The Carnot engine
The Carnot cycle
An engine works between a hot reservoir at T₁ and a cold one at T₂. The second law says it can never turn all the heat into work. So what is the best it can possibly do?
In 1824, the French engineer Sadi Carnot found the answer. The best engine must be reversible. Any friction, or any heat flowing across a finite temperature difference, wastes the chance to do work.
So heat may enter or leave only when the gas is at the same temperature as the reservoir: isothermally. And to change temperature between the two reservoirs, the gas must be insulated: adiabatically.
Step 1 to 2. The cylinder sits on the hot reservoir. The gas expands isothermally at T₁, absorbing heat Q₁ equal to μ R T₁ ln of V₂ over V₁, and doing the same amount of work.
Step 2 to 3. The cylinder moves to an insulating stand. The gas keeps expanding adiabatically. It does work μ R times T₁ minus T₂ over γ minus 1, and cools from T₁ down to T₂.
Step 3 to 4. The cylinder sits on the cold reservoir. The gas is compressed isothermally at T₂. Work is done on it, and it releases heat Q₂ equal to μ R T₂ ln of V₃ over V₄.
Step 4 to 1. Back on the insulating stand, the gas is compressed adiabatically. The work done on it raises its temperature from T₂ back to T₁, and the cycle is complete.
On the P-V diagram, the cycle is a closed loop of two isotherms and two adiabats, traced clockwise. The area inside the loop is the net work W done by the gas in one cycle.
Carnot efficiency
Over a full cycle the gas returns to its starting state, so ΔU is zero. The net work is W equals Q₁ minus Q₂, and the efficiency is η equals W over Q₁, which is 1 minus Q₂ over Q₁.
Notice that the work done in the two adiabatic steps is equal and opposite. They cancel out in the total.
Now use T V to the γ minus 1 equals constant on the two adiabats. On step 2 to 3, T₁ V₂ to the γ minus 1 equals T₂ V₃ to the γ minus 1. On step 4 to 1, T₁ V₁ to the γ minus 1 equals T₂ V₄ to the γ minus 1.
Divide one by the other, and V₂ over V₁ equals V₃ over V₄. So the two logarithms are equal, and Q₂ over Q₁ is simply T₂ over T₁.
This gives the Carnot efficiency: η equals 1 minus T₂ over T₁, with both temperatures in kelvin.
Carnot’s theorem says two things. No engine working between two temperatures can be more efficient than a Carnot engine. And the Carnot efficiency does not depend on the working substance: helium, air or steam, it is the same.
Example: a Carnot engine works between 500 K and 300 K. η equals 1 minus 300 over 500, which is 0.4, or 40 percent. If it takes 1000 joules from the hot reservoir, it does 400 joules of work and rejects 600 joules.
A coal power station near your city makes steam at roughly 800 K and cools it with water at about 300 K. Its Carnot limit is about 62 percent. Real plants reach only about 35 to 40 percent.
To reach 100 percent, T₂ would have to be absolute zero. That is impossible, so even a perfect engine always throws away some heat. That is the second law again.
Try it: run a Carnot cycle
- Press Run cycle (or Space), or step through it with Next step. Watch the base under the cylinder change: hot reservoir, insulating stand, cold reservoir, stand.
- The shaded loop is the net work W. After a full cycle, compare W/Q₁ with 1 − T₂/T₁: they match.
- Change the expansion ratio V₂/V₁. Q₁ and W change, but the efficiency does not.
- Lower T₂ or raise T₁ to raise the efficiency.
What sets the efficiency?
Two Carnot engines both work between 600 K and 300 K. Engine X uses helium and expands it to twice its volume in the hot isothermal step; engine Y uses air and expands it to three times its volume. How do their efficiencies compare?
- X is more efficient, because helium is monatomic
- Y is more efficient, because it expands more
- Both are 50% efficient
- It depends on how much gas each engine uses
Answer
C. The Carnot efficiency η = 1 − T₂/T₁ = 1 − 300/600 = 50% depends only on the reservoir temperatures. The working substance and the size of the expansion change how much work is done per cycle, but not the fraction W/Q₁.
Kelvin, not Celsius
A Carnot engine works between 527 °C and 127 °C. It absorbs 1600 J of heat from the hot reservoir in each cycle. How much work does it do per cycle?
- 400 J
- 800 J
- 1214 J
- 1600 J
Answer
B. Convert to kelvin first: T₁ = 527 + 273 = 800 K and T₂ = 127 + 273 = 400 K. η = 1 − 400/800 = 0.5, so W = ηQ₁ = 0.5 × 1600 = 800 J (and Q₂ = 800 J). Using Celsius, 1 − 127/527 ≈ 0.76, wrongly gives 1214 J.
Summary
- Heat absorbed at T₁:
Q₁ = μRT₁ ln(V₂/V₁) - Heat released at T₂:
Q₂ = μRT₂ ln(V₃/V₄) - Work per cycle:
W = Q₁ − Q₂ - Carnot efficiency:
η = 1 − Q₂/Q₁ = 1 − T₂/T₁
- A Carnot engine is a reversible engine working between two reservoirs. Its cycle has two isothermal and two adiabatic steps.
- Heat is exchanged only isothermally, so no heat ever flows across a finite temperature difference.
- Carnot’s theorem: no engine between T₁ and T₂ beats the Carnot engine, and its efficiency is independent of the working substance.
- η = 1 − T₂/T₁ uses kelvin. It reaches 1 only if T₂ = 0 K, which is impossible.