Carnot's Theorem and Absolute Temperature
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You should be comfortable with the definition of thermal efficiency \(\eta = W/Q_H\) and the concept of a reversible cycle. Familiarity with the Kelvin temperature scale as a theoretical construct is assumed.
Why this matters
Carnot's theorem establishes the universal upper limit for heat engine efficiency, independent of working substance.
While the Second Law of Thermodynamics tells us that perpetual motion machines of the second kind are impossible, it does not initially specify how efficient a real engine can be. Carnot's theorem resolves this by proving that the maximum possible efficiency depends solely on the temperatures of the heat reservoirs, not on the material used in the engine. This insight is profound because it decouples thermodynamic limits from material science, allowing us to define an absolute temperature scale based purely on thermal efficiency. It provides the rigorous foundation for the Kelvin scale and sets the benchmark against which all real-world engines, from steam turbines to internal combustion engines, are measured.
The Universal Limit
Imagine two heat engines operating between the same hot reservoir at temperature \(T_H\) and the same cold reservoir at temperature \(T_C\). One engine uses steam, the other uses an ideal gas. Intuitively, one might expect the steam engine to be more efficient due to the phase change, or the ideal gas engine to be superior due to its simplicity. The argument, which goes back to Carnot (1824) and was put in its modern form by Clausius and Kelvin, is that if one engine were more efficient than the other, we could couple them: let the more efficient engine run forward, producing work, and use that work to drive the less efficient engine in reverse, acting as a refrigerator. The net result would be a transfer of heat from the cold reservoir to the hot reservoir with no other change, violating the Clausius statement of the Second Law. Therefore, no engine can be more efficient than a reversible engine operating between the same two temperatures. This logical necessity implies that the maximum efficiency is a universal function of the reservoir temperatures alone.
Defining the Absolute Temperature Scale
Kelvin took this further: the ratio of heat rejected to heat absorbed by a reversible engine is a function of the reservoir temperatures only. Let \(Q_H\) be the heat absorbed from the hot reservoir and \(Q_C\) be the heat rejected to the cold reservoir. For a reversible engine, we can write:
By considering a third reservoir at temperature \(T_0\) and using the transitivity of the efficiency limit, Kelvin showed that this function must be separable. Specifically, the ratio depends on the ratio of two temperature functions, \(F(T_C)/F(T_H)\). By defining the absolute temperature \(T\) such that \(F(T) \propto T\), we arrive at the relation:
This equation is the cornerstone of the absolute temperature scale. It defines temperature not by the expansion of mercury or the resistance of platinum, but by the fundamental thermodynamic property of heat exchange in a reversible cycle. This scale is independent of any specific substance and is now known as the thermodynamic temperature scale, which aligns with the Kelvin scale.
Deriving the Carnot Efficiency
The efficiency \(\eta\) of any heat engine is defined as the work output divided by the heat input: \(\eta = W/Q_H\). From the First Law of Thermodynamics, for a cyclic process, the work done is the difference between the heat absorbed and the heat rejected: \(W = Q_H - Q_C\). Substituting this into the efficiency definition gives:
For a reversible engine, we substitute the Carnot relation \(Q_C/Q_H = T_C/T_H\). This yields the maximum possible efficiency:
Any irreversible engine operating between the same reservoirs will have a lower efficiency, meaning \(Q_C/Q_H > T_C/T_H\). Thus, the general inequality for any heat engine is:
This result is strikingly simple. It shows that efficiency approaches unity only if \(T_C\) approaches absolute zero or \(T_H\) approaches infinity. Since neither is physically achievable, all real engines have efficiencies strictly less than 100%.
Implications for Refrigerators and Heat Pumps
The logic of Carnot's theorem extends directly to refrigerators and heat pumps, which are heat engines run in reverse. For a refrigerator, the coefficient of performance (COP) is defined as the heat removed from the cold reservoir divided by the work input: \(\text{COP}_{\text{ref}} = Q_C/W\). Using the First Law, \(W = Q_H - Q_C\), so:
For a reversible refrigerator, \(Q_H/Q_C = T_H/T_C\). Therefore, the maximum COP is:
Similarly, for a heat pump, which delivers heat to the hot reservoir, the COP is \(\text{COP}_{\text{hp}} = Q_H/W\). The maximum value is:
These expressions demonstrate that the performance of cooling and heating devices is also bounded by the absolute temperatures of the reservoirs. A smaller temperature difference between \(T_H\) and \(T_C\) results in a higher COP, which is why heat pumps are more efficient in mild climates than in freezing conditions.
The derivation
Carnot's Theorem and Maximum Engine Efficiency
The second law forces every reversible engine between two reservoirs to share the same efficiency 1 - T_c/T_h, bounding all real engines.
A common misconception is that the Carnot efficiency depends on the working substance. It does not. The efficiency is a property of the reservoirs, not the engine. However, achieving this efficiency requires the engine to be perfectly reversible, which implies infinitely slow processes and zero friction. In practice, real engines operate far below this limit due to irreversibilities such as turbulence, friction, and finite temperature gradients during heat transfer. The Carnot limit is an idealization that serves as a benchmark, not a target for practical design. C
Worked example
Problem. A power station raises steam at 800 K and condenses it at 300 K, delivering 100 MW of electrical work. (a) What is the Carnot efficiency, and the least possible rate of heat rejection? (b) The real plant has an efficiency of 42%. How much heat does it actually reject? (c) A heat pump moves heat from outdoors at 273 K into a house at 293 K. What is its largest possible coefficient of performance?
- (a) \(\eta_{\mathrm{C}} = 1 - T_C/T_H = 1 - 300/800 = 0.625\).
- Least heat input: \(\dot Q_H = \dot W/\eta_{\mathrm{C}} = 100/0.625 = 160\ \mathrm{MW}\), so the least rejection is \(\dot Q_C = 160 - 100 = 60\ \mathrm{MW}\).
- (b) Real plant: \(\dot Q_H = 100/0.42 = 238\ \mathrm{MW}\), so \(\dot Q_C = 238 - 100 = 138\ \mathrm{MW}\), more than twice the reversible minimum.
- (c) \(\mathrm{COP_{hp,max}} = \dfrac{T_H}{T_H - T_C} = \dfrac{293}{293 - 273} = 14.65\).
- Interpret: an ideal heat pump would deliver almost 15 J of heat to the house for each joule of work. Real units achieve 3–4, because of finite temperature differences in the heat exchangers.
Units check. Efficiencies and COPs are ratios of energies (or powers) and are dimensionless; MW divided by a pure number stays in MW.
Three quick tests
- Can a heat engine operating between \(T_H = 400\ \mathrm{K}\) and \(T_C = 200\ \mathrm{K}\) have an efficiency of 50%?
Answer
Yes, this is exactly the Carnot limit. Any real engine would have an efficiency less than 50%. - If the cold reservoir temperature is held constant, how does the Carnot efficiency change as the hot reservoir temperature increases?
Answer
The efficiency increases, approaching 100% as \(T_H\) approaches infinity, but never reaching it. - Why is it impossible to build a refrigerator with a COP greater than \(T_C/(T_H - T_C)\)?
Answer
Such a device would violate the Second Law of Thermodynamics by transferring heat from cold to hot without sufficient work input, effectively creating a perpetual motion machine of the second kind.
Problem set for PU-203 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.