Heat engine efficiency, and how close it gets to the Carnot limit

Q: Why can't a heat engine be 100 per cent efficient?

Because 1 − Tc/Th reaches 1 only if the sink is at absolute zero or the source is at infinite temperature, and neither exists. In the entropy language it is sharper: the engine removes Qh/Th of entropy from the source and must put at least that much somewhere, and the only place available is the cold reservoir, which means rejecting at least QhTc/Th of heat. That rejected heat is the price of running a cycle at all, not a design flaw. NCERT's module states the engine form of the second law exactly this way: no cyclic heat engine can have 100 per cent efficiency because some heat must always go to the cold reservoir.

Q: What does a negative entropy generation mean?

That the inputs describe something impossible. ΔSgen = Qc/Tc − Qh/Th is zero for a reversible engine and positive for every real one; a negative value says the total entropy of the two reservoirs would fall, which the second law forbids. In practice it means one of four things: the rejected heat has been understated, the source temperature overstated, the sink temperature understated, or Celsius numbers have been used for one or both. It is always worth checking in that order, because the first is the commonest — real engines throw heat away through the exhaust, the block and the friction, not only through the condenser.

An efficiency on its own says nothing: 30 per cent is excellent between 100 °C and 20 °C and dreadful between 1500 and 20. This page gives the actual efficiency, the Carnot ceiling for your two temperatures, the fraction of that ceiling you reached, and the entropy generated per cycle — which goes negative if your engine is impossible.

Last updated 11 Nov 2026

Efficiency against the Carnot ceiling

The fraction of the Carnot limit actually achieved

Heat taken in from the hot reservoir, per cycleIn joules per cycle, or any consistent energy unit — only the ratio to the rejected heat matters to the efficiencies, though the entropy and work rows assume joules. This is the whole heat drawn from the source, not the useful work: NCERT’s own worked example has an engine absorbing 200 J and rejecting 150, which is 50 J of work and 25 per cent.

Heat rejected to the cold reservoir, per cycleIn the same unit. Every heat engine must reject some: the second law in its engine form says no cyclic engine can convert heat entirely into work, and the row below shows the minimum this one is allowed to throw away. Set it to zero to see what a 100 per cent engine would look like — the page will tell you flatly that it is impossible, and the entropy row will tell you why.

The two temperatures below are inThe Carnot limit uses ABSOLUTE temperatures in a ratio, so Celsius values are converted by adding 273.15. Getting this wrong is the standard way a Carnot question goes astray: 1 − 20/500 with Celsius numbers gives 96 per cent where the truth is 1 − 293.15/773.15, which is 62. The Celsius version is always the more flattering one, which is part of why it survives.

Temperature of the hot reservoir (the source)The temperature at which heat is TAKEN IN. For a steam plant it is the boiler or the superheated steam, not the flame: the flame may be at 1800 °C but the working fluid never sees that, and it is the working fluid’s temperature that sets the limit. This is why the practical ceiling on a power station is a materials problem long before it is a thermodynamics problem.

Temperature of the cold reservoir (the sink)The temperature at which heat is REJECTED — the condenser, the cooling tower, the river, the radiator, the outside air. NCERT’s own module makes the point that lowering the sink raises the efficiency more effectively than raising the source by the same number of degrees, and its table shows it: from a 500 K source, sinks of 350, 300, 250, 200 and 150 K give 0.3, 0.4, 0.5, 0.6 and 0.7, while from a 250 K sink, sources of 550, 600, 650, 700 and 750 K give only 0.54, 0.58, 0.61, 0.64 and 0.66. The catch is that nobody can choose the sink: it is the environment.

49.35%Example

An engine taking in 1000 J and rejecting 700 J per cycle between a 500 °C source and a 30 °C sink

Two efficiencies and one entropy, and the third tells you if the first two lie

η = W/Q_h = 1 − Q_c/Q_h · η_Carnot = 1 − T_c/T_h · second-law efficiency = η / η_Carnot · ΔS_gen = Q_c/T_c − Q_h/T_h ≥ 0

η: the actual efficiency: work out divided by heat in, which is a statement about the machine. On its own it is close to useless as a judgement, because it depends overwhelmingly on the two temperatures the machine was given
η_Carnot: the ceiling, 1 − T_c/T_h, which depends on NOTHING but the two absolute temperatures — not the working fluid, not the design, not the size. Carnot’s theorem: no engine between two given reservoirs beats a reversible one between the same two
the ratio: the second-law efficiency, which is the figure this page leads with because it is the one that can be compared across machines. 30 per cent absolute is 49 per cent of the limit between 500 °C and 30, and only about a third of the limit if the reservoirs were 1500 °C and −50
ΔS_gen: the entropy generated per cycle, Q_c/T_c − Q_h/T_h. Exactly zero for a reversible engine, positive for every real one, and never negative. It is the sharpest test available, because an engine that beats Carnot shows up here as a negative number rather than as a merely surprising percentage
absolute T: kelvin, always, because the limit is a RATIO of temperatures. 1 − 20/500 on Celsius numbers gives 96 per cent against the true 62, and the error always flatters — which is part of why it survives unchallenged in a student’s working

Worked example

An engine taking in 1000 J and rejecting 700 J per cycle between a 500 °C source and a 30 °C sink
THE ACTUAL EFFICIENCY, which is the easy part. W = Q_h − Q_c = 1000 − 700 = 300 J, so η = 300/1000 = 30.0 per cent. Equivalently 1 − 700/1000. This number is about the machine and says nothing about whether 30 per cent was a good result
THE TEMPERATURES, IN KELVIN. 500 °C = 773.15 K and 30 °C = 303.15 K. Do not be tempted by 1 − 30/500 = 94 per cent: that is a ratio of two Celsius numbers, which measure from an arbitrary zero, and it flatters the engine by more than thirty percentage points
THE CARNOT CEILING. η_C = 1 − 303.15/773.15 = 1 − 0.392 10 = 60.79 per cent. Nothing about the engine enters this: any engine at all, with any working fluid, between these two reservoirs is held under 60.79 per cent
THE FRACTION OF IT ACHIEVED. 30.00 / 60.79 = 49.35 per cent, so this engine takes a little under half the work that was thermodynamically available. The most work 1000 J could ever have given here is 1000 × 0.6079 = 607.9 J, and the engine delivered 300, so 307.9 J of available work was lost to irreversibility
THE ENTROPY CHECK, which is the row that cannot be argued with. Entropy out of the source: 1000/773.15 = 1.293 41 J K⁻¹. Entropy into the sink: 700/303.15 = 2.309 09. Generated: +1.016 J K⁻¹ per cycle, comfortably positive, so the engine is allowed to exist. A Carnot engine here would reject only 1000 × 303.15/773.15 = 392.1 J and generate exactly zero; this one throws away 307.9 J more heat than it had to. Enter 350 J of rejected heat instead of 700 and watch the entropy go negative and the verdict change to impossible

Why a bare efficiency means nothing: 30 per cent in four different engines

The actual efficiency is 30 per cent in every row and the verdict ranges from physically impossible to badly wasteful. That is the case for leading with the fraction of the limit rather than the efficiency: 30 per cent between 100 and 20 °C would be a violation of the second law, while 30 per cent between 1500 °C and −50 means more than half the available work was lost. Note also how the Carnot limit itself moves: the same engine gains 66 percentage points of ceiling from the first row to the last, entirely from the temperatures it was given. NCERT’s module makes the related point that lowering the sink buys more than raising the source does, with a table from a 500 K source showing efficiencies of 0.3 to 0.7 as the sink falls from 350 K to 150, against 0.54 to 0.66 as a source rises from 550 K to 750 above a 250 K sink. The practical trouble is that the sink is the environment and nobody gets to choose it.
Source	Sink	Carnot limit	Actual η	Fraction of the limit	Verdict
100 °C (373.15 K)	20 °C (293.15 K)	21.44%	30%	139.9%	impossible — beats Carnot
200 °C (473.15 K)	20 °C (293.15 K)	38.04%	30%	78.9%	near the ceiling; very good
500 °C (773.15 K)	30 °C (303.15 K)	60.79%	30%	49.3%	a realistic machine
1500 °C (1773.15 K)	−50 °C (223.15 K)	87.42%	30%	34.3%	poor — most of the available work is being thrown away

An efficiency is a judgement about temperatures, not about a machine

Thirty per cent is a meaningless number until you know what temperatures it was earned between. Between a 100 °C source and a 20 °C sink, 30 per cent is impossible: the Carnot ceiling there is 21.4 per cent and nothing may exceed it. Between 1500 °C and −50, the ceiling is 87.4 per cent and 30 per cent means two thirds of the available work was thrown away. Same engine figure, opposite verdicts. That is why this page leads with the FRACTION of the Carnot limit achieved — the second-law or exergy efficiency — and keeps the raw efficiency as a secondary row.

The Carnot limit depends on nothing but the two absolute temperatures. Not the working fluid, not the design, not the size or the century it was built in: η_Carnot = 1 − T_c/T_h, and Carnot’s theorem says no engine between two given reservoirs can beat a reversible engine between the same two. That is a strikingly strong claim for something so simple, and it is the reason the limit is worth computing before judging any machine. A Carnot engine itself is not a design: NCERT’s own module calls it a concept that cannot be built, needing no friction at all, quasi-static processes throughout, and heat exchange only between bodies at almost the same temperature — which, since heat flows slowly across a small temperature difference, means it would also take forever.

Where this goes wrong is almost always the temperature. Using Celsius in 1 − T_c/T_h gives 1 − 30/500 = 94 per cent where the truth is 1 − 303.15/773.15 = 60.79, and the error always flatters the engine, which is part of why it survives a student’s own sanity check. The second commonest is the source temperature itself: for a steam plant the limit is set by the steam, not by the flame. A furnace at 1800 °C feeding steam at 540 gives the engine a ceiling computed from 540, and the heat that crossed that enormous temperature drop on the way in destroyed available work before the engine ever saw it. The practical ceiling on a power station is a materials problem long before it is a thermodynamics one.

The entropy row is the one that cannot be argued with. ΔS per cycle is Q_c/T_c − Q_h/T_h: exactly zero for a reversible engine, positive for every real one, never negative. An engine that appears to beat the Carnot limit shows up here as a negative number, which is the second law in its sharpest form — the total entropy of system and surroundings never decreases. Try it: take the page’s default engine and reduce the rejected heat from 700 J to 350. The efficiency rises to 65 per cent, the headline goes past 100, and the entropy generated turns negative. The three statements are the same statement.

And the useful asymmetry, which NCERT’s module puts well. Lowering the sink raises the efficiency more effectively than raising the source by the same number of degrees. Its table shows a 500 K source with sinks at 350, 300, 250, 200 and 150 K giving efficiencies of 0.3, 0.4, 0.5, 0.6 and 0.7, while a 250 K sink with sources at 550, 600, 650, 700 and 750 K gives only 0.54, 0.58, 0.61, 0.64 and 0.66. The reason is in the formula: T_c is in the numerator of the fraction being subtracted and T_h in its denominator, so a given change does more at the cold end. The cruelty is that the sink is the environment — a river, a cooling tower, the outside air — and nobody gets to choose it, which is exactly why real engineering pushes the source temperature up against what the metal will take.

Frequently asked questions

Why can’t a heat engine be 100 per cent efficient?

Because 1 − T_c/T_h reaches 1 only if the sink is at absolute zero or the source is at infinite temperature, and neither exists. In the entropy language it is sharper: the engine removes Q_h/T_h of entropy from the source and must put at least that much somewhere, and the only place available is the cold reservoir, which means rejecting at least Q_hT_c/T_h of heat. That rejected heat is the price of running a cycle at all, not a design flaw. NCERT’s module states the engine form of the second law exactly this way: no cyclic heat engine can have 100 per cent efficiency because some heat must always go to the cold reservoir.

What is second-law or exergy efficiency, and why is it the headline here?

It is the actual efficiency divided by the Carnot efficiency for the same two reservoirs — the fraction of the thermodynamically available work that was actually taken. It is the headline because it is the only one of the two that can be compared between machines: 30 per cent absolute is impossible between 100 and 20 °C, excellent between 200 and 20, ordinary between 500 and 30, and poor between 1500 and −50. A bare efficiency mostly reports which temperatures the engine was handed, which is usually not the engineer’s achievement.

Do I have to use kelvin in the Carnot formula?

Yes, without exception, because the formula contains a ratio of temperatures and the Celsius zero is an arbitrary offset. 1 − 30/500 gives 94 per cent; 1 − 303.15/773.15 gives 60.79. The gap is thirty-three percentage points and it always points the same way, flattering the engine, which is why it is a mistake that passes a student’s own plausibility check. The quick test for any formula is whether adding 273.15 to every temperature leaves the answer unchanged; here it does not, so the scale must be absolute.

What does a negative entropy generation mean?

That the inputs describe something impossible. ΔS_gen = Q_c/T_c − Q_h/T_h is zero for a reversible engine and positive for every real one; a negative value says the total entropy of the two reservoirs would fall, which the second law forbids. In practice it means one of four things: the rejected heat has been understated, the source temperature overstated, the sink temperature understated, or Celsius numbers have been used for one or both. It is always worth checking in that order, because the first is the commonest — real engines throw heat away through the exhaust, the block and the friction, not only through the condenser.

Can I use this for a refrigerator or a heat pump?

Partly. The efficiency rows assume a device being driven BY heat to produce work, which is the wrong way round for a fridge, so the headline is not a useful figure of merit there. What is useful are the two COP rows: the refrigerator COP is Q_c/W and the heat pump COP is Q_h/W, and both are routinely greater than 1 because the device moves heat rather than making it. The Carnot COP row gives the corresponding ceiling, T_c/(T_h − T_c), which is why a heat pump working across a small temperature difference is so much better than one working across a large one.

Related calculators

References

NCERT / CIET, Physics-02 (Keph_201204): Thermodynamics, Unit 8 Module 4, “Second Law of Thermodynamics” (chapter 12, class XI), 2019, on the CIET e-content server (read 11 November 2026). The source of this page’s framing. It gives W = Q_H − Q_C and η = W/Q_H = 1 − Q_C/Q_H, with a worked example of 200 J absorbed and 150 rejected giving 25 per cent, and notes that the best steam engines reach only about 40 per cent. It states the second law three ways: that the total entropy of system and surroundings never decreases; Kelvin’s form, that “a process whose only net result is to transfer energy as heat from a cooler object to a hotter one is impossible”; and the engine form, that no cyclic engine can be 100 per cent efficient. It gives the Carnot efficiency as (T₁ − T₂)/T₁ = 1 − T₂/T₁, states Carnot’s theorem, calls the Carnot engine a concept that cannot be built, and is the source of the asymmetry table quoted on this page: from a 500 K source, sinks of 350, 300, 250, 200 and 150 K give 0.3, 0.4, 0.5, 0.6 and 0.7, while from a 250 K sink, sources of 550, 600, 650, 700 and 750 K give 0.54, 0.58, 0.61, 0.64 and 0.66.
The International System of Units (SI), 9th edition, BIPM, published May 2019; the version read for this batch is 4.01, revised June 2026 (read 11 November 2026). The source of the kelvin’s CURRENT definition: the kelvin is defined by fixing the Boltzmann constant at exactly k = 1.380 649 × 10⁻²³ kg m² s⁻² K⁻¹, which the brochure states as an exact relation. The degree Celsius is defined equal in magnitude to the kelvin, with T₀ = 273.15 K and t/°C = T/K − 273.15. Most school textbooks still define the kelvin as 1/273.16 of the triple point of water. The brochure records that as the PREVIOUS definition, and says the triple-point temperature must now be determined experimentally — it stood at 273.16 K with a relative standard uncertainty of 3.7 × 10⁻⁷ at the redefinition, and is a measured quantity from then on.
CODATA Recommended Values of the Fundamental Physical Constants: 2022, NIST SP 961 (the constants wall chart), May 2024 (read 11 November 2026). k = 1.380 649 × 10⁻²³ J K⁻¹ and N_A = 6.022 140 76 × 10²³ mol⁻¹, both marked exact, both fixed by definition and carrying no uncertainty. The molar gas constant is printed as R = 8.314 462 618… with an ellipsis and no “exact” tag, which is a typographical truncation and not an uncertainty: R = kN_A is the product of two exact decimals, so its digits terminate, and multiplying them out gives exactly R = 8.314 462 618 153 24 J mol⁻¹ K⁻¹. This batch did that multiplication rather than take the ellipsis on trust.
COULD NOT BE READ, and nothing was substituted for any of it. The NCERT textbook pages on ncert.nic.in, and four CIET modules in the same series as the ones cited here — Keph_201101 (temperature and its measurement), Keph_201102 (thermal expansion), Keph_201103 (specific heat capacity and calorimetry) and Keph_201105 (conduction and convection) — were all refused by the server’s robots policy on 11 November 2026. That is why this batch quotes no NCERT table number, and why the coefficients of expansion and the thermal conductivities are attributed to the reference actually read rather than to the textbook table you are more likely to be holding. If your table disagrees, use your table: these pages take the constant from you for exactly that reason.