Entropy and the second law

Two reservoirs, one observed direction

Two large thermal reservoirs stay at 400 K and 300 K. A thermal connection transfers 240 J from the hotter one to the colder one. The connection has no net energy or entropy storage over the interval. Enclose both reservoirs and the connection in an isolated boundary: no energy or matter crosses the outside. Energy lost by one reservoir equals energy gained by the other. Why does the reverse transfer, with no other change anywhere, fail to occur even though it would conserve energy too?

The first law tracks energy. The second law supplies an additional restriction on macroscopic change, expressed through a state property called entropy, S. For this transfer the hot reservoir loses 0.6 J/K of entropy, the cold reservoir gains 0.8 J/K, and the isolated combination gains 0.2 J/K. The difference is entropy produced by the irreversible transfer, not extra energy.

On a first reading, follow the reservoir question, the entropy balance, warming a finite body, and the comparison of reversible and free gas expansion. Sections marked “Further” return to the cyclic argument, equilibrium-state derivatives, maximum entropy and flowing matter. Published preparation is “Physical systems, states, boundaries, and environments”, “Work, kinetic energy, potential energy, and power”, and “Conservation of mass, momentum, and energy”. The thermodynamic definitions needed here are supplied locally; the neighboring first-law, temperature and heat-capacity articles need not already be available.

Define the system and the entropy change

A system is the selected matter or region; its boundary separates it from the surroundings. A closed system exchanges no matter, although heat and work may cross its boundary. An isolated system exchanges neither matter nor energy. An adiabatic boundary prevents heat transfer during the interval; work may still cross it. An equilibrium state has no macroscopic tendency to change under the constraints being imposed.

Heat Q is energy transferred because of a temperature difference. Take Q positive into the system and W_by positive for work output. The closed-system energy balance is Δ(U + K + Φ) = Q − W_by, where U is internal energy, K is bulk kinetic energy and Φ is gravitational potential energy in a consistent external-field model. The examples below have unchanged endpoint bulk stores unless stated otherwise.

Entropy S is an extensive state property, measured in joules per kelvin, J/K. Its change ΔS = S_final − S_initial depends on the complete endpoint material states, rather than on the history used to connect them. For weakly interacting macroscopic subsystems with negligible interface contributions, entropy is additive. Specific entropy s = S/m is entropy per mass, in J/(kg K), or commonly kJ/(kg K).

The temperatures in the entropy equations are positive absolute thermodynamic temperatures in kelvin. Celsius readings cannot replace them in Q/T or temperature ratios. A heat reservoir is an idealized body whose temperature remains fixed during the stated finite transfer. A finite body generally changes temperature and needs a material relation to calculate its entropy change.

What the second law forbids

The Clausius statement forbids a cyclic device whose sole effect is to transfer heat from a colder reservoir to a hotter one. The word “sole” matters: a refrigerator uses a work input and changes its surroundings, so cold-to-hot transfer can occur as part of a larger process.

The Kelvin–Planck statement forbids a cyclic device whose sole effect is to take heat from one reservoir and convert all of it into work. It does not forbid a gas from absorbing heat and doing an equal amount of work during one expansion: that gas has changed state and has not completed the required cycle.

The two statements express the same restriction. A hypothetical single-reservoir heat-to-work converter could power an ordinary refrigerator, leaving only cold-to-hot transfer. Conversely, a hypothetical work-free cold-to-hot pump could return an ordinary engine’s rejected heat to its hot reservoir, leaving single-reservoir heat-to-work conversion. The auxiliary devices can be cycled so their working substances return and the unwanted heat exchange cancels.

A reversible process is an ideal limit in which the system and all participating surroundings can be restored without an uncompensated change. Finite-temperature-difference heat transfer, friction, viscous dissipation and unrestrained expansion are sources of irreversibility. Moving slowly can keep a material close to equilibrium, but does not by itself eliminate those sources. Reversibility is a condition on the complete process.

Compute a state change with a reversible path

For a reversible transfer to a closed system, dS = δQ_rev/T_b. Here δQ_rev is an infinitesimal heat input and T_b is the temperature at that part of the boundary. When the reversible system has one uniform temperature, T_b equals that T. The symbol dS denotes a state differential; δQ denotes a transfer, whose value generally depends on the process.

Thus ΔS = ∫ from state A to state B δQ_rev/T_b along a reversible reference path. Different reversible reference paths give the same entropy difference. A convenient reference state fixes the additive constant; it does not change ΔS. The reversible path is a way to calculate the property difference, and need not be the process that actually occurred.

For a fixed-volume body with a known heat capacity C(T), use a reversible heating or cooling path: ΔS = ∫ from T_initial to T_final C(T)dT/T. If the extensive heat capacity is constant over that interval, ΔS = C ln(T_final/T_initial). C here has units J/K, and ln is the natural logarithm of a dimensionless temperature ratio.

The heat-capacity model must match the constraint and material regime. A constant-volume C_V and a constant-pressure C_p generally differ. A constant-C logarithm must not be extended through an unmodeled phase change. For an isothermal reversible equilibrium phase change, the general relation instead gives ΔS = Q_rev/T_transition for the specified amount.

Separate entropy transfer from production

The closed-system entropy balance is ΔS = ∫ δQ/T_b + S_gen, with S_gen ≥ 0. The integral includes every boundary heat transfer over the interval, using its local boundary temperature and the sign of heat into the system. S_gen is entropy produced inside the selected accounting boundary by irreversible effects. It is not a state property or an energy source.

For separate heat contacts at constant boundary temperatures, the transfer term is Σ_k Q_k/T_b,k. A system can lose entropy if the outward entropy transfer exceeds the entropy produced inside it. Only for an isolated system, where no heat or matter carries entropy across its boundary, does the balance reduce to ΔS = S_gen ≥ 0.

T_b is not automatically the system’s average temperature or a remote heater temperature. Moving the accounting boundary across a thermal contact layer moves the layer’s entropy production into or out of the system ledger. The warming example explicitly includes a negligible-storing contact layer and places its outer heat boundary at the reservoir temperature.

Work transfers energy without a separate entropy-transfer term in this classical macroscopic balance. Stirring can therefore raise a sealed insulated system’s entropy by producing entropy inside it, even though no entropy enters by heat or matter. A zero S_gen for one selected subsystem establishes no production there; calling the complete process reversible also requires the participating contacts and surroundings to produce none.

Solve the reservoir question

For an ideal internally equilibrated reservoir at constant temperature, ΔS_res = Q_into,res/T_res. The hot reservoir has Q_hot = −240 J, so ΔS_hot = −240/400 = −0.6 J/K. The cold one has Q_cold = +240 J, so ΔS_cold = +240/300 = +0.8 J/K.

The isolated combination includes their connection. Its entropy change and entropy production are ΔS_total = S_gen,total = −0.6 + 0.8 = +0.2 J/K. Each reservoir can remain internally equilibrated while the complete transfer is irreversible: the finite temperature difference across the connection produces entropy.

Reversing only that transfer would give ΔS_total = −0.2 J/K for an isolated combination, contrary to S_gen ≥ 0. Energy conservation permits both signs of the transfer; the second law eliminates the proposed sole reverse process. No heat-transfer rate or equilibration time follows from this entropy calculation.

Warming with a finite temperature difference

A fixed-volume body has constant extensive heat capacity C = 4 J/K. It warms from 300 K to 360 K using a reservoir at 400 K. No work, mass transfer or endpoint bulk-energy change occurs. The energy balance gives Q = C(360 − 300) = 240 J. The body is approximately uniform at each assigned intermediate temperature.

Its entropy change is found from a reversible reference path, not by dividing the actual heat by 400 K: ΔS_body = 4 ln(360/300) ≈ 0.729286 J/K. Include the negligible-storing contact layer in the selected system and put the outer heat boundary at 400 K. Entropy entering across that boundary is Q/400 = 0.6 J/K, so S_gen ≈ 0.729286 − 0.6 = 0.129286 J/K.

The reservoir entropy change is −0.6 J/K, and the reservoir–contact–body combination gains the same 0.129286 J/K. This provides an equivalent isolated-combination check. If the boundary instead enclosed the approximately uniform body alone, its heat-transfer temperature would track the body surface temperature; the excluded contact layer’s production would belong outside that ledger.

At an intermediate body temperature T, the cumulative quantities are ΔS_body(T) = C ln[T/(300 K)], entropy transferred into body + contact = C[T − 300 K]/(400 K), and S_gen(T) is their difference. These compare states along the stipulated uniform-body warming model. Temperature is the horizontal coordinate; no time law is implied.

Same endpoints, different entropy production

Consider a fixed amount of a hypothetical ideal monatomic gas with nR = 1.5 J/K, U = (3/2)nRT and C_V = (3/2)nR = 2.25 J/K over the stated range. Initially T = 300 K and V = 1 L. Compare two processes reaching T = 300 K and V = 2 L. The ideal equation pV = nRT gives endpoint absolute pressures of 450 kPa and 225 kPa, using 1 kPa L = 1 J.

First expand reversibly and isothermally with a frictionless controlled piston and infinitesimal thermal driving differences. ΔU = 0 and Q_rev = W_by,rev = nRT ln(2/1) ≈ 311.916 J. Therefore ΔS_gas = Q_rev/300 = nR ln 2 ≈ 1.039721 J/K. That entropy enters by heat; entropy production is zero in the reversible limit.

Now remove a negligible-energy partition inside a rigid insulated enclosure whose other compartment is evacuated. The gas freely expands from 1 L to 2 L. Here Q = W_by = 0 and ΔU = 0. For this ideal gas model, the final temperature remains 300 K, so the final gas state is the same as in the reversible comparison. Its entropy change is consequently the same 1.039721 J/K, entirely produced inside the isolated gas/enclosure accounting system.

No uniform equilibrium trajectory is assigned during free expansion. The curve S(T = 300 K, V) compares equilibrium states and may represent the controlled reversible route; joining the free-expansion endpoints by that curve would invent intermediate equilibrium states. A real gas need not retain the same temperature at constant U.

Adiabatic is not necessarily isentropic

Isentropic means constant entropy. A closed adiabatic system has ΔS = S_gen, so it is isentropic when no entropy is produced. Adiabatic alone does not make a process reversible or isentropic. Conversely, ΔS = 0 for a nonadiabatic system can result from outward entropy transfer balancing positive production.

Stir the same initially 300 K ideal gas in a rigid insulated vessel by supplying 225 J of shaft work. Include the paddle hardware or require its endpoint stored energy to be unchanged. With contents initially and finally at rest, ΔU = 225 J, so T_final = 300 + 225/2.25 = 400 K. At fixed volume the gas entropy change is ΔS = C_V ln(400/300) ≈ 0.647285 J/K, all produced inside. There was no entropy transfer by heat.

For comparison, a reversible adiabatic expansion of the initial gas from 1 L to 2 L keeps S constant and gives T_final ≈ 188.988 K under the stated caloric model. It reaches a different endpoint from the 300 K free expansion. The advanced entropy-state relation below derives that temperature condition; the named-process subject develops process families in more detail.

A cycle cannot hide entropy production

For a complete closed-device cycle, its entropy returns to its starting value: ΔS_cycle = 0. With all heat contacts included, the balance gives ∮δQ/T_b = −S_gen,cycle ≤ 0. This is the Clausius inequality. For constant-temperature contacts it becomes Σ_k Q_k/T_b,k ≤ 0, taking each Q_k positive into the device.

Consider a cycle receiving 240 J from a 400 K reservoir and rejecting 190 J to a 300 K reservoir. Include its contact layers inside the cycle accounting boundary, with no net entropy storage in them. The heat entropy sum is 240/400 − 190/300 = −0.0333333 J/K. Thus the cycle requires S_gen = +0.0333333 J/K. The energy balance gives W_by = 240 − 190 = 50 J.

A proposed variant rejects only 150 J while receiving the same 240 J. Energy bookkeeping would give 90 J of work output, but the heat entropy sum becomes +0.1 J/K and would require S_gen = −0.1 J/K. That proposal fails the second law. Nonnegative production is a necessary test; it does not by itself specify a realizable machine or its rates.

The working device can therefore return to its initial entropy after an irreversible cycle, while its surroundings gain entropy. This section uses cycle balances to test the second law, without assigning an engine design, efficiency or temperature-versus-time path.

Further: why the reversible integral is a state difference

The Clausius inequality applies to a closed cycle. If the entire cycle is reversible, it can also be run backward. The heat-transfer signs reverse, so an integral that were negative forward would be positive backward, violating the inequality. Therefore ∮δQ_rev/T_b = 0 for a reversible cycle.

Join two equilibrium states A and B by any reversible path 1, then return by the reverse of another reversible path 2. Their combined cyclic integral is zero. Hence ∫₁ δQ_rev/T_b = ∫₂ δQ_rev/T_b: the reversible integral depends only on the endpoints and can define S_B − S_A.

Now take an actual possibly irreversible path A → B and close it with a reversible return B → A. The cyclic inequality gives ∫actual δQ/T_b − (S_B − S_A) ≤ 0. Rearranging gives ΔS ≥ ∫actual δQ/T_b and defines the nonnegative remainder S_gen. This is why a fictitious reversible calculation of ΔS does not make the actual heat transfer reversible.

All reversible paths used in this construction lie within an appropriate connected equilibrium state domain of the material. Rapid spatially nonuniform processes may require local-equilibrium fields for an intermediate entropy description. The free-expansion calculation here requires only equilibrium endpoints and does not assume one global intermediate temperature.

Further: an equilibrium entropy relation

For a fixed amount and composition of a simple compressible equilibrium material with only pressure–volume reversible work, combine dU = δQ_rev − p dV with δQ_rev = T dS to obtain dU = T dS − p dV. This is a relation among neighboring equilibrium state properties. Its validity as a state relation does not permit replacing actual irreversible heat by T dS or actual work by p dV.

For the stated ideal gas with constant extensive C_V, dU = C_V dT and p/T = nR/V. Thus dS = C_V dT/T + nR dV/V. Integrating gives ΔS = C_V ln(T_final/T_initial) + nR ln(V_final/V_initial). The logarithms use ratios of positive absolute temperatures and positive volumes. A temperature-dependent C_V would remain inside its temperature integral.

Taking T_ref = 300 K, V_ref = 1 L and a relative S_ref = 0 gives S − S_ref = (2.25 J/K) ln[T/(300 K)] + (1.5 J/K) ln[V/(1 L)]. This function is a property model of the assigned fixed amount of gas. Setting a relative reference does not determine an absolute third-law entropy or justify extrapolating the classical model to T = 0.

For reversible adiabatic expansion, set ΔS = 0: ln(T_final/T_initial) = −(nR/C_V)ln(V_final/V_initial). Therefore T_final = T_initial(V_initial/V_final)^(nR/C_V). Here nR/C_V = 2/3, so doubling the volume gives about 188.988 K. This relation is not a replacement for entropy production in irreversible adiabatic processes.

Further: equilibrium as an entropy maximum

An isolated material system can increase its entropy by internal redistribution while keeping total energy fixed. Stable equilibrium is an entropy maximum among states allowed by those energy, volume, matter and other constraints. The constraint matters: an open system can exchange entropy, and a maximum under one set of constraints need not describe another.

Consider two internally uniform bodies with constant positive heat capacities C_hot = 3 J/K and C_cold = 6 J/K, initially at 450 K and 300 K. They exchange heat inside an isolated combination and do no work. Energy conservation gives a common final temperature T_eq = (3 × 450 + 6 × 300)/(3 + 6) = 350 K. The energy transferred from hot to cold is q_eq = 3(450 − 350) = 300 J.

For constrained state pairs, let q label energy transferred from the initially hot body to the cold one. Then T_hot(q) = 450 K − q/(3 J/K) and T_cold(q) = 300 K + q/(6 J/K). Their entropy change is ΔS_total(q) = (3 J/K) ln[T_hot(q)/(450 K)] + (6 J/K) ln[T_cold(q)/(300 K)]. This curve compares state pairs at fixed total energy, not their time history.

Differentiation gives dS_total/dq = 1/T_cold − 1/T_hot. It vanishes when the temperatures are equal. The second derivative is −1/(C_hot T_hot²) − 1/(C_cold T_cold²), which is negative for these positive heat capacities. Thus q = 300 J is the unique maximum in the displayed positive-temperature domain.

At this maximum, ΔS_hot ≈ −0.753943 J/K and ΔS_cold ≈ +0.924904 J/K, giving ΔS_total = S_gen ≈ 0.170961 J/K. Extending the state-comparison curve beyond q_eq would lower the total entropy from its maximum. It does not describe continued spontaneous hot-to-cold transfer after equilibrium. The second law alone gives no relaxation time or maximum-production-rate principle.

Further: entropy with flowing matter

A fixed open control volume can receive entropy with matter as well as with heat. For uniform ports and local-equilibrium boundary heat transfer, dS_CV/dt = Σ_in ṁ s − Σ_out ṁ s + Σ_k Q̇_k/T_b,k + Ṡ_gen. Here S_CV is stored entropy, ṁ is mass flow rate, s is the port’s specific entropy, Q̇ is heat-transfer rate into the volume and Ṡ_gen ≥ 0 is entropy-production rate inside it.

At steady state the stored entropy does not accumulate, so dS_CV/dt = 0. This does not require each transfer or production term to vanish. The numerical model uses one unchanged material and thermal heat transfer; it does not specify species-diffusion, chemical-reaction or nonthermal-radiation physics. Ideal macroscopic work has no separate entropy flux in this equation.

A steady device carries 0.02 kg/s through one inlet and one outlet. Their common-reference specific entropies are 1.00 and 1.12 kJ/(kg K). A 0.6 kW heat loss crosses the chosen boundary at 300 K. Mass carries entropy inward at 20 W/K and outward at 22.4 W/K. The heat entropy term is −600/300 = −2 W/K.

The required entropy-production rate is Ṡ_gen = 22.4 − 20 − (−2) = 4.4 W/K. Equivalently, 20 + 4.4 = 22.4 + 2 W/K. A common additive shift of the inlet and outlet specific-entropy reference cancels for equal steady mass flow of the same material. These entropy data do not determine enthalpy or work power; additional first-law data would be needed.

Use the law with its conditions

For a new problem, name the boundary and endpoint states, identify matter and heat transfers, and retain the temperature associated with each heat crossing. Calculate the state difference using valid material properties or a reversible reference path. Then solve the entropy balance for production and check its sign. Use the energy balance independently; neither balance replaces the other.

Entropy does not measure visible untidiness, and a system’s positive ΔS is not automatically its S_gen. Statistical mechanics explains entropy through microscopic state descriptions, while the classical definitions and balances here already allow macroscopic calculations. The current examples assume macroscopic equilibrium endpoints, ordinary positive temperatures and the stated constitutive models.

The neighboring “Heat capacities and calorimetry” article develops heat-capacity and phase-change measurements. “Engines, refrigerators, and heat pumps” develops performance limits and cycle designs. “Microstates, macrostates, and multiplicity” and “Probability and statistical entropy” provide the statistical interpretation; “Irreversible thermodynamics and heat and mass transfer” extends local production and transport. These follow-on topics are map links until their actual articles become available.

Worked examples

Try it yourself

1. Reverse only the 240 J reservoir transfer, with no external work, matter or other change. What total entropy production would be required?

   Show the answer and reasoning

   −0.2 J/K; the proposed sole reverse process is impossible.

   The reservoir entropy changes reverse signs. The isolated total would decrease, requiring negative production even though energy still balances.

2. A closed system has ΔS = −0.3 J/K while Q = −120 J crosses its boundary at 300 K. Find S_gen. Is the entropy decrease alone a second-law violation?

   Show the answer and reasoning

   S_gen = +0.1 J/K; the entropy decrease alone is not a violation.

   Entropy transfer is −120/300 = −0.4 J/K. Thus S_gen = −0.3 − (−0.4) = +0.1 J/K. The system exports more entropy than it produces.

3. A 4 J/K fixed-volume body cools from 360 K to 300 K using a 300 K reservoir. Include a negligible-storing contact layer inside the body’s accounting boundary. Find Q, ΔS and S_gen.

   Show the answer and reasoning

   Q = −240 J; ΔS ≈ -0.7292862 J/K; S_gen ≈ 0.07071377 J/K.

   Use Q = C(T_final − T_initial) and ΔS = C ln(T_final/T_initial). Entropy transfer is Q/300 = −0.8 J/K, so production is the state change minus that transfer. A negative system entropy change is compatible with positive production.

4. A proposed closed adiabatic process has ΔS = −0.2 J/K and no matter transfer. Is it compatible with the stated classical entropy balance?

   Show the answer and reasoning

   No. It would require S_gen = −0.2 J/K.

   With zero heat and matter entropy transfer, ΔS equals the nonnegative production. Work input or output does not provide an omitted entropy-transfer term.

5. Does ΔS = 0 by itself prove that a process is adiabatic and reversible?

   Show the answer and reasoning

   No.

   A nonadiabatic system can export entropy while producing an equal amount inside. To infer zero production from constant entropy, the relevant heat and matter transfers must also be zero. Global reversibility additionally concerns the complete participating surroundings.

6. Reversibly compress the assigned gas at 300 K from 2 L to 1 L. Find ΔS, Q and W_by.

   Show the answer and reasoning

   ΔS ≈ -1.039721 J/K; Q = W_by ≈ -311.9162 J; S_gen = 0.

   Reversing the reversible isothermal path reverses both transfer signs and the entropy change. The ideal gas has ΔU = 0; its entropy loss is exported by heat.

7. Add 5 J/K to every entropy value for the fixed gas amount. What happens to its isothermal expansion ΔS and the inferred production?

   Show the answer and reasoning

   The endpoint values become 5 and approximately 6.039721 J/K. Their difference stays 1.039721 J/K; production does not change.

   The same additive constant cancels from the state difference. A reference change does not transfer or produce entropy.

8. The finite-body comparison curve includes q = 450 J, where the initially hot body is at 300 K and the other at 375 K. Does that point describe further spontaneous forward heat transfer after the 350 K equilibrium?

   Show the answer and reasoning

   No.

   That constrained state pair is allowed as a comparison, but lies below the entropy maximum at q = 300 J. Continuing from the maximum toward it would reduce the isolated total entropy. The q axis is not elapsed time.

9. A cycle absorbs 240 J at 400 K and rejects 150 J at 300 K, with no other exchanges. Energy balance gives 90 J work output. What additional test rejects it?

   Show the answer and reasoning

   Σ Q/T = +0.1 J/K, requiring S_gen = −0.1 J/K for a cycle.

   A cyclic state has ΔS = 0, so production equals the negative heat-entropy sum. The proposed heat exchanges violate nonnegative production despite satisfying an energy ledger.

10. Why is the steady device’s heat entropy term −2 W/K rather than −2 W, and what does zero entropy accumulation mean?

    Show the answer and reasoning

    Dividing −600 W by 300 K gives −2 W/K. Zero accumulation means all entropy flux and production terms balance, not that each is zero.

    Specific entropy times kg/s also has units W/K after converting kJ to J. The steady ledger has inward mass entropy plus production balancing outward mass and heat entropy.

What to carry forward

• Entropy is a state property; heat entropy transfer and entropy production describe a process and its chosen boundary.
• The entropy of an isolated system cannot decrease. A nonisolated system can lose entropy by exporting it.
• Use the actual local heat-crossing temperature in an entropy-transfer term, and account for finite contact layers consistently.
• A reversible reference path computes ΔS; it does not turn an irreversible actual process into a reversible one.
• Adiabatic does not mean isentropic, and a cyclic device can produce entropy while returning to its initial entropy.
• Entropy maximization requires explicit constraints; entropy balance alone supplies no rate or complete machine design.