<0$ and $Q_{\mathrm{H}}>0$, so $Q_{\mathrm{C}}/Q_{\mathrm{H}}<0$ and $\eta <1$: the work supplied cannot exceed the heat received, which the first law alone is enough to guarantee. A coefficient of performance, by contrast, can exceed $1$: a heat pump does not create the heat it delivers; it moves it from cold to hot, and the work received is only part of it. In fact, we always have ${\mathrm{COP}}_{\mathrm{HP}}>1$, since ${\mathrm{COP}}_{\mathrm{R}}>0$.

Question 3. The fluid undergoes a cycle while exchanging heat with only the two reservoirs. The Clausius inequality reads

$$\frac{Q_{\mathrm{H}}}{T_{\mathrm{H}}}+\frac{Q_{\mathrm{C}}}{T_{\mathrm{C}}}\le 0,$$

with equality if and only if the cycle is reversible. Isolate the cold term, then multiply by $T_{\mathrm{C}}>0$, which preserves the direction of the inequality:

$$\frac{Q_{\mathrm{C}}}{T_{\mathrm{C}}}\le -\frac{Q_{\mathrm{H}}}{T_{\mathrm{H}}}⟹Q_{\mathrm{C}}\le -Q_{\mathrm{H}}\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}.$$

As an engine, $Q_{\mathrm{H}}>0$: dividing by $Q_{\mathrm{H}}$ again preserves the direction of the inequality,

$$\frac{Q_{\mathrm{C}}}{Q_{\mathrm{H}}}\le -\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}.$$

Substituting into the expression from Question 2,

$$\eta =1+\frac{Q_{\mathrm{C}}}{Q_{\mathrm{H}}}\le 1-\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}.$$

Equality holds if and only if the Clausius inequality is an equality, that is, for a reversible cycle: we recover Carnot's first and third theorems without any assumption about the working fluid. Finally, $\eta =1$ would require $Q_{\mathrm{C}}=0$; the Clausius inequality would then become $Q_{\mathrm{H}}/T_{\mathrm{H}}\le 0$, which contradicts $Q_{\mathrm{H}}>0$. This is the Kelvin–Planck statement: an engine must necessarily give up heat to the cold reservoir.

Question 4. This time, start from the Clausius inequality by isolating the hot term, then multiply by $T_{\mathrm{H}}>0$:

$$\frac{Q_{\mathrm{H}}}{T_{\mathrm{H}}}\le -\frac{Q_{\mathrm{C}}}{T_{\mathrm{C}}}⟹Q_{\mathrm{H}}\le -Q_{\mathrm{C}}\frac{T_{\mathrm{H}}}{T_{\mathrm{C}}}⟹-Q_{\mathrm{H}}\ge Q_{\mathrm{C}}\frac{T_{\mathrm{H}}}{T_{\mathrm{C}}},$$

the direction of the inequality being reversed when taking the negatives. Add $-Q_{\mathrm{C}}$ to both sides and use $W=-Q_{\mathrm{H}}-Q_{\mathrm{C}}$:

$$W\ge Q_{\mathrm{C}}\left(\frac{T_{\mathrm{H}}}{T_{\mathrm{C}}}-1\right)=Q_{\mathrm{C}}\frac{T_{\mathrm{H}}-T_{\mathrm{C}}}{T_{\mathrm{C}}}.$$

Both sides are positive. Divide by $W>0$, then multiply by $T_{\mathrm{C}}/(T_{\mathrm{H}}-T_{\mathrm{C}})>0$:

$$1\ge \frac{Q_{\mathrm{C}}}{W}\frac{T_{\mathrm{H}}-T_{\mathrm{C}}}{T_{\mathrm{C}}}⟹{\mathrm{COP}}_{\mathrm{R}}=\frac{Q_{\mathrm{C}}}{W}\le \frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}.$$

With the relation from Question 2,

$${\mathrm{COP}}_{\mathrm{HP}}=1+{\mathrm{COP}}_{\mathrm{R}}\le 1+\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}=\frac{T_{\mathrm{H}}-T_{\mathrm{C}}+T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}=\frac{T_{\mathrm{H}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}.$$

As for the engine, these bounds are reached only by a reversible machine.

Question 5. As $T_{\mathrm{C}}\to T_{\mathrm{H}}$, the efficiency bound $1-T_{\mathrm{C}}/T_{\mathrm{H}}$ tends to $0$: without a temperature difference, no work can be produced. The bounds on the two COPs, whose denominator $T_{\mathrm{H}}-T_{\mathrm{C}}$ tends to $0$, tend instead to infinity: moving heat between two bodies at the same temperature costs no work in the reversible limit. As $T_{\mathrm{C}}\to 0$, the efficiency bound tends to $1$, and that of ${\mathrm{COP}}_{\mathrm{R}}$ tends to $0$: extracting heat from an ever colder body costs more and more work per joule extracted, so cooling towards absolute zero becomes increasingly expensive. Question 6. For the refrigerator,

$${\mathrm{COP}}_{\mathrm{R},\max }=\frac{278}{293-278}=\frac{278}{15}\approx 18,5.$$

For the freezer,

$${\mathrm{COP}}_{\mathrm{R},\max }=\frac{255}{293-255}=\frac{255}{38}\approx 6,7.$$

The work expended per joule of heat extracted is $W/Q_{\mathrm{C}}=1/{\mathrm{COP}}_{\mathrm{R}}$. The ratio of the two minimum expenditures is therefore $18,5/6,7\approx 2,8$: even in the ideal case, the freezer expends nearly three times as much work as the refrigerator to extract the same heat, because the temperature difference to be bridged is more than twice as large. Real, irreversible appliances remain well below these bounds.