A cyclic engine can conserve energy and still violate the second law. Reservoir temperatures set a stricter limit.
By the end: derive the Carnot bound from two reservoir entropy changes and distinguish a bound from an operating design.
Close the cycle before comparing flows
A heat engine returns its working substance to the same state each cycle. Its internal-energy change is zero over the cycle. Let Qh be positive heat taken from a hot reservoir, Qc positive heat rejected to a cold reservoir, and W positive work output. Conservation requires the heat input to equal rejected heat plus work. Thermal efficiency is work output divided by heat input.
At first this seems to allow turning all input heat into work. But returning the working substance to its starting state also makes its entropy change zero. The hot reservoir loses Qh divided by Th in entropy; the cold reservoir gains Qc divided by Tc. Their sum cannot be negative if these are the only reservoirs and the full device is cyclic.
Explain why an engine must complete its return
A single expansion is not yet a heat-engine cycle. Imagine the working gas pushing a piston outward while taking heat from a hot reservoir. That stroke may deliver positive work, but the gas finishes at a larger volume. To repeat the same stroke indefinitely, the working substance must return to its initial equilibrium state. Compression requires work input. The useful cycle output is expansion work minus compression work, not the largest positive stroke reported alone. The return must be included.
A pressure-volume picture explains how positive net work remains possible. In a constructed rectangular quasistatic loop, expansion from 10 L to 20 L against 100 kPa gives 1,000 J. Compression over the same volume interval against 50 kPa gives work output of negative 500 J. The fixed-volume connecting legs do no boundary work. Net output equals the enclosed area. A kilopascal times litre is one joule, so 50 kPa height times 10 L width gives 500 J.
Subtract compression from expansion
Scaled quasistatic pressure-volume rectangle, not a Carnot cycle. Work uses opposing pressures equal to the gas pressures in the reversible mechanical limit. Clockwise traversal gives +1,000 − 500 = +500 J. Vertical legs do zero boundary work.
This rectangle illustrates subtraction of work, not a Carnot engine. The working gas's temperature changes on its legs, so reversible heat exchange would not occur solely at two fixed reservoir temperatures. Drawing a loop does not establish how its thermal contacts are arranged. For any real boundary-work account, use the opposing external pressure. The equilibrium gas-pressure curve gives that work only when the mechanical conditions justify it. Slow piston movement with friction may remain quasistatic while still generating entropy and failing reversibility.
For the cyclic working substance, internal energy returns to its starting value. With heat entering positive and work output positive, the first law then requires net Q to equal net W. Naming positive heat magnitudes clarifies the subtraction. Heat absorbed from the hot reservoir is Qh; heat delivered to the cold reservoir is Qc. Net heat input is Qh minus Qc. Both can be large even when their difference, the useful work, is small.
Entropy must also return to its starting value because it is a state property. A cyclic engine can generate entropy, but cannot accumulate it while returning to the same state. It must export enough entropy through its thermal contacts. The cold reservoir is therefore not simply a destination for energy that mechanical parts failed to convert. Under the two-reservoir cyclic assumptions, entropy export requires some heat rejection even in a perfectly reversible engine with no friction or other dissipative loss.
A cyclic substance must export its generated entropy
Schematic thermal contacts for the permitted 300 J work proposal. Reservoir changes determine generation for the complete isolated account. The working substance returns to its initial entropy; work is not counted as an entropy transfer.
Contrast the cycle with a finite hot body cooling while driving an ideal reversible engine against cooler surroundings. The body supplies energy while its own state changes. It can support a temporary output, but it is not a fixed-temperature reservoir indefinitely. Restoring it would require an additional process. A proposed one-reservoir cyclic engine cannot use an unnoticed cooling store as a permanent explanation of its output. Include the changing store in the boundary account before applying a bound derived for two fixed-temperature reservoirs.
Turn the entropy condition into a bound
Nonnegative total entropy requires rejected heat to be at least the input heat multiplied by the cold-to-hot kelvin temperature ratio. Substitute that requirement into the first-law work account. Maximum efficiency is one minus that temperature ratio. Equality describes a reversible limit. For 600 K and 300 K the limit is 50%, so 1,000 J of heat can yield at most 500 J of work.
The model also lets you propose outputs above the limit. A 700 J work claim leaves 300 J to reject, so its energy account balances. Yet the reservoir entropy change is negative, identifying the impossible combination of assumptions. These are proposed cycle accounts, not a simulation that physically runs a forbidden engine. At 300 J work, rejected heat is 700 J and entropy generation is positive.
Derive the rejected heat before calculating efficiency
Derive a new numerical bound before using the fixed-temperature engine controls. Suppose a cyclic device takes 900 J from a 450 K reservoir and rejects heat to a 300 K reservoir. The hot reservoir loses 900 divided by 450, or 2 J/K. Because the working substance finishes with no entropy change, the cold reservoir must gain at least 2 J/K. At 300 K, that requires at least 600 J of rejected heat. Energy conservation leaves at most 300 J of net work.
The efficiency bound in this example is therefore 300 divided by 900, or one third. Keeping the algebra in the same order helps explain the general formula. Nonnegative total entropy requires Qc/Tc to be at least Qh/Th. Multiplying by the positive cold temperature gives a minimum rejection Qh times Tc/Th. Subtract that amount from heat input, then divide by heat input to find the largest allowed work fraction. The entropy denominators require absolute temperatures.
Test a proposed 450 J work output for that same 900 J input. Conservation demands rejection of 450 J. The cold reservoir then gains only 1.5 J/K, while the hot reservoir still loses 2 J/K. The total falls by 0.5 J/K. The first law passes, but the second fails. Reducing proposed work to 180 J instead makes rejection 720 J and cold entropy gain 2.4 J/K. The resulting 0.4 J/K increase permits that account without demonstrating a working device.
Check the new 450 K and 300 K example
Proposed work
Rejected heat
Cold ΔS
Total ΔS
180 J
720 J
+2.4 J/K
+0.4 J/K
300 J, reversible bound
600 J
+2.0 J/K
0 J/K
450 J, forbidden
450 J
+1.5 J/K
−0.5 J/K
Every row takes 900 J from 450 K, so hot-reservoir ΔS is −2 J/K. The rows conserve energy; only the nonnegative total-entropy rows pass the second-law test. Passing is not device validation.
The difference between maximum and proposed work has a useful algebraic meaning at fixed Qh, Th, and Tc. Let generated entropy be the sum of reservoir entropy changes per cycle. That sum times Tc gives rejection above its reversible minimum. By conservation, this also equals the shortfall below reversible work. In the 180 J proposal, 300 K times 0.4 J/K equals 120 J, exactly the difference between the 300 J bound and the proposed 180 J output.
Relate additional rejection to the work shortfall
Both accounts absorb 900 J at 450 K and reject at 300 K. Bar lengths use 1.8 SVG units per joule, beginning at zero. The 120 J shortfall equals 300 K × 0.4 J/K of generated entropy for this fixed task.
This relation depends on the stated comparison. It holds here because the reservoir temperatures and heat input stay fixed, the device is cyclic, and there are no other transfers. It is not a universal rule that multiplying every entropy change by an arbitrary temperature finds lost work. A subsystem's entropy change can include transfer rather than generation. If heat input changes between proposals, comparing work outputs alone also changes the task. Always distinguish a reversible reference for the same task from an unrelated higher-output account.
A Celsius shortcut would fail even on the simple 450 K and 300 K example. Those temperatures are 176.85 °C and 26.85 °C. Substituting their numerical Celsius ratio into one minus cold over hot would give roughly 84.8%, rather than 33.3%. The two physical states did not change; only the numerical zero changed. Temperature differences are equal in kelvin and Celsius, but this bound uses absolute reciprocal temperatures in its derivation. A reported temperature difference alone cannot determine its efficiency limit.
Work through the mechanism
Conservation over a full cycle
Qh and Qc are positive heat magnitudes in J per cycle. The working substance returns to its initial U and S.
Nonnegative entropy generation
No other reservoirs, matter flows, or energy stores are included.
Carnot efficiency bound
η is a dimensionless fraction, not a power. Equality requires reversibility. Use kelvin.
Test a proposed cyclic energy account
Efficiency 30%; reversible bound 50%. Reservoir entropy change 0.67 J/K. Permitted by these balances; not verified device performance.
All proposals take 1,000 J from 600 K and reject heat at 300 K. Arrows show directions, not scaled flow widths; the bottom strip divides the same 1,000 J into work and rejected heat by length. Controls replace the proposed work immediately. An impossible proposal is labeled, not simulated as a functioning engine.
Proposed W per 1,000 J input
Rejected Qc
Reservoir ΔS
Assessment
300 J
700 J
+0.667 J/K
Below reversible limit
500 J
500 J
0 J/K
Reversible limit
700 J
300 J
−0.667 J/K
Forbidden under these assumptions
Both reservoirs stay at 600 K and 300 K; the device returns to its initial state.
Explore the proposed outputs without assuming operation
The engine controls hold the reservoirs at 600 K and 300 K and the absorbed heat at 1,000 J per cycle. Only the proposed work output changes. First select the no-work preset. All 1,000 J goes to the cold reservoir, so the account reduces to direct transfer across the temperature difference. The hot entropy loss is about 1.667 J/K and the cold gain about 3.333 J/K. Their difference, about 1.667 J/K, is positive despite zero useful work.
Before selecting 300 J work, predict both changes from the no-work case. Rejection falls to 700 J while the hot heat input stays fixed. The hot reservoir's entropy change therefore stays fixed too. The cold entropy gain falls to about 2.333 J/K, leaving about 0.667 J/K generation. The useful work fraction is 30%. Confirm that the lower strip partitions the same 1,000 J total between output and rejection. It shows energy by length; the directional arrows do not encode heat or entropy magnitude.
The energy strip balances even for a forbidden proposal
All bars represent the same 1,000 J heat input, at 0.6 SVG units per joule. Blue is work; gold is rejected heat. Reservoir temperatures are 600 K and 300 K. The last row fails entropy, not energy conservation.
Now select the 500 J reversible-limit preset. Rejected heat is also 500 J, so the cold entropy gain exactly cancels the hot loss. The working substance has zero entropy change in both cases, but total generation differs. That distinction prevents a common error: returning to the same state does not make every cycle reversible. It makes the working substance's state changes zero over the cycle, while irreversibility remains detectable in the reservoir changes.
Finally select the forbidden 700 J proposal and identify the first quantity that becomes inconsistent. Its 300 J rejection gives only 1 J/K to the cold reservoir, producing total change of about negative 0.667 J/K. The strip still sums to 1,000 J. The display evaluates an account rather than running an impossible engine. If an observed report contained these numbers, some stated flow, temperature, cyclic condition, or omitted interaction would have to differ from the claimed account.
Use the slider for a less familiar case, 400 J work. Predict 600 J rejection, 40% efficiency, and about 0.333 J/K generation before reading the result. The reversible maximum remains 500 J. Multiply the generation by the 300 K cold temperature to recover the 100 J work shortfall. Then compare with the 600 J proposal. It rejects 400 J with total entropy change about negative 0.333 J/K. Equal distances from the bound give opposite signs.
Entropy changes sign at the reversible work limit
For 1,000 J absorbed at 600 K and rejection at 300 K, total ΔS = (500 J − W)/(300 K). Axes are linear; proposals to the right of 500 J have negative total entropy change and are forbidden under the stated assumptions.
These comparisons keep both energy input and the task fixed. A 40% account would produce more work than a 50% account if it received sufficiently more heat, so efficiency alone cannot rank output quantities. Neither gives power without a cycle duration. For a separate hypothetical timing example, 300 J per cycle at two cycles per second would mean 600 W; 500 J per cycle at one cycle per second would mean 500 W. The controls contain no duration setting and establish neither of those operating rates.
Separate efficiency, power, and available conditions
A reversible Carnot cycle uses isothermal heat exchange at each reservoir temperature and reversible adiabatic connections between those temperatures. Reversibility requires avoiding friction and finite driving differences. The ideal bound is not a construction plan for an engine delivering finite power. Real engines lose performance through finite-temperature transfers, friction, pressure losses, and other effects.
Higher efficiency does not necessarily mean higher power or lower total cost. Power is work per time; this model has no cycle duration. Changing the hot or cold reservoir changes the bound, but a real material may not survive the proposed temperature. Combustion and fuel-cell efficiencies also depend on the chosen chemical-energy reference and boundary. Do not apply this two-reservoir heat-engine bound to every device with a percent label.
Connect reversible heat flow to a complete ideal cycle
An ideal-gas Carnot cycle gives a mechanism consistent with the equality case. During reversible isothermal expansion at the hot temperature, heat enters while work leaves. Internal energy stays fixed because ideal-gas internal energy depends only on temperature. Reversible insulated expansion then lowers temperature as more work leaves. At the cold temperature, reversible isothermal compression rejects heat while receiving work. Finally reversible insulated compression restores the hot temperature and the starting state. The net work includes the negative compression contributions, not merely both expansion strokes.
A temperature-entropy diagram shows why the heat ratio matches the temperature ratio. Entropy rises by some amount during the hot isothermal leg and falls by that same amount during the cold isothermal leg. The reversible insulated connections hold entropy constant. The enclosed shape is therefore a rectangle. Heat absorbed equals hot temperature times the entropy increase; heat rejected equals cold temperature times the same entropy decrease magnitude. Dividing cancels the common entropy width. This area construction requires reversible heat along equilibrium states.
For a worked rectangle, choose 600 K and 300 K with an entropy width of 2 J/K. The upper rectangle has area 1,200 J and the lower rectangle 600 J. Their difference gives 600 J of net work and 50% efficiency. The entropy origin is arbitrary because only differences matter. The reversible insulated sides have zero width and contribute no heat. An irreversible insulated step generally changes entropy and would not follow those vertical sides.
Reversible heat is temperature times entropy width
Scaled temperature-entropy rectangle with width 2 J/K. The entropy origin S₀ is arbitrary. Upper heat area is 1,200 J; rejected heat area is 600 J. The difference, shown blue, is 600 J net work. Reversible insulated legs are vertical.
The ideal-gas construction helps explain the bound, but the two-reservoir entropy derivation did not assume a monatomic gas. Different working substances cannot evade that bound merely by storing energy in different microscopic modes. For detailed gas calculations, however, heat capacities and the reversible adiabatic pressure-volume shape depend on molecular structure and temperature. The classical monatomic capacity model is one approximation. Quantum suppression of excitations or a phase change alters state relationships without removing the nonnegative total-generation requirement.
Reversible heat transfer requires infinitesimal working-substance and reservoir temperature differences. Finite-conductance transfer at useful rates needs a finite difference and produces entropy. Friction introduces another source even when internal states remain nearly uniform. Thus the reversible cycle is a limit used to compare accounts, not proof of a finite-power device. Raising the hot temperature or lowering the cold temperature raises its efficiency bound, but the inequality contains no material-survival model, heat-exchanger sizing rule, or permitted operating specification.
As a final transfer problem, double both reservoir temperatures while keeping their ratio fixed. The efficiency bound is unchanged. With the same absorbed heat, reversible work and rejected heat stay unchanged too, although their associated entropy transfers halve. By contrast, raising only the hot temperature increases the efficiency bound. This separates absolute entropy amounts from the ratio controlling reversible efficiency. Passing these comparisons establishes thermodynamic consistency, not a validated engine rate or evidence that a particular mechanism can execute the cycle.
Equal temperature ratios give equal reversible efficiency
Hot / cold
Minimum rejection
Maximum work
Entropy transferred hot → cold
600 K / 300 K
500 J
500 J, 50%
1.667 J/K
1,200 K / 600 K
500 J
500 J, 50%
0.833 J/K
1,200 K / 300 K
250 J
750 J, 75%
0.833 J/K
Separate hypothetical reversible accounts, each taking 1,000 J. Doubling both reservoir temperatures halves the entropy transferred without changing the heat split. Raising only the hot temperature changes the efficiency bound.
Exercises
Change the reservoir temperatures
A two-reservoir engine absorbs 1,200 J per cycle at 500 K and rejects heat at 300 K. What is its maximum work output?
Hint
Find the minimum rejected fraction before subtracting from input.
Worked answer
Step
Result
Minimum rejection
1,200 × 300 / 500 = 720 J
Maximum work
1,200 − 720 = 480 J
Efficiency bound
480 / 1,200 = 40%
All temperature ratios use kelvin.
At least 300/500 = 60% of the input must be rejected. That is 720 J. Maximum work is 480 J and maximum efficiency is 40%. A real irreversible engine under the same conditions has less work output for the same heat input.
Locate the failure in a balanced claim
A cyclic device takes 1,000 J from 600 K, produces 700 J of work, and rejects 300 J at 300 K. It has no other transfers. Which law rejects the claim?
Hint
Check both the energy sum and the reservoir entropy sum.
Worked answer
The energy sum closes, but the stated 700 J work output exceeds the 600 K / 300 K two-reservoir limit. Arrow directions are schematic; the bottom strip divides the energy to scale.
Check
Calculation
Result
Energy
1,000 − 700 − 300
0 J, balanced
Entropy
300/300 − 1,000/600
−0.667 J/K, forbidden
Energy conservation is necessary but not sufficient.
The first law passes: 1,000 = 700 + 300. The second law fails. The hot reservoir loses 1.667 J/K while the cold gains only 1 J/K. The total decreases by about 0.667 J/K. Either the reported flows, temperatures, cyclic behavior, or completeness of the boundary is wrong.
Sources and further reading
These optional references support the stated laws and model assumptions. The numerical examples and diagrams are authored teaching models, not experimental measurements or engineering specifications. Selected relevant passages were reviewed, not entire textbooks or a comprehensive research literature.