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The energy and entropy budget of a heat engine, the Carnot efficiency $1 - T_c/T_h$ as the second law's limit, the Carnot cycle that reaches it, and why waste heat is unavoidable.
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By the end of this lesson you will be able to complete an engine's energy budget, find its efficiency and its Carnot limit, derive the limit from entropy, follow a Carnot cycle with an ideal gas, and judge whether a claimed engine is possible.
You know the first law as energy accounting, that heat $Q$ entering a reservoir at temperature $T$ changes its entropy by $Q/T$, and that the total entropy of an isolated system never falls. You can also find the work in isothermal and adiabatic processes of an ideal gas. This lesson puts all of that to work on the question that founded thermodynamics: how much work can heat do?
| Term | What it means |
|---|---|
| Heat engine | A device that runs in a cycle, taking heat $Q_h$ from a hot reservoir, doing work $W$, and rejecting heat $Q_c$ to a cold reservoir. |
| Efficiency | $e = W/Q_h$, the fraction of the heat taken in that becomes work. |
| Cycle | A sequence of processes that returns the engine to its starting state, so its own energy and entropy are unchanged. |
| Carnot efficiency | $e_{\max} = 1 - T_c/T_h$, the greatest efficiency any engine between two reservoirs can have. |
| Carnot cycle | Two isothermal and two adiabatic steps, all reversible; an engine that reaches the Carnot efficiency. |
| Waste heat | The rejected heat $Q_c$, required by the second law. |
| Kelvin-Planck statement | No cyclic device can turn heat from a single reservoir entirely into work. |
A heat engine runs in a cycle: it takes heat $Q_h$ from a hot reservoir at $T_h$, does work $W$, rejects heat $Q_c$ to a cold reservoir at $T_c$, and returns to where it started. Over a full cycle its own energy is unchanged, so the first law gives
$$Q_h = W + Q_c, \qquad e = \frac{W}{Q_h} = 1 - \frac{Q_c}{Q_h}.$$
The first law alone would allow $Q_c = 0$ and $e = 1$. The second law does not. The engine's own entropy is also unchanged over a cycle, so the entropy it takes from the hot reservoir, $Q_h/T_h$, must end up somewhere — and the only place it can go is the cold reservoir, which gains $Q_c/T_c$. For the total not to fall,
$$\frac{Q_c}{T_c} \geq \frac{Q_h}{T_h} \quad\Rightarrow\quad \frac{Q_c}{Q_h} \geq \frac{T_c}{T_h}.$$
Substituting into the efficiency:
$$e \leq 1 - \frac{T_c}{T_h}.$$
This is the Carnot efficiency, found by Sadi Carnot in 1824 before either law had been stated. It holds for every engine — steam, gasoline, jet, thermoelectric — because nothing in the argument mentions how the engine works. The only way to raise it is to raise $T_h$ or lower $T_c$. And some heat must always be rejected: the cold reservoir has to carry off the entropy that came in with the heat.
Another way: picture
Picture heat as water held behind a dam at a height set by the temperature. A water wheel can extract work only by letting water fall from the high reservoir to a lower one, and the water must still leave at the bottom. Entropy is what cannot be destroyed along the way: every bit of it that came in with the heat must flow out with the rejected heat. The lower the bottom reservoir, the less heat it takes to carry that entropy away, and the more is left over as work.
Another way: steps
The limit is reached only by an engine that creates no entropy, which means transferring heat only across zero temperature difference. Carnot's cycle does this with a gas in a cylinder in four reversible steps.
For an ideal gas the isothermal heats are $Q_h = nRT_h\ln(V_2/V_1)$ and $Q_c = nRT_c\ln(V_3/V_4)$, and the two adiabats force $V_3/V_4 = V_2/V_1$. So $Q_c/Q_h = T_c/T_h$ exactly, and the efficiency is the Carnot limit. The catch is that transferring heat across a vanishing temperature difference takes forever: a Carnot engine produces its perfect efficiency at zero power. Real engines trade efficiency for speed, and a useful estimate for an engine run at maximum power is $1 - \sqrt{T_c/T_h}$, which for many real power plants is close to what they achieve.
Checks. The efficiency is between $0$ and the Carnot limit. The rejected heat is never zero. A larger temperature ratio gives a larger limit, and the limit approaches one only as $T_c/T_h \to 0$. For a steam plant rejecting heat near $300$ K, real efficiencies of $0.33$ to $0.45$ and limits of about $0.6$ are the right scale.
An engine that took heat from a single reservoir — the ocean, say — and turned all of it into work would violate no energy conservation. The oceans hold about $10^{25}$ J of thermal energy above freezing, enough to run civilization for tens of thousands of years. Such a machine is called a perpetual motion machine of the second kind, and the second law forbids it: in the Kelvin-Planck form, no cyclic device can convert heat from one reservoir entirely into work. With a single reservoir, $T_c = T_h$ and the Carnot limit is zero.
The United States Patent and Trademark Office treats claims of perpetual motion as inoperable and, unusually, may ask for a working model before examining them. The entropy argument is why: a device that claims more than $1 - T_c/T_h$ does not need to be tested to be refuted, because it would make the entropy of an isolated system fall.
A Carnot engine reaches its limit only by running infinitely slowly, since heat must cross zero temperature difference. Curzon and Ahlborn asked in 1975 what efficiency an engine achieves if it is instead run for maximum power output, with heat flowing across finite differences at each reservoir. Their answer is strikingly simple:
$$e_{\text{max power}} = 1 - \sqrt{\frac{T_c}{T_h}}.$$
For a coal plant with steam at $838$ K and cooling at $298$ K, Carnot gives $0.64$ and the maximum-power formula $0.40$; the measured efficiency is about $0.36$. For a nuclear plant between $573$ K and $298$ K the formula gives $0.28$ against a measured $0.30$. Real plants are designed to make money, which means delivering power, not to approach an ideal that delivers none, and they land close to the formula rather than to Carnot.
A typical American coal plant generates $600$ MW of electricity at an efficiency of about $33$ percent. It therefore takes in $600/0.33 \approx 1800$ MW of heat and rejects about $1200$ MW — twice its electrical output — to the cooling water.
That heat has to go somewhere. Carried away by a river, $1200$ MW warms the water: with a flow of $300$ m³ per second, a moderate American river, the temperature rises by $1.2 \times 10^{9}/(300\,000 \times 4186) \approx 1$ K. Plants on smaller rivers or lakes use cooling towers instead, which evaporate water and carry the heat off as latent heat: at $2.4$ MJ per kilogram, $1200$ MW evaporates about $500$ kg of water every second, some $40$ million liters a day. Thermoelectric power is the largest user of water withdrawn in the United States.
Combined-cycle gas plants attack the limit directly. They burn gas in a turbine at over $1600$ K, then use its $850$ K exhaust to raise steam for a second turbine rejecting at about $300$ K. The overall Carnot limit, $1 - 300/1600 = 0.81$, is far higher than a steam plant's, and the best such plants reach $60$ to $64$ percent, cutting both fuel and waste heat per unit of electricity.
The tropical ocean is a vast heat engine waiting to be used: surface water at about $25$ °C sits above deep water at about $5$ °C. Ocean thermal energy conversion, OTEC, runs an engine between them, and a demonstration plant has operated in Hawaii since 2015.
The Carnot limit is brutal: $1 - 278/298 = 0.067$. Even a perfect engine would turn only one fifteenth of the heat into work, and real OTEC systems manage about $3$ percent, after the pumps that lift cold water from a kilometer down take their share. To deliver $100$ kW of net power the Hawaii plant must move several tons of seawater every second through its heat exchangers.
The same arithmetic explains why OTEC works only in the tropics, where the surface is warm all year: at $15$ °C on the surface the limit falls to $1 - 278/288 = 0.035$. It is a clean demonstration that a temperature difference is a resource, and that the second law prices it by the ratio, not the difference, of the absolute temperatures.
It is natural to see the heat a power plant dumps into a river as a failure of engineering that better machines would eliminate. The second law says otherwise. The heat taken from the hot reservoir brings entropy $Q_h/T_h$ with it, and an engine that returns to its starting state every cycle cannot keep that entropy; it must deliver at least as much to the cold reservoir, which takes heat $Q_c \geq Q_hT_c/T_h$. A perfect engine between $800$ K and $300$ K still rejects three eighths of its heat.
A second misconception is that efficiency depends on the fuel or the working substance — that a better fuel or a cleverer gas could beat the limit. The Carnot argument never mentions either. A better fuel matters only by allowing a higher flame temperature; everything else about the limit is the two temperatures. And a 100 percent efficient electric heater is no contradiction: it turns work into heat, which the second law allows freely. Only the reverse is limited.
Steam enters a turbine at $527$ °C and the condenser is at $27$ °C. Convert the hot temperature to kelvin.
$T_h = 527 + 273 = 800\ \text{K}$
Carnot's formula needs absolute temperatures.
Convert the cold temperature.
$T_c = 27 + 273 = 300\ \text{K}$
The condenser rejects heat to cooling water near room temperature.
Write the Carnot limit and substitute.
$e_{\max} = 1 - \dfrac{300}{800}$
The greatest efficiency any engine between these temperatures can have.
Evaluate the fraction.
$e_{\max} = 1 - 0.375 = 0.625$
At best five eighths of the heat can become work.
Compare with a real plant.
$e_{\text{real}} \approx 0.40 \approx 0.64\,e_{\max}$
Real steam plants reach about two thirds of the limit; the rest is lost to friction and to heat crossing large temperature differences in the boiler.
An engine takes $2000$ J from a reservoir at $600$ K and does $600$ J of work each cycle. Find the rejected heat.
$Q_c = Q_h - W = 2000 - 600 = 1400\ \text{J}$
The first law over a cycle.
Find the efficiency.
$e = \dfrac{600}{2000} = 0.30$
Work out over heat in.
Find the Carnot limit with a cold reservoir at $300$ K.
$e_{\max} = 1 - \dfrac{300}{600} = 0.50$
The engine is below the limit, so it is possible.
Find the entropy the hot reservoir loses.
$\dfrac{Q_h}{T_h} = \dfrac{2000}{600} = 3.33\ \text{J/K}$
Heat over temperature.
Find the entropy the cold reservoir gains.
$\dfrac{Q_c}{T_c} = \dfrac{1400}{300} = 4.67\ \text{J/K}$
More than the hot reservoir lost.
Find the entropy created each cycle.
$\Delta S = 4.67 - 3.33 = 1.33\ \text{J/K}$
Positive, as the second law requires; a Carnot engine would create none.
One mole of helium runs a Carnot cycle between $500$ K and $300$ K, first expanding isothermally from $10$ L to $20$ L. Find the heat taken in.
$Q_h = nRT_h\ln\dfrac{V_2}{V_1} = 8.31 \times 500 \times \ln 2 = 2880\ \text{J}$
At constant temperature an ideal gas's energy is unchanged, so the heat in equals the work out.
Find the volume at the end of the adiabatic expansion.
$V_3 = V_2\left(\dfrac{T_h}{T_c}\right)^{3/2} = 20 \times \left(\dfrac{5}{3}\right)^{3/2} = 43.0\ \text{L}$
Along an adiabat $TV^{2/3}$ is constant for a monatomic gas.
Find the volume at the start of the adiabatic compression.
$V_4 = V_1\left(\dfrac{T_h}{T_c}\right)^{3/2} = 10 \times 2.152 = 21.5\ \text{L}$
The same adiabatic factor, from the starting volume.
Find the ratio of the cold isothermal volumes.
$\dfrac{V_3}{V_4} = \dfrac{43.0}{21.5} = 2 = \dfrac{V_2}{V_1}$
The two adiabats force the same ratio on both isotherms.
Find the heat rejected.
$Q_c = nRT_c\ln\dfrac{V_3}{V_4} = 8.31 \times 300 \times \ln 2 = 1728\ \text{J}$
The isothermal compression at $300$ K.
Find the work per cycle.
$W = Q_h - Q_c = 2880 - 1728 = 1152\ \text{J}$
The adiabatic steps exchange no heat, and their works cancel.
Find the efficiency.
$e = \dfrac{1152}{2880} = 0.40$
Work over heat in.
Compare with the Carnot limit.
$1 - \dfrac{300}{500} = 0.40$
Exactly equal: the reversible cycle reaches the limit, as the entropy argument says it must.
Write the Carnot limit.
$e_{\max} = 1 - \dfrac{T_c}{T_h}$
The temperatures are already in kelvin.
Substitute the temperatures.
$e_{\max} = 1 - \dfrac{300}{450}$
Cold over hot.
Evaluate the fraction.
An inventor claims an engine that takes heat from a reservoir at $500$ K, rejects heat to one at $300$ K, and has an efficiency of $0.45$. What should an engineer conclude?
Complete the worked solution: what is the Carnot efficiency of an engine running between $227$ °C and $27$ °C?
Add $273$ to the hot temperature.
$T_h = 227 + 273 =$ h K
The limit uses absolute temperatures.
Add $273$ to the cold temperature.
$T_c = 27 + 273 =$ c K
The same scale for both.
Subtract the ratio of the temperatures from one.
$e_{\max} = 1 - \dfrac{T_c}{T_h} =$ e
The entropy taken from the hot reservoir must all reach the cold one.
For an engine taking heat from a reservoir at $900$ K, match each idea to its equation.
| $e = W/Q_h$ | $Q_c = Q_h - W$ | $Q_c/T_c \geq Q_h/T_h$ | $e \leq 1 - T_c/T_h$ | |
|---|---|---|---|---|
| the definition of efficiency | ||||
| the first law for a cycle | ||||
| the second law for a cycle | ||||
| the Carnot limit |
Three engines each run one cycle. Fill in the heat each rejects and its efficiency.
| heat in (J) | work out (J) | heat rejected (J) | efficiency | |
|---|---|---|---|---|
| engine A | 9000 | 2250 | ||
| engine B | 18000 | 7200 | ||
| engine C | 4500 | 900 |
What is the greatest possible efficiency of an engine running between $527$ °C and $27$ °C?
Answer:
A power plant delivers $400$ MW of electricity with an overall efficiency of $0.2$. At what rate does it reject heat to its cooling water?
Answer: unit: W / MW / kW
In a geothermal plant, heat enters at about $453$ K and is rejected at about $300$ K. What is the greatest efficiency the second law allows it? Give it to three decimal places.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Three engines each run one cycle. Fill in the heat each rejects and its efficiency.
| heat in (J) | work out (J) | heat rejected (J) | efficiency | |
|---|---|---|---|---|
| engine A | 2000 | 500 | ||
| engine B | 4000 | 1600 | ||
| engine C | 1000 | 200 |
You can find the efficiency, waste heat and Carnot limit of any heat engine. Explain to someone why even a perfect power plant must dump heat into a river.
15. Your turn: a geothermal plant draws heat from water at $450$ K and rejects it at $300$ K. What is its greatest possible efficiency?, step 3
$e_{\max} = 1 - 0.667 = 0.333$
A modest limit, because the hot water is not very hot.