The Carnot cycle & the efficiency ceiling

Heat that crosses a finite temperature difference cannot be pushed back without extra work, so a reversible engine may exchange heat with a reservoir only while its working substance is at that reservoir's temperature. With two reservoirs there are two such temperatures: heat can enter only along an isotherm at T_H and leave only along an isotherm at T_C, and every stroke joining them must be adiabatic. That fixes the cycle. From state a the gas expands isothermally at T_H to b, taking in Q_H; it expands adiabatically to c, cooling to T_C; it is compressed isothermally at T_C to d, rejecting Q_C; and it is compressed adiabatically back to a. Take 1.00 mol of monatomic gas (γ = 5/3) with Vₐ = 1.00 L, V_b = 2.00 L, T_H = 600 K and T_C = 300 K. Along each adiabat TV^(γ−1) is constant, so V_c = V_b (T_H/T_C)³⁄² = 2.00 × 2.83 = 5.66 L and V_d = Vₐ (T_H/T_C)³⁄² = 2.83 L. On the p-V diagram the cycle is a thin curved quadrilateral hugging the axes, and that thinness is the first hint that a Carnot engine does little work per litre swept.

For an ideal gas U depends on T alone, so ΔU = 0 along an isotherm and the first law gives Q = W = ∫ p dV = nRT ∫ dV/V = nRT ln(V_f/Vᵢ). On a→b: Q_H = nRT_H ln(V_b/Vₐ) = 1.00 × 8.314 × 600 × ln 2 = 3458 J, positive, taken in. On c→d the gas is compressed, so the integral is negative; the heat rejected is Q_C = nRT_C ln(V_c/V_d) = 8.314 × 300 × ln(5.66/2.83) = 1729 J. The adiabats exchange no heat, but they do work: on the expansion b→c the gas does W_bc = −ΔU = nC_V(T_H − T_C) = 1.5 × 8.314 × 300 = 3741 J, and the compression d→a takes exactly the same 3741 J back, because it spans the same temperature interval. The adiabatic works cancel identically, so the net work per cycle is W = Q_H − Q_C = 3458 − 1729 = 1729 J, the area the cycle encloses, and the efficiency is W/Q_H = 1729/3458 = 0.500.

Why did the two logarithms match? Along an adiabat TV^(γ−1) is constant, so b→c gives T_H V_b^(γ−1) = T_C V_c^(γ−1) and d→a gives T_H Vₐ^(γ−1) = T_C V_d^(γ−1). Divide one by the other: (V_b/Vₐ)^(γ−1) = (V_c/V_d)^(γ−1), hence V_b/Vₐ = V_c/V_d and the two logarithms are equal. Then Q_H/Q_C = nRT_H ln(V_b/Vₐ) ÷ nRT_C ln(V_c/V_d) = T_H/T_C, and e = 1 − Q_C/Q_H = 1 − T_C/T_H. Everything about the gas has cancelled: n, R, γ and all four volumes. The expansion ratio scales Q_H, Q_C and W by the same factor and leaves their ratios alone. Check the result as you would any other. Kelvin over kelvin is dimensionless. T_C → 0 gives e → 1, the case where nothing need be rejected; T_C = T_H gives e = 0, and the two isotherms coincide so the cycle encloses no area. Temperatures must be absolute: 600 K and 300 K give 0.500, while the Celsius values 327 and 27 would give the nonsense 1 − 27/327 = 0.917.

Suppose an inventor claims engine X between 600 K and 300 K with efficiency 0.60, above the Carnot value 0.50. Let X take 1000 J from the hot reservoir per cycle: it delivers 600 J of work and rejects 400 J to the cold reservoir. Being reversible, a Carnot engine can run backwards as a refrigerator with every heat and work reversed in sign but not in ratio, Q_H/Q_C = T_H/T_C = 2. Feed it the 600 J: it draws Q_C′ from the cold reservoir and delivers Q_H′ = 2Q_C′ to the hot one, with Q_H′ − Q_C′ = 600 J, so Q_C′ = 600 J and Q_H′ = 1200 J. Audit the pair. Hot reservoir: −1000 + 1200 = +200 J. Cold reservoir: +400 − 600 = −200 J. Net work: zero. So 200 J per cycle has passed from cold to hot with no other change, which the Clausius statement forbids. Hence no engine beats e_C. Run it with two reversible engines, each in turn as the rival, and both directions are forbidden: every reversible engine between the same reservoirs has the same efficiency, whatever its working substance. That is what lets Q_H/Q_C = T_H/T_C define temperature itself, independent of any thermometer substance.

Heat flows between a reservoir and the working substance at a rate proportional to their temperature difference, Q̇ = K(Tᵣₑₛₑᵣᵥₒᵢᵣ − T_gas) for some conductance K in W K⁻¹. Reversibility demands that this difference vanish, so the heat flows at zero rate, each isotherm takes infinite time, and the power W/τ of a strictly reversible engine is zero. A real engine must run its gas at some Tₕ below T_H while absorbing and at some T_c above T_C while rejecting. The cycle between Tₕ and T_c can still be a Carnot cycle with efficiency 1 − T_c/Tₕ, but that is below 1 − T_C/T_H, and the closer the gas temperatures sit to the reservoirs, the less heat flows per second. Maximising the power over the choice of Tₕ and T_c, with linear conduction on both sides, gives the Curzon–Ahlborn efficiency e_CA = 1 − √(T_C/T_H). For a steam plant with T_H = 800 K and T_C = 300 K the Carnot ceiling is 1 − 0.375 = 0.625, while e_CA = 1 − √0.375 = 1 − 0.612 = 0.388, close to what such plants achieve. The gap between the two is not sloppiness; it is the price of finishing a cycle in finite time.

Because e_C = 1 − T_C/T_H depends on two temperatures, ask how each moves it. The partial derivatives are ∂e_C/∂T_H = T_C/T_H² and ∂e_C/∂T_C = −1/T_H, so the magnitude of the second exceeds the first by the factor T_H/T_C, which is always greater than one. Lowering the sink by one kelvin buys T_H/T_C times as much efficiency as raising the source by one kelvin. At T_H = 800 K and T_C = 300 K the gains are 300/800² = 4.69 × 10⁻⁴ per kelvin of hotter source against 1/800 = 1.25 × 10⁻³ per kelvin of colder sink, a factor of 2.67. Engineers still chase T_H rather than T_C: the sink is the atmosphere or a river, fixed near 290 K by the planet, while the source is limited only by turbine blades and boiler steel, so a 50 K rise in T_H is available and a 50 K drop in T_C is not. The derivative also says where the last percentage points are hardest: at T_H = 800 K a further kelvin on the source moves e_C by only 0.047 of a percentage point.