Physics · Thermodynamics

Carnot efficiency

Q: How do you calculate the Carnot efficiency?

For a reversible cyclic heat engine between two constant-temperature reservoirs use $\eta_C=1-\frac{T_C}{T_H}$. Both temperatures must be absolute kelvin values satisfying $0<T_C<T_H$. With 625 K and 350 K, $\eta_C=0.44$, or 44%. Receiving 2400 J per cycle allows at most 1056 J net work and requires 1344 J rejected heat. Check energy and reservoir entropy: $-\frac{2400}{625}+\frac{1344}{350}=0\,\mathrm{J\,K^{-1}}$. Actual irreversible engines fall below this bound at identical reservoirs. Temperature difference alone does not determine efficiency; the ratio of absolute temperatures matters. The result is a bound, not a power rating.

Q: Why must you insert the temperature in kelvin?

The derivation compares reversible entropy transfers $\frac{Q_H}{T_H}$ and $\frac{Q_C}{T_C}$, using absolute thermodynamic temperatures. Celsius has a shifted zero, so ratios of Celsius readings produce different and incorrect results here. Convert with $T[\mathrm K]=\vartheta[{}^\circ\mathrm C]+273.15$ before forming the ratio. Temperatures of 600 °C and 30 °C become 873.15 K and 303.15 K, giving $\eta_C\simeq65.281\,\%$. Only temperature differences have the same numerical value in kelvin and Celsius. Equal differences do not guarantee equal efficiencies: $\frac{600}{300} K$ gives 50%, whereas $\frac{900}{600} K$ gives 33.3%. Always record both reservoir temperatures and their temperature scale.

Q: Why can the efficiency never reach 100 percent?

This bound concerns cyclic heat engines between finite positive reservoir temperatures, not every device. With $T_C>0$ and finite TH, $\frac{T_C}{T_H}$ stays positive, so $\eta_C<1$. The second law requires enough heat rejection to the cold reservoir to compensate entropy lost by the hot reservoir. Complete cyclic conversion of heat from a single reservoir into work is forbidden. Friction is therefore not the sole reason for the limit: even a reversible engine rejects heat. Electrical work in a motor and heat delivery by a heat pump need different measures. Applying the statement to every efficiency would be too broad.

Q: How do you rearrange the Carnot efficiency for the cold temperature?

Move the ratio to one side: $\frac{T_C}{T_H}=1-\eta_C$. Multiply by TH to obtain $T_C=(1-\eta_C)T_H$. With $T_H=600\,\mathrm K$ and $\eta_C=0.50$, $T_C=300\,\mathrm K$. Convert a percentage to a fraction first: 50% means 0.50. Substitution into $1-\frac{300}{600}$ returns 0.50. This equality describes a reversible limit. A desired actual efficiency requires a Carnot bound at least as high, but that condition is not sufficient: heat transfer, friction and engineering constraints introduce further departures. The resulting temperature remains an absolute value in kelvin, not a Celsius reading.

Q: What is the difference between Carnot efficiency and real efficiency?

ηC is the reversible upper bound for specified hot and cold reservoirs. Actual thermal efficiency instead equals net work divided by absorbed heat using a consistent system boundary. With $T_H=750\,\mathrm K$, $T_C=300\,\mathrm K$ and $Q_H=6000\,\mathrm J$, maximum reversible work is 3600 J. An engine delivering 1800 J has $\eta=30\,\%$ and $Q_C=4200\,\mathrm J$. Reservoir entropy $-\frac{6000}{750}+\frac{4200}{300}$ gives $\frac{6 J}{K}$ generation. General power-plant percentages do not follow from the equation; they require operating data. Multiple heat sources or finite cooling bodies need broader accounting, and auxiliary electrical consumption changes the chosen plant boundary.

Carnot efficiency bounds cyclic heat-to-work conversion between two constant-temperature reservoirs. The reversible limit has $\eta_C=1-\frac{T_C}{T_H}$ .

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Formula

Calculate

η_{C a r n o t} = 1 - \frac{T _{ka lt}}{T _{w a r m}}

η dimensionless (0 to 1) · T in kelvin [K]

Variables & units

Variables & units
Symbol	Meaning	Unit
$η$	Efficiency	dimensionless
$T_{\mathrm{warm}}$	Temperature of the hot reservoir	K
$T_{\mathrm{kalt}}$	Temperature of the cold reservoir	K

LaTeX

\eta_{Carnot} = 1 - \frac{T_{kalt}}{T_{warm}}

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Carnot bound for a 300-kelvin cold reservoir: zero at hot-reservoir temperature 300 kelvin, 0.52 at 625 kelvin and 0.8 at 1500 kelvin; the curve remains below 1. — Reversible bound at fixed cold-reservoir temperature TC = 300 K. ηC = 1 − TC/TH rises with TH in kelvin. At TH = 625 K, ηC = 0.52; a different cold reservoir gives a different curve.

Carnot efficiency calculator

Find the highest possible efficiency of a heat engine working between two temperatures, or one of the temperatures from the efficiency.

Find

Hot-side temperature Th

Temperature of the hot side, e.g. 800 K (527 °C) in a steam boiler; °C is converted to kelvin for you.

Cold-side temperature Tc

Temperature of the cold side, e.g. 300 K (27 °C) at the cooling water; no warmer than the hot side.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for η, substitute, evaluate

1. Rearrange for η
$\eta=1-\frac{T_c}{T_h}$
2. Substitute values (in SI units)
$\eta=1-\frac{300}{800}=0.625$
3. Result
η = 62.5 %

Result: Efficiency η

η = 62.5 %

With Th = 800 K and Tc = 300 K, η = 62.5 %.

Quick tips

Always use kelvin; °C gives nonsense: 527 °C and 27 °C are about 800 K and 300 K, so η = 1 − 300/800 = 62.5%.
Equivalent form: η = (Th − Tc) / Th. The efficiency rises when the hot side gets hotter or the cold side gets colder.

Assumptions and limits

Reversible heat engine between reservoirs with 0 < Tcold ≤ Thot. Temperature ratios use kelvin; °C inputs are converted. The value is an ideal upper bound, not a promise for a real engine.

Derivation & background

After a cycle the working medium has $\Delta U=0$ , so $W_{\mathrm{net}}=Q_H-Q_C$ , with both heat magnitudes positive. Reversible entropy balance $-\frac{Q_H}{T_H}+\frac{Q_C}{T_C}=0$ gives $\eta_C=1-\frac{T_C}{T_H}$ . Irreversibility generates entropy and reduces work yield. Both temperatures are absolute kelvin values. This reservoir bound does not require an ideal gas. Changing heat sources and additional energy inputs need their own balances.

Exam blueprint

Validity range

Reversible upper bound of a cyclic heat engine between exactly two fixed reservoir temperatures in kelvin, $0<T_C<T_H$ . No additional energy or material sources. Thermal efficiency is $\eta=\frac{W_{\mathrm{net}}}{Q_H}$ ; irreversible engines satisfy $\eta\le\eta_C$ . Heat pumps and refrigerators use another metric.

Derivation steps

Reversible entropy accounting of two reservoirs bounds work from a full heat-engine cycle.

1Consider a cyclic engine between fixed reservoir temperatures $0<T_C<T_H$ in kelvin.
2 $Q_H>0$ is absorbed heat and $Q_C>0$ rejected magnitude. After the cycle the working medium has $\Delta U=0$ .
3Energy conservation gives $W_{\mathrm{net}}=Q_H-Q_C$ , hence $\eta=\frac{W_{\mathrm{net}}}{Q_H}=1-\frac{Q_C}{Q_H}$ .
4In the reversible limit, total entropy change is $-\frac{Q_H}{T_H}+\frac{Q_C}{T_C}=0$ .
5Thus $\frac{Q_C}{Q_H}=\frac{T_C}{T_H}$ and $\eta_C=1-\frac{T_C}{T_H}$ .
6With irreversibility the reservoir sum is nonnegative, giving $\eta\le\eta_C$ .
7Additional sources, material streams or changing temperatures require their own energy and entropy balances.

Rearrangements

Cold temperature from efficiency

T_C=(1-\eta_C)T_H

This equality uses reversible limiting efficiency between two fixed reservoirs. Both temperatures are kelvin values; actual irreversible efficiency cannot simply replace it.

Task variant

A cyclic engine works only between $T_H=500\,\mathrm K$ and $T_C=300\,\mathrm K$ . Find its reversible efficiency bound.

Approach: $\eta_C=1-\frac{T_C}{T_H}$ .
Substitute: $\eta_C=1-\frac{300}{500}=0.40$ , or 40%.
This is the reversible bound between these two fixed reservoirs.

Common mistakes

Substituting temperatures in °C.

Always convert to kelvin first.

Exam context

Typical exams compare maximum and real efficiency.

Formula cluster

Thermodynamic limits

Connects heat engines, heat capacity and the gas law.

Q = m·c·ΔT

Heat Capacity: Q = m·c·ΔT

pV = nRT

Ideal Gas Law

ΔG = ΔH − TΔS

Gibbs Energy (Free Enthalpy)

Worked example

Ideal reservoirs at 600 °C and 30 °C: $T_H=873.15\,\mathrm K$ , $T_C=303.15\,\mathrm K$ . Approach: $\eta_C=1-\frac{T_C}{T_H}$ . Substitution gives $\eta_C=1-\frac{303.15}{873.15}\simeq0.652809$ , or 65.281%. This is a theoretical bound, not a power plant measurement.

Applications

Check thermodynamic limits of cyclic heat engines at specified reservoir temperatures. Refrigerators and heat pumps use COP rather than this heat-to-work efficiency.

How do you calculate with Carnot Efficiency?

Here is how to work through a typical Carnot Efficiency (η = 1 − T_k/T_w) task step by step:

1
Task
A cyclic engine works only between $T_H=500\,\mathrm K$ and $T_C=300\,\mathrm K$ . Find its reversible efficiency bound.
Solution path
1. Approach: $\eta_C=1-\frac{T_C}{T_H}$ .
2. Substitute: $\eta_C=1-\frac{300}{500}=0.40$ , or 40%.
3. This is the reversible bound between these two fixed reservoirs.

Understand and apply the equation

A heat engine receives heat and converts only part of it into work. Even an ideal engine between two fixed-temperature reservoirs must reject heat to the cold side. Carnot efficiency gives the greatest possible work yield for that reservoir pair. Start with energy accounting, then examine the additional entropy constraint and use both to assess an actual engine.

What efficiency compares

Thermal efficiency compares net work output with heat input. Consider a complete cycle. The working medium returns to its initial state, so its stored energy has no net change. It absorbs positive heat magnitude $Q_H$ from the hot reservoir at $T_H$ and rejects positive magnitude $Q_C$ to the cold reservoir at $T_C$ . The remainder is positive net work $W_{\mathrm{net}}$ . Here $Q_C$ denotes a rejected magnitude; the working medium’s signed heat input on that stage is $-Q_C$ .[1]

\begin{aligned}W_{\mathrm{net}}&=Q_H-Q_C\\\eta&=\frac{W_{\mathrm{net}}}{Q_H}=1-\frac{Q_C}{Q_H}\end{aligned}

The denominator is supplied heat, not net heat or fuel heating value without additional accounting.

A reservoir maintains its temperature despite giving or receiving heat. This idealisation is part of the formula. A small tank cooling during operation is not a reservoir at its initial temperature. Its process must be divided into temperature stages or integrated. The third example makes precisely this distinction visible.

Derive the temperature limit from entropy

Energy conservation alone would allow all supplied heat to become work. The second law adds the missing constraint: in the reversible limit, entropy of the engine and reservoirs remains unchanged. During reversible isothermal heat transfer, the hot reservoir loses $\frac{Q_H}{T_H}$ , the cold one gains $\frac{Q_C}{T_C}$ , and the working medium returns to its initial entropy after the cycle. The reservoir contributions must therefore cancel. This argument does not require an ideal-gas working substance.[2]

\begin{aligned}-\frac{Q_H}{T_H}+\frac{Q_C}{T_C}&=0\\\frac{Q_C}{Q_H}&=\frac{T_C}{T_H}\end{aligned}

Temperatures refer to the constant hot and cold reservoirs.

\eta_C=1-\frac{T_C}{T_H},\qquad0<T_C<T_H

Finite positive reservoir temperatures give 0 < ηC < 1.[1]

Irreversibility instead generates entropy. The inequality $\frac{Q_C}{T_C}-\frac{Q_H}{T_H}\ge0$ requires at least as much heat rejection as in the reversible case. For equal heat input, less work remains: $\eta\le\eta_C$ . Equality is the reversible limit. This comparison is meaningful only if the engine exchanges heat exclusively with these two reservoirs and no additional material, heat or energy sources are hidden.[5]

How the Carnot cycle describes the bound

The Carnot process illustrates this limit as an idealised cycle: two reversible isotherms are joined by two reversible adiabats. At high temperature the medium absorbs heat; at low temperature it rejects heat. No heat flows on the adiabatic links, and their reversibility means entropy remains constant there. A usual gas representation follows hot expansion, adiabatic expansion, cold compression and adiabatic compression.[1]

Absence of friction alone does not make an engine reversible. Heat transfer across a finite temperature difference also generates entropy. Reversible heat exchange requires that difference to vanish in the limit. The formula therefore describes a yield limit, not a promised power output in watts or a specific device size. Engineering design additionally has to account for finite heat exchangers and transfer rates.[3]

Why does Carnot efficiency require kelvin?

\begin{aligned}T[\mathrm K]&=\vartheta[{}^\circ\mathrm C]+273.15\\T_C&=(1-\eta_C)T_H\\T_H&=\frac{T_C}{1-\eta_C}\end{aligned}

Rearrangements describe the reversible limit; a desired real efficiency requires at least an equally high bound.

The Carnot formula requires absolute temperatures in kelvin because the Celsius offset does not cancel in a ratio. It does cancel in a temperature difference. At fixed $T_C$ , higher $T_H$ raises the bound. At fixed $T_H$ , higher $T_C$ lowers it. The same difference of 300 K gives a 50% bound for the hot/cold pair 600 K and 300 K, but only 33.3% for 900 K and 600 K. The decisive quantity is $\frac{T_C}{T_H}$ .

Reservoirs at 600 °C and 30 °C correspond to 873.15 K and 303.15 K. Their reversible bound is about 65.281%. This does not yet determine an actual power plant’s efficiency. Heat input, net work output and the chosen accounting boundary must also be known.

View diagram: compare two reservoir pairs

A warmer cold reservoir lowers the bound

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Original limiting curves for two fixed reservoirs in each case. At hot temperature 625 K the bounds are 52% and 44%. These are not measurements of actual engines.[1]

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Example: heat and work in a reversible cycle

Check every energy transfer against entropy

Given: $T_H=625\,\mathrm K$ and $T_C=350\,\mathrm K$ ; each reversible cycle absorbs $Q_H=2400\,\mathrm J$ .
Find: Find ηC, QC and Wnet.

Approach: Use absolute reservoir temperatures for the reversible comparison. $\eta_C=1-\frac{T_C}{T_H},\quad\frac{Q_H}{T_H}=\frac{|Q_C|}{T_C}$ . Form the reservoir temperature ratio.
$\eta_C=1-\frac{350}{625}=0.440$
Obtain rejected heat from the reversible heat ratio.
$Q_C=\frac{2400\cdot350}{625}=1344\,\mathrm J$
The difference gives net work.
$W_{\mathrm{net}}=2400-1344=1056\,\mathrm J$

Result: 44.0% efficiency, 1056 J work and 1344 J heat rejected per cycle.

Check: $\frac{1056}{2400}=0.44$ reconstructs efficiency. Reservoir entropy changes $-\frac{2400}{625}=-3.84\,\mathrm{J\,K^{-1}}$ and $\frac{1344}{350}=3.84\,\mathrm{J\,K^{-1}}$ cancel. Delivered work adds no heat term $\frac WT$ to this entropy balance.

Is Carnot efficiency the actual engine efficiency?

No. Carnot efficiency is the reversible upper bound for exactly the same reservoir pair. Actual thermal efficiency follows from net work output and heat input; the example below checks both quantities and entropy generation.[1][3]

Compare the bound at identical reservoirs

Given: A cyclic engine between $T_H=750\,\mathrm K$ and $T_C=300\,\mathrm K$ absorbs $Q_H=6000\,\mathrm J$ and delivers $W_{\mathrm{net}}=1800\,\mathrm J$ . Only these two heat reservoirs and no other energy sources.
Find: Find actual efficiency, rejected heat, entropy generation and reversible maximum work.

Approach: Compare the actual energy balance with the Carnot bound. $\eta=\frac W{Q_H},\quad |Q_C|=Q_H-W$ . Balance energy and form actual efficiency.
$\begin{aligned}Q_C&=6000-1800=4200\,\mathrm J\\{} \eta&=\frac{1800}{6000}=0.300\end{aligned}$
Evaluate Carnot efficiency at the same temperatures.
$\begin{aligned}\eta_C&=1-\frac{300}{750}=0.600\\{} W_{\mathrm{rev,max}}&=3600\,\mathrm J\end{aligned}$
Sum both reservoir entropy changes.
$S_{\mathrm{gen,total}}=-\frac{6000}{750}+\frac{4200}{300}=6.00\,\mathrm{J/K}$

Result: The engine reaches 30% against a 60% bound. Total entropy generation is 6 J/K per cycle.

Check: $W_{\mathrm{rev,max}}-W_{\mathrm{net}}=1800\,\mathrm J=T_C S_{\mathrm{gen,total}}$ checks the work difference. This applies here at equal $Q_H$ with two fixed reservoirs; it is not a universal loss formula for arbitrary machines. Also, $Q_C+W_{\mathrm{net}}=6000\,\mathrm J$ confirms energy balance.

Explore further: refrigerator and heat pump

\begin{aligned}\mathrm{COP}_{R,\mathrm{rev}}&=\frac{T_C}{T_H-T_C}\\\mathrm{COP}_{HP,\mathrm{rev}}&=\frac{T_H}{T_H-T_C}\end{aligned}

COP compares moved heat with supplied work and is not heat-engine η.[4]

In reverse operation, work is supplied to move heat from the cold reservoir to the hot one. The useful output is transported heat rather than delivered work. The appropriate metric is therefore $\mathrm{COP}$ , which can exceed one because heating output also includes heat extracted from a reservoir. At $T_H=320\,\mathrm K$ and $T_C=280\,\mathrm K$ , reversible COP values are seven for cooling and eight for heating. This violates neither energy conservation nor the Carnot limit for engine operation. These reversible COP values also assume fixed reservoir temperatures; the separate cooling-source example accounts for the source temperature changing.

Explore further: finite heat source

A finite hot body is not a reservoir. Its reversible work bound can be found by imagining many small Carnot cooling stages, each using the current temperature. This retains the local limiting relation without valuing all heat at the initial temperature. Its heat capacity determines how much heat it releases during a small temperature step.[1][3]

Work from a cooling source

Given: Hot body of constant total heat capacity 20.0 J/K cools reversibly from 600 K to 400 K. The only cold reservoir is at 300 K. No phase change or other energy sources; an ideal intervening engine operates cyclically.
Find: Find maximum work and the average heat-to-work fraction.

Approach: each small released heat element $-C\,dT$ has its instantaneous reversible yield. Integrate along cooling.
$W_{\max}=\int_{T_f}^{T_i}C\left(1-\frac{T_C}{T}\right)\,dT$
Evaluate the integral, then substitute the inputs.
$\begin{aligned}W_{\max}&=C\left[T_i-T_f-T_C\ln\frac{T_i}{T_f}\right]\\&=20.0\left[200-300\ln(1.5)\right]\,\mathrm J\\&\simeq1567.21\,\mathrm J\end{aligned}$
Calculate heat released by the source and the work fraction.
$\begin{aligned}Q_H&=C(T_i-T_f)=4000\,\mathrm J\\\overline\eta&=\frac{W_{\max}}{Q_H}\simeq0.391802\\Q_C&=Q_H-W_{\max}\simeq2432.79\,\mathrm J\end{aligned}$

Result: Maximum work is about 1.567 kJ; on average, 39.18% of supplied heat becomes work.

Check: The source loses $C\ln\frac{T_f}{T_i}\simeq-8.10930\,\mathrm{J\,K^{-1}}$ . The cold reservoir gains $\frac{Q_C}{T_C}\simeq8.10930\,\mathrm{J\,K^{-1}}$ . Their sum is zero. The average fraction lies between the final 25% bound and the initial 50% bound; applying 50% to all 4000 J would be incorrect.

Where the bound does not give a complete answer

The bound applies to cyclic heat-to-work conversion with exactly two fixed-temperature reservoirs. A battery and an electric motor receive energy differently; their efficiency is not assessed by $1-\frac{T_C}{T_H}$ . Multiple heat sources, material streams or a changing source temperature require a broader balance. Equal reservoir temperatures give a zero bound: heat exchange with that pair alone cannot produce positive cyclic net work.[5]

Five solved bound calculations

Calculate · Foundation

$T_H=500\,\mathrm K$ , $T_C=300\,\mathrm K$ and $Q_H=900\,\mathrm J$ . Find reversible limiting work.

Show hint

First find ηC.

Learning goal: Find reversible work limit.

Show solution and calculation

Approach: $\eta_C=1-\frac{T_C}{T_H}$ , then $W_{\mathrm{rev}}=\eta_C Q_H$ .
Substitution gives 40% and 360 J.
Energy balance gives 540 J rejected heat.

Check: the reservoir entropy changes calculated below cancel; both heat magnitudes are positive and work is smaller than heat input.

\begin{aligned}\eta_C&=1-\frac{300}{500}=0.40\\W_{\mathrm{rev}}&=0.40\cdot900\,\mathrm J=360\,\mathrm J\\Q_C&=900\,\mathrm J-360\,\mathrm J=540\,\mathrm J\\\Delta S_{\mathrm{res}}&=-\frac{900}{500}+\frac{540}{300}=0\,\mathrm{J\,K^{-1}}\end{aligned}

Rearrange · Application

An actual engine should reach 45%; $T_C=315\,\mathrm K$ . What is the minimum possible TH?

Show hint

A necessary Carnot bound is not an actual design guarantee.

Learning goal: Bound necessary hot-reservoir temperature.

Show solution and calculation

Approach: target efficiency cannot exceed $1-\frac{T_C}{T_H}$ , hence $T_H\ge\frac{T_C}{1-\eta_{\mathrm{target}}}$ .
$T_H\ge\frac{315}{1-0.45}\,\mathrm K=572.727\,\mathrm K\text{,}$ about 299.58 °C.
Equality is reversible; irreversibility requires a higher temperature bound for that target.

T_H\ge572.727\,\mathrm K

Error analysis · Application

An engine between 800 K and 400 K receives 1000 J and claims 700 J work per cycle. Is it possible?

Show hint

Check energy and entropy separately.

Learning goal: Apply energy and entropy checks separately.

Show solution and calculation

Approach: check conservation of energy first, then the second law.
Energy accounting alone gives 300 J rejected heat and is satisfied.
However, the claimed 70% efficiency exceeds the 50% bound.
Negative total reservoir entropy directly exposes the contradiction.
Under the stated conditions, this cyclic engine is impossible.

\begin{aligned}Q_C&=1000-700=300\,\mathrm J\\\eta&=\frac{700}{1000}=0.70>\eta_C=1-\frac{400}{800}=0.50\\\Delta S_{\mathrm{res}}&=-\frac{1000}{800}+\frac{300}{400}=-0.50\,\mathrm{J\,K^{-1}}<0\end{aligned}

Interpret · Application

Two reversible cyclic heat engines operate between 600 and 400 K and between 400 and 200 K. Both differences are 200 K. Each receives 1000 J per cycle. Compare efficiencies and work; only two ideal reservoirs.

Show hint

Carnot depends on absolute-temperature ratio.

Learning goal: Avoid treating temperature difference alone as the Carnot bound.

Show solution and calculation

Approach: The first ratio gives one-third efficiency and about 333.333 J work; the second gives one-half and 500 J.
Equal temperature differences do not imply equal bounds.
At equal heat input the smaller cold-to-hot ratio is decisive.
Real engines between those reservoirs cannot exceed these limits.

\begin{aligned}\eta_1&=1-\frac{400}{600}=\frac13,&W_1&=\frac{1000}3\,\mathrm J\\{}\eta_2&=1-\frac{200}{400}=\frac12,&W_2&=500\,\mathrm J\end{aligned}

[1]

Error analysis · Application

A heat pump works between 300 and 600 K and supplies 600 J heat to the hot reservoir using 300 J work. Someone calls this 200% and a violation of the 50% Carnot bound. Check energy and the appropriate reversible metric.

Show hint

Heating delivered per work is a coefficient of performance, not engine efficiency.

Learning goal: Avoid using Carnot efficiency as a general device metric.

Show solution and calculation

Approach: The pump additionally extracts 300 J from the cold reservoir, conserving energy.
Its heating COP is two, equal to the reversible limit for these temperatures.
Reservoir entropy changes $-\frac{300}{300}\,\mathrm{J\,K^{-1}}$ and $\frac{600}{600}\,\mathrm{J\,K^{-1}}$ cancel.
The 50% bound applies to the reverse heat-engine role, not this COP.

\begin{aligned}Q_C&=Q_H-W=300\,\mathrm J\\{}\mathrm{COP}_{H,\mathrm{rev}}&=\frac{T_H}{T_H-T_C}=\frac{600}{300}=2\\{}\Delta S_{\mathrm{res}}&=-\frac{300}{300}+\frac{600}{600}=0\,\mathrm{J\,K^{-1}}\end{aligned}

[4]

In brief

First define the boundary and operating role. A heat engine produces work from heat, and a complete cycle stores no additional net energy. The first law gives the energy balance; the second gives the Carnot bound. Use kelvin and fixed reservoir temperatures. The bound is not an actual power specification; changing sources, additional streams and heat pumps require the appropriate balance.

Sources for this explanation

Quanta · Updated October 1, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Reworked the introduction and explanations; added a third complete example, an original quantitative diagram and a short summary. Retained existing exercises, models and sources; clarified mathematical typography and model conditions.
October 1, 2026Answered kelvin and actual-efficiency questions directly, distinguished COP and connected efficiency, entropy transfer and heat capacity in the text.

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Quanta. Carnot Efficiency. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/wirkungsgrad-carnot

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Frequently asked questions about Carnot Efficiency

How do you calculate the Carnot efficiency?+

For a reversible cyclic heat engine between two constant-temperature reservoirs use $\eta_C=1-\frac{T_C}{T_H}$ . Both temperatures must be absolute kelvin values satisfying $0<T_C<T_H$ . With 625 K and 350 K, $\eta_C=0.44$ , or 44%. Receiving 2400 J per cycle allows at most 1056 J net work and requires 1344 J rejected heat. Check energy and reservoir entropy: $-\frac{2400}{625}+\frac{1344}{350}=0\,\mathrm{J\,K^{-1}}$ . Actual irreversible engines fall below this bound at identical reservoirs. Temperature difference alone does not determine efficiency; the ratio of absolute temperatures matters. The result is a bound, not a power rating.

Why must you insert the temperature in kelvin?+

The derivation compares reversible entropy transfers $\frac{Q_H}{T_H}$ and $\frac{Q_C}{T_C}$ , using absolute thermodynamic temperatures. Celsius has a shifted zero, so ratios of Celsius readings produce different and incorrect results here. Convert with $T[\mathrm K]=\vartheta[{}^\circ\mathrm C]+273.15$ before forming the ratio. Temperatures of 600 °C and 30 °C become 873.15 K and 303.15 K, giving $\eta_C\simeq65.281\,\%$ . Only temperature differences have the same numerical value in kelvin and Celsius. Equal differences do not guarantee equal efficiencies: $\frac{600}{300} K$ gives 50%, whereas $\frac{900}{600} K$ gives 33.3%. Always record both reservoir temperatures and their temperature scale.

Why can the efficiency never reach 100 percent?+

This bound concerns cyclic heat engines between finite positive reservoir temperatures, not every device. With $T_C>0$ and finite TH, $\frac{T_C}{T_H}$ stays positive, so $\eta_C<1$ . The second law requires enough heat rejection to the cold reservoir to compensate entropy lost by the hot reservoir. Complete cyclic conversion of heat from a single reservoir into work is forbidden. Friction is therefore not the sole reason for the limit: even a reversible engine rejects heat. Electrical work in a motor and heat delivery by a heat pump need different measures. Applying the statement to every efficiency would be too broad.

How do you rearrange the Carnot efficiency for the cold temperature?+

Move the ratio to one side: $\frac{T_C}{T_H}=1-\eta_C$ . Multiply by TH to obtain $T_C=(1-\eta_C)T_H$ . With $T_H=600\,\mathrm K$ and $\eta_C=0.50$ , $T_C=300\,\mathrm K$ . Convert a percentage to a fraction first: 50% means 0.50. Substitution into $1-\frac{300}{600}$ returns 0.50. This equality describes a reversible limit. A desired actual efficiency requires a Carnot bound at least as high, but that condition is not sufficient: heat transfer, friction and engineering constraints introduce further departures. The resulting temperature remains an absolute value in kelvin, not a Celsius reading.

What is the difference between Carnot efficiency and real efficiency?+

ηC is the reversible upper bound for specified hot and cold reservoirs. Actual thermal efficiency instead equals net work divided by absorbed heat using a consistent system boundary. With $T_H=750\,\mathrm K$ , $T_C=300\,\mathrm K$ and $Q_H=6000\,\mathrm J$ , maximum reversible work is 3600 J. An engine delivering 1800 J has $\eta=30\,\%$ and $Q_C=4200\,\mathrm J$ . Reservoir entropy $-\frac{6000}{750}+\frac{4200}{300}$ gives $\frac{6 J}{K}$ generation. General power-plant percentages do not follow from the equation; they require operating data. Multiple heat sources or finite cooling bodies need broader accounting, and auxiliary electrical consumption changes the chosen plant boundary.

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Related formulas

P = \sigma \cdot A \cdot T^4

Stefan-Boltzmann Law

More formulas: Heat and thermodynamics

\Delta U = Q - W

First Law of Thermodynamics

\Delta L = \alpha L_0 \Delta T

Linear thermal expansion

Q = m L

Latent heat during a phase change

\Delta S=\int_{i}^{f}\frac{\delta Q_{rev}}{T},\qquad T=\mathrm{const.}:\ \Delta S=\frac{Q_{rev}}{T}

Entropy change along a reversible path

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Create a curated FSRS exam set for η = 1 − T_k/T_w: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Carnot efficiency

Variables & units

Carnot efficiency calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Thermodynamic limits

Worked example

Applications

How do you calculate with Carnot Efficiency?

What efficiency compares

Derive the temperature limit from entropy

How the Carnot cycle describes the bound

Why does Carnot efficiency require kelvin?

A warmer cold reservoir lowers the bound

Example: heat and work in a reversible cycle

Check every energy transfer against entropy

Is Carnot efficiency the actual engine efficiency?

Compare the bound at identical reservoirs

Explore further: refrigerator and heat pump

Explore further: finite heat source

Work from a cooling source

Where the bound does not give a complete answer

Five solved bound calculations

TH=500 KT_H=500\,\mathrm KTH​=500K, TC=300 KT_C=300\,\mathrm KTC​=300K and QH=900 JQ_H=900\,\mathrm JQH​=900J. Find reversible limiting work.

An actual engine should reach 45%; TC=315 KT_C=315\,\mathrm KTC​=315K. What is the minimum possible TH?

An engine between 800 K and 400 K receives 1000 J and claims 700 J work per cycle. Is it possible?

Two reversible cyclic heat engines operate between 600 and 400 K and between 400 and 200 K. Both differences are 200 K. Each receives 1000 J per cycle. Compare efficiencies and work; only two ideal reservoirs.

A heat pump works between 300 and 600 K and supplies 600 J heat to the hot reservoir using 300 J work. Someone calls this 200% and a violation of the 50% Carnot bound. Check energy and the appropriate reversible metric.

In brief

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Subject references and further reading

Related formulas

More formulas: Heat and thermodynamics

Retain Carnot Efficiency for exams

$T_H=500\,\mathrm K$ , $T_C=300\,\mathrm K$ and $Q_H=900\,\mathrm J$ . Find reversible limiting work.

An actual engine should reach 45%; $T_C=315\,\mathrm K$ . What is the minimum possible TH?