core25 min studyLesson 314 in Physics

Carnot cycles and absolute temperature

Derive a reversible two-reservoir benchmark, relate heat ratios to kelvin temperatures, and distinguish engine efficiency from refrigerator performance.

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What you’ll understand

  • Account for heat and work over a reversible two-reservoir cycle.
  • Explain the working-substance independence of Carnot efficiency.
  • Calculate an engine efficiency and a refrigerator coefficient of performance.

Energy conservation alone would permit an engine to convert every joule of absorbed heat into work. The second law places a stronger restriction on a cyclic device interacting with thermal reservoirs. Carnot’s reversible construction provides the benchmark and makes temperature ratios independent of a particular thermometer material. We will treat the hot and cold heat amounts as positive magnitudes, while retaining negative work on an operating engine. This separates the device’s engineering output from the module’s consistent energy-input sign convention.

A complete cycle must restore the working substance

A heat engine absorbs magnitude Qh>0Q_h>0 from a hot reservoir and rejects magnitude Qc>0Q_c>0 to a cold reservoir. Reservoirs are ideal bodies whose temperatures remain fixed during the specified exchanges. Over a cycle the working substance returns to its initial state, giving zero net internal-energy change. The net heat into it is Qh−QcQ_h-Q_c, so work done on it is the negative of that quantity. Work output is positive when the device operates as an engine.

A Carnot cycle has two reversible isothermal exchanges and two reversible adiabatic connections. During each isotherm the working substance differs only infinitesimally in temperature from its reservoir; the adiabatic stages exchange no heat. Every step must be reversible, including mechanical boundaries and thermal contacts. A cycle drawn with two constant-temperature curves is not automatically Carnot if friction or finite thermal gaps remain. The device must also complete the cycle; a one-time expansion that consumes a changing resource is a different situation.

ΔUcycle=0,Won=−(Qh−Qc),η=WbyQh=1−QcQh\Delta U_{\mathrm{cycle}}=0,\qquad W_{\mathrm{on}}=-(Q_h-Q_c),\qquad \eta=\frac{W_{\mathrm{by}}}{Q_h}=1-\frac{Q_c}{Q_h}
Q h and Q c are positive heat magnitudes in joules, while actual signed heat into the engine is Q h minus Q c. Work by the engine is minus work on it.↔ Scroll wide equations horizontally.

Reversibility makes a universal benchmark

Suppose an engine exceeded the efficiency of a reversible engine operating between the same two reservoirs. Use its work output to run the reversible device backward, scaling the devices so their work magnitudes cancel. The combined cyclic apparatus would transfer heat from cold to hot without external work or another compensating effect. That violates the Clausius statement of the second law. Reversibility is essential because the benchmark must be able to run backward with exactly reversed exchanges.

Applying the argument in both directions shows that all reversible engines between the same reservoirs have the same efficiency, independent of working substance. Cascading reversible engines through an intermediate reservoir then makes their heat ratios multiply. This multiplicative structure permits a thermodynamic temperature scale with heat ratio equal to the cold-to-hot temperature ratio. For the ordinary positive-temperature reservoirs considered here, use kelvin values; Celsius offsets destroy temperature ratios even when temperature differences are correctly represented.

QcQh=TcTh(reversible),η≤1−TcTh,COPR,rev=TcTh−Tc\frac{Q_c}{Q_h}=\frac{T_c}{T_h}\quad\text{(reversible)},\qquad \eta\le1-\frac{T_c}{T_h},\qquad \mathrm{COP}_{\mathrm{R,rev}}=\frac{T_c}{T_h-T_c}
The efficiency inequality concerns cyclic engines exchanging heat only with two fixed reservoirs at positive kelvin temperatures. COP R is cold heat extracted divided by positive work input for a reversible refrigerator.↔ Scroll wide equations horizontally.

Bounds are not performance predictions

A practical engine can lose useful work to friction, finite-temperature heat transfer, leakage, or other irreversibilities. Carnot efficiency is an upper bound under the stated reservoir conditions, not the predicted efficiency of any particular apparatus. A real heat source that cools substantially is not one fixed-temperature reservoir; its extraction process must be analyzed with changing temperatures or an appropriate availability calculation. Increasing hot temperature or decreasing cold temperature raises the ideal engine bound, subject to the physical system’s other constraints.

Running a reversible cycle backward produces a refrigerator. It absorbs heat from the cold reservoir, requires positive work on the device, and rejects their sum to the hot reservoir. Its coefficient of performance can exceed one because the numerator includes heat transported rather than only work energy created. This does not violate conservation. For a fixed positive temperature gap the refrigerator COP is finite; as the ideal gap tends to zero the formal COP diverges while the finite-rate engineering problem remains outside the reversible model.

An ideal two-reservoir engine

A reversible engine operates between 600 K and 300 K and absorbs 900 J from the hot reservoir per cycle. Find rejected heat, work output, work on the engine, and efficiency.

  1. Reversibility requires Qc/Qh=300/600=1/2Q_c/Q_h=300/600=1/2, so the rejected heat magnitude is Qc=450 JQ_c=450\,\mathrm J. This ratio uses absolute temperatures.

  2. The net heat into the device is 900−450=450 J900-450=450\,\mathrm J. Zero cyclic internal-energy change therefore gives Won=−450 JW_{\mathrm{on}}=-450\,\mathrm J, or work output Wby=450 JW_{\mathrm{by}}=450\,\mathrm J.

  3. Efficiency is 450/900=0.50450/900=0.50, agreeing with 1−300/6001-300/600. The working substance returns to its initial state even though the reservoirs and mechanical work receiver have changed.

The result

Each cycle rejects 450 J, delivers 450 J of work, and has 50 percent efficiency. The work-on value is −450 J under the declared convention.

Explore the model

Interactive lab

Interactive lab · function

A plotted curve is not a heat engine

Use the instrument for the limited idea that an area changes with a protocol shape. Reset, set Coefficient a = 0.5 and Position x = 1, and predict area 1/6. For a fresh run Reset, set a = 1.5 and x = 1, and predict area 1/2.

A plotted curve is not a heat engineThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe parabola is an integration analogy. There are no reservoir temperatures or cycle controls here, so Carnot efficiency and refrigerator COP must be calculated from the written equations.

Check your understanding

Which temperature values belong in the Carnot ratio?

Try it yourself

A reversible refrigerator moves 600 J per cycle out of a 300 K reservoir and rejects heat to a 400 K reservoir. Find positive work input, rejected heat, and refrigerator COP.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the reversible heat ratio, then energy conservation; the larger heat magnitude is on the hot side.

Compare with the solution

The hot rejection is Qh=600(400/300)=800 JQ_h=600(400/300)=800\,\mathrm J. Work input is Won=800−600=200 JW_{\mathrm{on}}=800-600=200\,\mathrm J. The refrigerator COP is 600/200=3600/200=3, agreeing with 300/(400−300)300/(400-300). The hot reservoir gains both the extracted cold heat and the work supplied to the device, so a COP above one is consistent with energy conservation.

Take it with you

  • Carnot universality follows from a reversible device that can be run backward.
  • The temperature ratio uses kelvin and assumes fixed thermal reservoirs.
  • Engine efficiency and refrigerator COP use different useful outputs and are not interchangeable.

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