IB Physics HLThermodynamicsPaper 1 & 2Maximum Efficiency~11 min read
The Carnot Cycle
Every real engine wastes heat — but how little can it possibly waste? Is there a “best” engine that no design can ever beat? There is: the Carnot cycle, a perfect, reversible engine that sets the absolute ceiling on efficiency. Beautifully, it’s built entirely from the four processes you already know — two isothermal, two adiabatic — and it’s the grand finale of the whole topic.
📚 What you need to know
The Carnot cycle is the most efficient possible cycle between two reservoirs — an idealised, fully reversible engine
It runs four stages: isothermal expansion (heat in QH at TH) → adiabatic expansion (cools TH→TC) → isothermal compression (heat out QC at TC) → adiabatic compression (heats TC→TH)
On a p–V graph it’s a loop (area = work); on a T–S graph it’s a neat rectangle
The two adiabatic steps have ΔS = 0, so over a full cycle net ΔS = 0, giving QC/QH = TC/TH
Maximum (Carnot) efficiency: ηC = 1 − TC/TH (temperatures in kelvin)
A bigger temperature gap → higher maximum efficiency, and no real engine can beat the Carnot limit
The perfect engine
Back on the Heat Engines page we saw that every engine dumps some waste heat. Sadi Carnot asked the natural next question: what’s the best you could ever do? His answer was an idealised engine — perfectly reversible, with no friction and no wasted entropy — that squeezes the maximum possible work out of a given temperature difference. No real machine reaches it, but it tells us the ceiling that nothing can exceed.
Treat the Carnot engine as a thought experiment, like a frictionless surface in mechanics. You’ll never build one, but it’s incredibly useful: it tells you instantly whether a claimed efficiency is possible, and it shows exactly what makes an engine good — a big temperature gap.
The four stages
The cycle strings together four of our processes, alternating isothermal and adiabatic, and traces a closed loop clockwise on a p–V graph. The area inside is the net work per cycle.
Two isothermals (heat in at the top, heat out at the bottom) joined by two adiabatics (which only change the temperature). The gas ends exactly where it began, ready to go again.
🍽️ The Carnot cycle, step by step
Isothermal expansion (A→B, at TH): the gas absorbs heat QH from the hot reservoir and expands, doing work — temperature stays constant.
Adiabatic expansion (B→C): sealed off from heat, the gas keeps expanding and cools from TH down to TC.
Isothermal compression (C→D, at TC): the gas is compressed and releases waste heat QC to the cold reservoir — temperature stays constant.
Adiabatic compression (D→A): sealed off again, the gas is compressed further and heats back up from TC to TH, returning to the start.
Notice heat only crosses the boundary during the isothermal stages. The adiabatic stages are purely about changing temperature to bridge between the hot and cold isotherms.
The T–S diagram and the efficiency formula
Here’s where the Carnot cycle becomes truly elegant. Draw it on a temperature–entropy graph instead, and the loop turns into a perfect rectangle: the isothermals are horizontal (constant T), and the adiabatics are vertical (constant S, because ΔQ = 0 means ΔS = 0).
A rectangle! The top edge sits at TH (heat in), the bottom at TC (heat out), and the two sides are the adiabatics where entropy doesn’t change. This picture hands us the efficiency formula almost for free.
Watch how simply the famous result drops out. Using ΔS = ΔQ/T on the two isothermal stages, where the entropy changes by the same width ΔS:
Now just feed that into the ordinary efficiency formula from the last page, η = 1 − QC/QH, and the heats turn into temperatures:
Maximum (Carnot) efficiency
ηC = 1 − TC⁄TH
This is one of the most beautiful results in physics. It says the best possible efficiency depends on nothing but the two temperatures — not the gas, not the design, not the size. And it comes straight from “entropy returns to where it started after a full cycle”. Everything you learned about entropy pays off right here.
What the formula tells us
Because TC and TH are always positive, TC/TH is always more than zero — so ηC is always less than 1. You’d only reach 100% if the cold reservoir were at absolute zero, which is impossible. The way to improve an engine is to widen the gap: a hotter hot side or a colder cold side.
The maximum efficiency depends only on the temperature ratio. Push TC/TH toward zero (a huge gap) and efficiency climbs toward 1; let the temperatures get close and it falls toward zero.
WE 1
A power station runs between a boiler at 800 K and a cooling river at 300 K. What is its maximum possible efficiency?
Use ηC = 1 − TC/TH (both already in kelvin)ηC = 1 − 300/800 = 1 − 0.375ηC = 0.625 = 62.5%No engine between these two temperatures can ever beat 62.5% — and a real one, with friction, will fall short of it.
WE 2
An engineer claims to have built an engine that is 60% efficient while running between 600 K and 300 K. Is this possible?
Find the Carnot ceiling for these temperaturesηC = 1 − 300/600 = 0.50 = 50%Compare with the claimImpossible — 60% > the 50% Carnot limitNo engine can beat Carnot, so a claim above the ceiling breaks the second law. The engineer is mistaken.
WE 3
A Carnot engine takes 1000 J from a hot reservoir at 500 K and rejects heat to a cold reservoir at 400 K. Find (a) its efficiency, (b) the work done per cycle, and (c) the heat rejected.
Heat only moves in the isothermal steps; the adiabatic steps just change the temperature.
ηC = 1 − TC/TH — temperatures in kelvin, always.
Bigger temperature gap = higher ceiling. 100% would need TC = 0 (absolute zero).
Use it as a test: any claimed efficiency above the Carnot value is impossible.
⚠ Common mistakes
Using temperatures in °C — ηC = 1 − TC/TH needs kelvin
Thinking a real engine can reach the Carnot efficiency — it’s an unreachable ideal
Forgetting heat is exchanged only in the two isothermal stages
Confusing ηC = 1 − TC/TH (Carnot only) with η = 1 − QC/QH (any engine)
Getting the four-stage order wrong, or swapping expansion and compression
Quick recap: The Carnot cycle is the most efficient possible engine — a reversible loop of isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. It’s a loop on a p–V graph and a rectangle on a T–S graph. Since net entropy change is zero, QC/QH = TC/TH, giving the maximum efficiency ηC = 1 − TC/TH — a ceiling no real engine can beat.
And that completes Thermodynamics — congratulations! You’ve travelled the whole way: from what internal energy really is, through the first law’s energy bookkeeping, into entropy and the direction of time, up to the second law, the gas processes, and finally the perfect engine that ties it all together. Every idea in this last page rests on the ones before it — which is exactly why it feels so satisfying when it clicks. Go back over the diagrams, re-do the worked examples from a blank page, and you’ll be more than ready for the exam.
Want this to actually click before the exam?
Book a free meeting and let’s work through the tricky bits together.
