What is the formula for Carnot engine efficiency?

The Carnot efficiency is η = 1 − Tc/Th, where Tc and Th are the absolute (kelvin) temperatures of the cold and hot reservoirs the engine works between. It is the maximum efficiency any heat engine can reach between those two temperatures — a limit set by the second law of thermodynamics rather than by any engineering shortfall. Equivalently, η = W/Qh = (Qh − Qc)/Qh, where W is the net work per cycle and Qh, Qc are the heat absorbed from the hot reservoir and rejected to the cold one. For example, between 500 K and 300 K the ceiling is 1 − 300/500 = 40%.

Why can't a heat engine be 100% efficient?

Because of the second law of thermodynamics: turning heat into work in a repeating cycle always requires dumping some waste heat into a colder reservoir. An engine converting 100% of its absorbed heat into work would reject zero heat, which is exactly what the Kelvin–Planck statement of the second law forbids — the entropy of the universe would have to fall. The Carnot formula only reaches 100% in the limit Tc = 0 K, and the third law says absolute zero is unreachable. On top of that, every real engine also falls short of its own Carnot limit because real processes involve friction and finite-rate heat transfer, both of which generate entropy.

What is the difference between the Carnot cycle and a real engine cycle?

The Carnot cycle is a theoretical, perfectly reversible cycle: every step happens infinitely slowly, with no friction, and heat only ever crosses an infinitesimal temperature difference. Real cycles — the Otto cycle in a petrol engine, the Diesel cycle, the Rankine cycle in a steam plant — run fast, involve friction and turbulence, and transfer heat across finite temperature gaps, all of which are irreversible and generate entropy. That is why every real engine falls below the Carnot limit computed from the same two reservoir temperatures. The deepest difference is that a Carnot engine would produce zero power, because an infinitely slow engine delivers no work per unit time.

What is the second law of thermodynamics in simple terms?

Heat flows downhill, from hot to cold, on its own — never the other way without outside help. Applied to engines, that means you cannot build a machine that takes heat from a single warm reservoir and turns all of it into work; some has to be dumped somewhere colder. The formal version says the entropy of an isolated system never decreases, and every forbidden engine is forbidden for the same reason: running it would require the entropy of the universe to drop, which does not happen even by a thousandth of a joule per kelvin.

Can the Carnot cycle run in reverse as a refrigerator or heat pump?

Yes. Run every step backwards and you get a Carnot refrigerator: work is put in to pull heat out of a cold space and dump it somewhere warmer, which is what a fridge, an air conditioner and a heat pump all do. Performance is measured by the coefficient of performance rather than efficiency — COP = Qc/W for cooling — with a reversible ceiling of Tc/(Th − Tc). A fridge holding its interior at 275 K in a 295 K kitchen (roughly 2 °C in a 22 °C room) has a Carnot COP of 275/20 = 13.75, so in principle one joule of electricity moves nearly fourteen joules of heat. That is not free energy: the heat is being moved, not created, and the COP collapses as the temperature gap widens.

What is the maximum possible efficiency of a power plant?

The Carnot ceiling for a coal-fired steam plant running at about 550 °C (823 K) and rejecting heat near 25 °C (298 K) is about 64%, but no plant is designed to reach it, because a Carnot engine produces zero power — it has to run infinitely slowly. The number real plants actually track is the maximum-power efficiency, 1 − √(Tc/Th), which for those temperatures is about 40%. Real coal plants deliver 35–42%, sitting right on that maximum-power point. Combined-cycle gas plants reach 55–62% by running a much hotter gas turbine (inlet near 1500 °C) and raising steam from its exhaust, which lifts their own Carnot ceiling far above 64%.

Does the Carnot efficiency formula work with Celsius temperatures?

No — Th and Tc must be absolute temperatures in kelvin. Using Celsius gives a badly wrong answer: an engine between 227 °C and 27 °C might tempt you to compute 1 − 27/227 ≈ 88%, while converting first to 500 K and 300 K gives the correct 1 − 300/500 = 40%. The formula contains a ratio of temperatures, and a ratio only means anything when both numbers are counted from the same true zero. Celsius counts from the freezing point of water, an arbitrary landmark; kelvin counts from absolute zero, which is where the entropy derivation behind the formula actually starts.

Carnot Engine Efficiency Calculator: The Laws of Thermodynamics Explained

Q: What are the laws of thermodynamics?

The first law is energy conservation: energy moves between heat and work but the total is fixed, ΔU = Q − W. The second law says heat does not flow from cold to hot on its own, and no engine can convert heat into work with 100% efficiency in a repeating cycle — some heat must always be rejected to a colder reservoir. The zeroth law defines temperature itself (two systems each in equilibrium with a third are in equilibrium with each other), and the third law says entropy approaches a minimum as temperature approaches absolute zero, which is itself unreachable. The Carnot efficiency formula follows directly from the second law.

Every engine ever built is, mostly, a machine for throwing heat away. The petrol engine in a car sheds around 70% of the energy in its fuel as hot exhaust and warm coolant. The finest power stations on Earth, running steam at temperatures that would soften aluminium, still throw away close to 40%. That is not a workshop problem waiting on a better alloy or tighter tolerances — nobody has dented it in two centuries of trying. A young French army engineer named Sadi Carnot worked out why in 1824, a quarter of a century before anyone wrote down the first law of thermodynamics, using a theory of heat that later turned out to be wrong. He was right anyway, and the ceiling he found has never been crossed.

There are three things on this page to push on until that stops sounding like a rule and starts sounding obvious: a four-stroke Carnot cycle you can watch run in a real cylinder, an auditor that takes any engine you can specify and stamps the impossible ones FORBIDDEN, and a power-plant throttle that explains why real stations land on 40% rather than the number the formula promises them. All three ask something the ideal gas law calculator never has to. That page relates pressure, volume, moles and temperature for a gas sitting in one state. This one drags the gas around a closed cycle of states and asks what the trip pays.

What Are the Laws of Thermodynamics?

There are four of them, numbered zero to three, and two do nearly all the work where engines are concerned.

What Is the First Law of Thermodynamics? (Conservation of Energy)

The first law is energy conservation, written for a system that can exchange both heat and work:

$\Delta U = Q - W$

Here $Q$ is heat added to the system and $W$ is work done by the system on its surroundings — the convention engine problems use, because the whole point of an engine is the work coming out. So a gas that absorbs 100 J of heat while pushing the piston out with 30 J of work ends up 70 J richer inside. Heat that flows in either raises the internal energy or leaves again as work. There is no third destination, and nothing is created or destroyed anywhere in the transaction.

What Is the Second Law of Thermodynamics? (Why Every Engine Wastes Heat)

The second law is where engines get their bad news. Heat never flows on its own from cold to hot — you have to pay work to push it uphill, which is what a refrigerator is for. And no engine can take heat from a single reservoir and turn all of it into work in a repeating cycle; some heat always has to go somewhere colder.

Those two statements sound unrelated and are the same law. Both say that a quantity called entropy never decreases for an isolated system. A perfectly efficient engine, running in a loop back to its own starting state, would need entropy to drop somewhere for free — and the universe declines. The first law says the books have to balance. The second says which direction the ledger is allowed to move, and that turns out to be the binding constraint on every engine anyone has ever built.

What Is a Heat Engine and How Does It Turn Heat Into Work?

A heat engine is any device that absorbs heat $Q_h$ from something hot, turns part of it into work $W$ , dumps the remainder as $Q_c$ into something cold, and then returns to exactly the state it started in, ready to go again. That last clause is the one doing the heavy lifting. Because the engine is back where it began, its internal energy has not changed over the lap — $\Delta U = 0$ — and the first law collapses to a single line of bookkeeping:

$W = Q_h - Q_c$

Every engine is some physical arrangement built to do exactly that. A car's four-stroke motor and a power station's steam turbine share almost no parts and no design assumptions, and they are the same machine at this level of description. The Carnot cycle is the idealised, perfectly reversible version, and its whole importance is that it sets the best-case answer for all of them at once.

How Does the Carnot Cycle Work? The Four Strokes Explained

The cycle runs in four legs, alternating between letting heat cross the cylinder wall and blocking it completely:

Isothermal expansion. The gas sits on the hot plate at $T_h$ and expands slowly, absorbing $Q_h = nRT_h\ln(V_2/V_1)$ as it goes. The temperature does not budge, because the reservoir keeps replacing whatever energy leaves as work.
Adiabatic expansion. The cylinder is lifted onto an insulating stand. The gas keeps expanding, but now the only way it can pay for that work is out of its own internal energy, so it cools from $T_h$ down to $T_c$ with no heat exchanged at all.
Isothermal compression. Onto the cold plate at $T_c$ , where the gas is squeezed and rejects $Q_c = nRT_c\ln(V_3/V_4)$ . That volume ratio is the same $V_2/V_1$ from the first leg — the two adiabatic legs have to connect the two isotherms, and the only way the loop closes is if $V_3/V_4 = V_2/V_1$ .
Adiabatic compression. Back on the insulating stand, squeezed the rest of the way home, heating from $T_c$ back to $T_h$ on its own compression, and the cycle is ready to repeat.

The plate under the cylinder is doing something subtler than it looks. Heat crossing a finite temperature difference is irreversible and destroys work, which is why the Carnot version insists that the gas and the plate be at the same temperature and the heat crawl across on an infinitesimal gradient. Push a hot plate hard in the other direction and the physics changes character altogether: a droplet dropped onto a plate far above its boiling point stops boiling and starts levitating, on a vapour cushion of its own making.

Try it yourself

It loads running, one lap every 7 seconds — or paused, if your device asks for reduced motion, in which case the Step button in step 5 is your way around the loop. The banner reads "Isothermal expansion — absorbing Qh = 2.88 kJ from the hot plate at 500 K". Watch one full lap and match each stroke to the furniture under the cylinder: a red plate captioned "hot plate · Th = 500 K", then a hatched "insulating stand · Q = 0", then "cold plate · Tc = 300 K", then the stand again. The plate is the physics; the caption just tells you which one is there.
Press Pause somewhere in the middle of the adiabatic expansion — the second leg, on the dashed grey adiabat. Now press Play and watch the particles inside the cylinder as the piston retreats. They visibly slow down, because nothing is feeding them: the gas is paying for that stroke out of its own kinetic energy, which is what "the temperature falls from 500 K to 300 K" means at the level of the molecules.
Set the Work areas toggle to "Shaded" if it isn't already. Each expansion leg sweeps a solid green region under itself as the state point crosses it, and each compression leg claws back a red hatched one. Green is work coming out, red is work being paid back in, and the green is visibly the fatter of the two — the gas pushes while it is hot and high-pressure, and gets pushed back while it is cold and low. At the end of every lap the whole loop floods green for a beat with a pill reading "W = area ≈ 1.15 kJ".
Now drag the Cold reservoir Tc slider up to 380 K and leave it there. The loop visibly shrinks in front of you — the two isotherms close on each other and the area between them closes with them. That enclosed area is the work, so the ledger falls with it: η = 24.0%, W = 691.9 J. Nothing about the cylinder changed. You moved the cold end 80 K closer, and 40% of the loop's area — 40% of the work — went with it.
If seven seconds a lap is too fast to follow, press "Step one stroke" instead. It pauses the engine and eases it to the end of whichever leg it is on, so you can walk the loop a corner at a time and read the four corner states — marked 1, 2, 3, 4 on the P–V plane — as you go.

Isothermal expansion — absorbing Qh = 2.88 kJ from the hot plate at 500 K

V = 1.00 L · P = 41.1 atm · T = 500 K · this leg's work 2.88 kJ out · η = 1 − 300/500 = 40.0%

This is the quasi-static idealisation: a genuinely reversible Carnot cycle would have to run infinitely slowly, so the 7-second lap here is pedagogical pacing, not a physical rate — a real engine run this fast would not be reversible and could not reach η = 1 − Tc/Th. The particles inside the cylinder are a cartoon of ⟨v⟩ ∝ √T: 36 dots sharing one speed set from the current temperature, not a Maxwell–Boltzmann sample, and their bounces are canvas kinematics rather than a molecular-dynamics solve — they illustrate the temperature, they do not compute it. The plates and the insulating stand that swap under the cylinder are Carnot’s conceptual furniture: nothing here models the machinery that would actually move a cylinder between two reservoirs, or the irreversibility that doing so would cost. The cylinder is drawn with a small clearance volume at the bottom of the piston’s travel so the gas stays visible when V₁ is a sliver of V₃; piston travel itself is exactly proportional to ΔV. Every number on the canvas — the four corner states, both isotherms, both adiabats, Qh, Qc, W and η — is solved for 1 mol of monatomic ideal gas (γ = 5/3) at a fixed expansion ratio V₂/V₁ = 2, but η = 1 − Tc/Th itself does not depend on any of those choices.

Hot reservoir Th

Th500 K

Cold reservoir Tc

Tc300 K

The two sliders stay at least 20 K apart — at Th = Tc the cycle has no legs left to run.

One lap takes 7 s. Step one stroke pauses the engine and eases it to the end of the current leg.

Work areas

Qh — heat in, per cycle2.88 kJ

Qc — heat out, per cycle1.73 kJ

W = Qh − Qc1.15 kJ

η = 1 − Tc/Th40.0%

The green flood at the end of each lap is not decoration. On a pressure–volume diagram, work is $\int P\,dV$ — pressure times the volume swept — so the area beneath each expansion leg is work the gas pays out, and the area beneath each compression leg is work paid back. Subtract the second from the first and everything outside the loop cancels, leaving exactly the region the loop encloses. At the default temperatures the engine draws $Q_h = 2.88$ kJ from the hot plate and returns $Q_c = 1.73$ kJ to the cold one. The 1.15 kJ that never comes back as heat left as work, and the shape on the screen is where it went. Squeeze the loop flat and you have not built a bad engine; you have built no engine.

What Is the Carnot Efficiency Formula? (η = 1 − Tc/Th)

The Carnot efficiency is the largest fraction of absorbed heat that any engine can turn into work while running between a hot reservoir at $T_h$ and a cold one at $T_c$ , both in kelvin:

$\eta = 1 - \frac{T_c}{T_h} = \frac{W}{Q_h}$

Look at what is not in that formula. There is no mass in it, no gas constant, no compression ratio, and no hint of whether the machine is a piston engine or a thermoelectric wafer with no moving parts at all. Two temperatures, and the answer. An engineer who has spent a career on valve timing and a chemist who thinks the answer lies in a better working fluid are both told the same thing: it does not matter, and here is your ceiling.

At the simulation's defaults, $T_h = 500$ K and $T_c = 300$ K, that ceiling is $1 - 300/500 = 0.40$ . The engine draws $Q_h \approx 2.88$ kJ per cycle, rejects $Q_c \approx 1.73$ kJ, and keeps $W \approx 1.15$ kJ — and $1.15/2.88$ is 40%, arrived at by an entirely different route.

How Do You Derive the Carnot Efficiency Formula?

The cleanest derivation never touches a gas law. Entropy is a state function, so over one complete reversible cycle it comes back to where it started:

$\oint \frac{dQ}{T} = 0$

Heat only crosses the boundary on the two isothermal legs, and on each of those $T$ is constant, so the integral collapses to two terms:

$\frac{Q_h}{T_h} - \frac{Q_c}{T_c} = 0 \quad \Longrightarrow \quad \frac{Q_c}{Q_h} = \frac{T_c}{T_h}$

Combine that with $W = Q_h - Q_c$ from the first law:

$\eta = \frac{W}{Q_h} = \frac{Q_h - Q_c}{Q_h} = 1 - \frac{Q_c}{Q_h} = 1 - \frac{T_c}{T_h}$

This is Carnot's theorem: no engine working between two reservoirs can beat a reversible engine working between the same two, and every reversible engine between them hits precisely the same number, whatever it is made of. The gas never appeared in the derivation, which is exactly why the result does not care what you build the engine out of.

Why Can't Any Heat Engine Be 100% Efficient?

Two reasons, and they get run together constantly even though they are completely different complaints.

The first is a wall in the formula itself. $\eta = 1$ needs $T_c = 0$ K, absolute zero, and the third law of thermodynamics says you can approach it but never arrive. Even a flawless, infinitely patient Carnot engine is stopped here. It is not an engineering shortfall — there is simply nowhere cold enough to dump the waste.

The second is why real engines fall short of even their own Carnot limit. The reversible cycle assumes friction has been abolished, and that every step is infinitely slow and driven by an infinitesimal temperature difference. Real pistons move fast, and real heat crosses real temperature gaps rather than infinitesimal ones. Both of those generate entropy nobody gets to un-generate, and every scrap of entropy generated is work quietly deleted from the output.

The simulation below is an auditor rather than an engine. The heat input is nailed down at 1000 J per cycle and the only thing you control is how much of it leaves as work — which is exactly the knob anyone who believes waste heat is a manufacturing defect thinks they should be allowed to turn.

Try it yourself

It opens with the split at W = 250 J: 1000 J in at 500 K, 250 J out as work, 750 J dumped at 300 K, and the entropy of the universe up by 0.500 J/K. The needle on the gauge sits in the green "allowed" region. The banner calls this "Legal, and typical" — which is a fair description of every engine you have ever ridden in.
Grab the handle on the rotor's rim and drag it slowly upward. Up asks for more work, so the work stream to the flywheel fattens and the waste stream to the cold block starves. Watch the needle: as you take more, ΔS shrinks and the needle walks steadily left, out of the green and toward the red "forbidden" band on the other side of the knife-edge tick marked "reversible".
Keep going and find where the stamp drops. It lands at W = 405 J — one 5 J notch past the limit, with ΔS at −0.017 J/K — and the streams go grey and broken, the flywheel stops, and "FORBIDDEN · ΔS < 0 — this machine cannot exist" falls across the rotor. Seventeen thousandths of a joule per kelvin is enough. The second law does not do close enough.
Back off and press "Snap to the limit". W lands on 400 J exactly, the needle parks dead on the knife edge, and a quiet pill appears: "You built a Carnot engine — the best the universe permits." It is the only setting on the whole slider where the entropy ledger comes out at exactly zero.
Now the argument. Push W up to 500 J — forbidden, stamp down, ΔS = −0.333 J/K. Leave it there and drag the Cold reservoir Tc slider down to 250 K instead. The stamp lifts and the badge comes back: the same 500 J of work, out of the same 1000 J of heat, from a machine you did not touch. The limit was never a property of the engine.
Finally press "W = 0 (heat leak)". Nothing turns, the flywheel stops, and ΔS rockets to +1.333 J/K — the needle swings hard into the green. This is the most legal machine on the page and the most useless, which is the second law's actual shape: it is not the work that costs entropy, it is refusing to waste anything that is forbidden.

1000 J in at 500 K, 250 J of work out, 750 J dumped at 300 K — the universe's entropy grew by 0.500 J/K. Legal, and typical.

η = 25.0% against a Carnot ceiling of 40.0% — 150 J of work still on the table before the audit fails.

This is single-cycle bookkeeping with ideal reservoirs: two bodies so large that taking 1000 J out of one and putting 750 J into the other leaves both temperatures exactly where they were, which is why ΔS is just Q/T for each of them. The working substance inside the rotor never appears in the ledger — over a complete cycle it returns to the state it started in, so its own entropy change is exactly zero and only the two reservoirs are left holding anything. “Forbidden” therefore means forbidden for a cycle: a one-shot process, such as a gas expanding isothermally once and never coming back, really can turn heat into work with nothing dumped anywhere, and no law is broken because the gas does not end where it began. Stream widths are energy per cycle, not power — nothing on this canvas says how long a cycle takes, and the dash drift and the spinning wheel are pace, not rate.

Work extracted W

W250 J

Or grab the handle on the rotor’s rim and drag it up and down: up sends more of the heat out as work and swings the handle toward the flywheel, down sends it on into the cold block. Qc = Qh − W, always.

At W = 0 the heat simply falls from hot to cold and nothing turns — maximally legal, maximally useless. The limit is 400 J here, and it is rarely a round number, so the button lands on it exactly.

Hot reservoir Th

Th500 K

Cold reservoir Tc

Tc300 K

Neither slider will let the gap close below 20 K: at Th = Tc there is no engine left to audit.

Qh — heat in, per cycle1000 J

W — work out250 J

Qc = Qh − W — heat dumped750 J

ΔS hot = −Qh/Th−2.000 J/K

ΔS cold = +Qc/Tc+2.500 J/K

ΔS universe+0.500 J/K

η = W/Qh vs η Carnot25.0% vs 40.0%

What you just did by hand was rediscover $W_{max} = Q_h\left(1 - T_c/T_h\right)$ , from the wrong end. You did not derive it; you pushed on the second law until it pushed back, and the place where it pushed back was 400 J. Move either temperature and the wall moves with it, always to exactly $Q_h(1 - T_c/T_h)$ , because the audit is only ever comparing $Q_h/T_h$ against $Q_c/T_c$ — two numbers about the reservoirs, with nothing about the machine in them.

Why Do Real Power Plants Get 40%, Not 64%?

Take a coal-fired station: steam at 550 °C, waste heat into a river at 25 °C. In kelvin that is 823.15 K and 298.15 K, and the Carnot formula hands back $1 - 298.15/823.15 = 63.8\%$ . Real coal plants deliver 35–42%. That is a wide gap for something built by people who are very good at their jobs, and the usual explanation — friction and leaky pipework — is real but nowhere near large enough to account for it.

Most of the gap is not waste at all. It is speed, bought deliberately.

A reversible engine is infinitely slow. That is not a figure of speech: heat has to cross the boiler wall on an infinitesimal temperature difference, and driving a finite amount of heat through an infinitesimal gradient takes forever. A Carnot power station would reach 63.8% efficiency and generate exactly zero watts, which is not a business. Real plants therefore spend temperature to buy rate — they fire hard enough that the steam inside the boiler is meaningfully colder than the furnace and the condenser is meaningfully warmer than the river, and the engine works across the smaller gap that is left. Power is work per unit time, the same watts the work–energy theorem hands you, and here it is bought with efficiency.

Push that trade to its optimum — as much power as the plant can possibly deliver — and there is a second formula waiting, published by Curzon and Ahlborn in 1975 and by Novikov in 1957 before them:

$\eta_{CA} = 1 - \sqrt{\frac{T_c}{T_h}}$

For 823.15 K and 298.15 K that gives 39.8%. Real plants sit at 35–42%. The maximum-power point, not the Carnot point, is the number reality tracks.

Try it yourself

It loads at 15% throttle with the banner reading "Pushing harder: η falls to 59.9%, power climbs to 36% of max." Ease the "Throttle — firing rate" slider down to 0% and read the other extreme: "η = 63.8% — nearly Carnot. Power: 0% of max. Beautiful efficiency, dark city." The Carnot number is real. It is also worth nothing.
Now sweep the throttle up slowly and watch the two thermometer columns. The boiler's working temperature sags down from the 823 K furnace tick and the condenser's climbs up from the 298 K river tick, and the hatched gaps between tick and level are the temperature each exchanger threw away to move heat at that rate. The engine core only ever gets what is left between them.
Keep sweeping and watch the city above the core instead of the numbers. Its windows light in proportion to power. Try to stop on the brightest skyline by eye alone before you press anything — then press "Snap to max power" and see how close you got. The crest is at 62.4% throttle, and the banner there reads "Max power. η = 39.8% ≈ 1 − √(Tc/Th) = 39.8% — this is where real plants live."
Push past the crest toward 100%. Efficiency keeps falling, which you expect — but the power falls too, and the city dims a second time. Watch the waste band running into the river fatten the whole way. At full throttle the two internal temperatures have met in the middle, the core has no gap left to work across, and every joule from the furnace goes straight through into the water. The banner's phrase for this is "boiling the river for nothing".
Last one: drag the River Tc slider down to 275 K, a winter river. The whole hill rebuilds and the crest badge resets, because it is no longer the same curve. The Carnot flag at the right-hand edge climbs from 63.8% to 66.6%, and the ledger's η CA row from 39.8% to 42.2%. Press "Snap to max power" again and the new crest stands just clear of the top of the shaded "real coal plants" band. Nothing about the plant changed; 23 K of colder water did that — which is also why real stations de-rate in a heatwave.

Pushing harder: η falls to 59.9%, power climbs to 36% of max.

Boiler drop 39 K, condenser drop 16 K: of the 525 K between furnace and river, the core only ever sees 470 K. Q̇h is 15% of the most these exchangers can pass, and the Carnot ceiling for this pair — 63.8% — pays exactly zero watts.

This is the endoreversible model: every irreversibility in the plant has been pushed into the two heat exchangers, and the core between them is taken to be a perfect, reversible Carnot engine. Real stations also lose work to friction in the bearings, turbulence in the steam path, incomplete combustion and heat leaking out of the pipework — none of which appears here. So the fact that η_CA = 1 − √(Tc/Th) tracks real station efficiencies as closely as it does is partly a celebrated coincidence rather than a derivation of them (Curzon & Ahlborn, 1975; the same result had been published by Novikov in 1957). Both exchangers are also given the same constant conductance K, which is what puts the crest at η_CA exactly. K sets nothing but the absolute size of the power number, which is why power appears here only as a percentage of this plant’s own maximum and never in megawatts.

Furnace Th

Th823.15 K

River Tc

Tc298.15 K

Those two set the plant. Move either one and the whole hill below is reshaped, so the trail you have swept and the crest badge both reset.

Throttle — firing rate

throttle15 %

This is the one that walks the hill. Ease it toward 0 and η climbs toward the Carnot ceiling while the city goes dark. Floor it and the two internal temperatures meet in the middle, η falls to zero, and the city goes dark again — this time with the river taking every joule.

The slider moves in whole percent, and the crest — throttle 62.4% here — hardly ever falls on one of those notches. Press this to park on it. Sweep the throttle across it — or press this — and the chart will mark it.

Thw — boiler steam784 K

Tcw — condenser steam314 K

η = 1 − Tcw/Thw59.9%

η Carnot = 1 − Tc/Th63.8%

η CA = 1 − √(Tc/Th)39.8%

P — % of max power36%

Q̇h — % of max heat draw15%

Engine or power plant	Typical real-world thermal efficiency
Early Newcomen steam engine (1712)	~0.5%
Watt steam engine (late 1700s)	~2–4%
Gasoline car engine (Otto cycle)	~25–30%
Modern coal-fired steam power plant	~35–42%
Diesel engine	~35–45%
Combined-cycle gas power plant	~55–62%*
Max-power (Curzon–Ahlborn) point at those temperatures	~40%
Carnot limit at typical steam-plant temperatures (550 °C hot, 25 °C cold)	~64%

* Combined-cycle plants are not bounded by that 64% figure, because they are not working between those temperatures. A gas turbine's inlet runs near 1500 °C, which puts its own Carnot ceiling above 80%; the exhaust, still hot, then raises steam for a second cycle underneath. Hotter hot end, higher ceiling — the formula was never violated, only fed different numbers.

One honest caveat about $\eta_{CA}$ . The model behind it pushes every irreversibility in the plant into the two heat exchangers and treats the core between them as a perfect Carnot engine, so friction in the bearings and incomplete combustion never appear in the derivation at all. That it lands inside the real-plant band anyway is partly luck — those missing losses happen to be small enough, and to pull in the same direction, that the formula absorbs them without noticing. It is a very good rule of thumb with a slightly overqualified pedigree, and worth quoting for the physical point it makes rather than as a prediction anyone would commission a station on.

Can a Heat Engine Run in Reverse? Refrigerators and Heat Pumps

Run every leg of the Carnot cycle backwards and you get a refrigerator. Work goes in, heat comes out of the cold space and gets dumped somewhere warmer — which is what the compressor in your kitchen has been doing since you bought it, and what the Clausius statement of the second law says cannot happen unaided. Unaided is the operative word. Heat will not climb on its own; pay it in work and it will climb all day.

Efficiency is the wrong word for the reversed machine, because the thing you want is not work out but heat moved. The figure of merit is the coefficient of performance:

$\mathrm{COP}_{\text{fridge}} = \frac{Q_c}{W}$

and, being the Carnot cycle run backwards, it has its own reversible ceiling:

$\mathrm{COP}_{\text{fridge, max}} = \frac{T_c}{T_h - T_c}$

A domestic fridge holding its interior at 275 K in a 295 K kitchen — roughly 2 °C in a room at 22 °C — has a ceiling of $275/20 = 13.75$ : in principle, one joule of electricity moves nearly fourteen joules of heat out of the salad drawer. Real fridges manage a fraction of that, but the number is startling the first time you meet it, and it looks for a moment like a free lunch.

It isn't, and the reason is the same denominator. That 13.75 exists because 20 K is a very small gap to lift heat across. Try to lift it across 200 K instead and the COP falls by an order of magnitude. The energy is not being created — it is being moved, and the work you pay buys the lift, not the cargo. It is exactly the engine formula seen from the other side: a big temperature gap is what makes an engine good and a heat pump expensive, and a small one does the reverse.

Worked Examples for Physics Exams

Worked Example

Example 1 — Efficiency and energy flows for a 500 K / 300 K engine

A heat engine runs between reservoirs at 500 K and 300 K. Find its maximum possible efficiency, and the heat and work per cycle for 1 mol of gas with a volume ratio of 2 on each isothermal leg.

The efficiency needs nothing but the temperatures:

$\eta = 1 - \frac{T_c}{T_h} = 1 - \frac{300}{500} = 0.40$

For the energies, only the isothermal legs exchange heat, and on those $\Delta U = 0$ so $Q = W = nRT\ln(V_b/V_a)$ :

$Q_h = (1)(0.0821)(500)\ln 2 = 28.45\ \text{L·atm} = 2883\ \text{J} \approx 2.88\ \text{kJ}$

$Q_c = (1)(0.0821)(300)\ln 2 = 17.07\ \text{L·atm} = 1730\ \text{J} \approx 1.73\ \text{kJ}$

$W = Q_h - Q_c = 1153\ \text{J} \approx 1.15\ \text{kJ}$

Cross-check the efficiency the long way round: $W/Q_h = 1153/2883 = 0.40$ , matching $1 - T_c/T_h$ exactly.

Check it in the machine above: those are its default sliders, and the ledger reads Qh = 2.88 kJ, Qc = 1.73 kJ, W = 1.15 kJ, η = 40.0%. The lap pill reads the same 1.15 kJ, because the loop's area is that work.

Worked Example

Example 2 — The Celsius mistake, and why kelvin is not optional

The same engine's reservoirs are quoted as 227 °C and 27 °C. A student substitutes directly:

$\eta = 1 - \frac{27}{227} = 0.881$

That says 88.1%, and it is wrong. Convert first: 227 °C = 500 K and 27 °C = 300 K, so

$\eta = 1 - \frac{300}{500} = 0.40$

and the honest answer is 40%.

The Celsius version more than doubles the answer, and the reason is worth more than the arithmetic. The formula contains a ratio of temperatures, and a ratio is only meaningful when both numbers are measured from the same true zero. Celsius counts from the freezing point of water, an arbitrary landmark with no thermodynamic significance whatsoever. Kelvin counts from the point at which the entropy derivation's $Q_c/Q_h = T_c/T_h$ genuinely holds. If a formula has $T$ inside a ratio or under a square root — this one and the Curzon–Ahlborn result alike — it wants kelvin, and it will not tell you when you get it wrong.

Worked Example

Example 3 — A coal-fired plant: Carnot limit vs real output vs max power

A station raises steam at 550 °C and condenses against river water at 25 °C. Find the Carnot limit and the maximum-power efficiency, and compare both to what such plants actually deliver.

In kelvin, $T_h = 823.15$ K and $T_c = 298.15$ K.

$\eta_{Carnot} = 1 - \frac{298.15}{823.15} = 0.6378 \approx 63.8\%$

$\eta_{CA} = 1 - \sqrt{\frac{298.15}{823.15}} = 1 - 0.6018 = 0.3982 \approx 39.8\%$

Real coal plants deliver 35–42%. The Carnot figure overshoots by more than 20 percentage points; the maximum-power figure lands inside the band. The difference between them is not a list of losses — it is the price of producing power at a finite rate at all, since the Carnot number describes an engine whose output is exactly zero.

Check it in the throttle above: these are its default reservoirs. Press "Snap to max power" and the ledger shows η = 39.8% against an η Carnot of 63.8%, with the operating dot standing on the crest of the hill, inside the shaded band where real coal plants live.

Worked Example

Example 4 — The entropy ledger: proving 400 J is the ceiling

An engine takes in $Q_h = 1000$ J per cycle from a 500 K reservoir and rejects the rest to one at 300 K. Show that no more than 400 J of work can come out, using entropy alone.

Over one complete cycle the working substance returns to its starting state, so its own entropy change is zero and only the two reservoirs are left holding anything:

$\Delta S_{univ} = -\frac{Q_h}{T_h} + \frac{Q_c}{T_c}, \qquad Q_c = Q_h - W$

Test three settings:

$W$	$Q_c$	$\Delta S_{univ}$	Verdict
0 J	1000 J	$-2 + 3.333 = +1.333$ J/K	Allowed — and useless
400 J	600 J	$-2 + 2.000 = 0$ J/K	Reversible — the ceiling
500 J	500 J	$-2 + 1.667 = -0.333$ J/K	Forbidden

The reversible row is the boundary, and setting $\Delta S_{univ} = 0$ solves to the familiar result:

$W_{max} = Q_h\left(1 - \frac{T_c}{T_h}\right) = 1000(0.40) = 400\ \text{J}$

Check it in the auditor above: those are its defaults, and dragging W through 400 J walks the needle from green through the knife edge into red, with the badge appearing at exactly 400 J and the FORBIDDEN stamp landing at the next 5 J notch above it.

Frequently Asked Questions

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Carnot Engine Efficiency Calculator: The Laws of Thermodynamics Explained

What Are the Laws of Thermodynamics?

What Is the First Law of Thermodynamics? (Conservation of Energy)

What Is the Second Law of Thermodynamics? (Why Every Engine Wastes Heat)

What Is a Heat Engine and How Does It Turn Heat Into Work?

How Does the Carnot Cycle Work? The Four Strokes Explained

What Is the Carnot Efficiency Formula? (η = 1 − Tc/Th)

How Do You Derive the Carnot Efficiency Formula?

Why Can't Any Heat Engine Be 100% Efficient?

Why Do Real Power Plants Get 40%, Not 64%?

Can a Heat Engine Run in Reverse? Refrigerators and Heat Pumps

Worked Examples for Physics Exams

Example 1 — Efficiency and energy flows for a 500 K / 300 K engine

Example 2 — The Celsius mistake, and why kelvin is not optional

Example 3 — A coal-fired plant: Carnot limit vs real output vs max power

Example 4 — The entropy ledger: proving 400 J is the ceiling

Frequently Asked Questions

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