Irreversible thermodynamics and heat and mass transfer

A wall that keeps passing heat

A 10 mm layer and a 20 mm layer separate reservoirs held at 330 K and 300 K. Their shared area is 0.010 m². A real contact between the layers resists heat transfer, and heat must also cross the two outer boundary regions. After the wall reaches a steady state, its stored energy no longer changes, yet about 1.03448 J crosses it every second. Where is entropy being produced? Why can the two faces of one contact have different temperatures?

We will answer those questions, then ask the corresponding question about matter: what drives two species through a mixture in opposite directions, even when their net molar current is zero? The same entropy balance will lead from an ordinary wall to coupled heat and mass transfer.

For a first reading, follow the local terms, wall calculation, counterdiffusion example, and cooling comparison. The sections marked “Further” develop the current conventions, matrix restrictions and reciprocal relations. They can be read on a second pass without losing the physical examples.

Published preparation is “Physical systems, states, boundaries, and environments”, “Pressure, buoyancy, and fluid statics”, and “Conservation of mass, momentum, and energy”. Heat flux, entropy production and chemical potential are introduced here. The neighboring entropy, thermodynamic-potential, stability and phase-equilibrium lessons provide fuller return paths; they are not assumed live prerequisites. All material and transfer coefficients in our numerical examples are assigned illustrative values, rather than measurements of a particular wall or membrane.

Local terms and a direction to measure

Temperature T is an absolute temperature in kelvin. Heat transfer rate Q̇ measures energy crossing per second, in watts. Heat flux q measures the rate per area, in W/m²; in a uniform one-dimensional wall Q̇ = A q. We choose +x from the hot reservoir toward the cold one, so positive q means heat travels in that direction.

Entropy measures a thermodynamic state property. Entropy can cross a boundary, and irreversible processes produce it. The production rate Ṡ_gen of a whole region has units W/K. Its volume density σ has units W/(m³ K); a contact production per area has units W/(m² K). A negative signed entropy flow is possible. Negative total entropy production is not allowed in the passive local-equilibrium models considered here.

A flux is signed relative to a direction or oriented area. A gradient describes how a field changes with position. If temperature decreases with x, dT/dx is negative, while d(1/T)/dx is positive. Calling a thermodynamic “force” a gradient does not mean it has units of newtons: its units are chosen so that its product with the conjugate flux is entropy production.

A balance comes before a transfer law

For a control volume, accumulation equals inflow minus outflow plus production. Energy is conserved. Entropy satisfies dS_CV/dt = entropy inflow − entropy outflow + Ṡ_gen, with Ṡ_gen ≥ 0 for the complete selected region. An entropy transfer Q̇/T uses the temperature at the particular thermal boundary crossed, not an unrelated bath temperature.

An open fluid control volume also carries entropy with matter: a stream contributes its molar flow times molar entropy, or mass flow times specific entropy. Its energy balance includes stream enthalpy, and kinetic and potential energy when relevant, plus heat and shaft or electrical work. Flow work is already inside enthalpy; adding it again double counts it. Do not substitute the closed-wall balance for an open exchanger or membrane feed stream.

Steady means that the macroscopic fields and stored amounts are time independent. It does not mean equilibrium: reservoirs can maintain gradients and currents indefinitely. In the wall, the entropy carried out exceeds the entropy carried in by exactly the production inside the solid and contact. The unchanged stored wall entropy does not imply zero production.

Local equilibrium and the continuum window

Local equilibrium assumes that a small cell can be assigned thermodynamic variables such as T, energy density u and chemical potentials μᵢ, and a local equilibrium equation of state. The cell must contain enough microscopic degrees of freedom and relax internally fast enough, while remaining small compared with the length over which the macroscopic fields vary. Thus different neighboring cells can have different temperatures even though each cell has a useful local state.

Local equilibrium is weaker than global equilibrium and does not by itself require linear response. Conversely, a straight-looking temperature profile does not prove local equilibrium. Large gradients can destroy a local description if the variation length approaches a mean free path, an interface thickness or a relevant microscopic relaxation scale. A zero-thickness contact in our model has separate face states and a jump law; it is not a bulk cell with an infinite ordinary gradient.

Conservation balances are more general than Fourier, Fick or a constant boundary coefficient. A constitutive law supplies the mechanism and rate. We must check the scale, frame, material branch, boundary conditions and driving range before choosing it.

Fourier conduction and the composite wall

For a stationary isotropic solid with scalar conductivity k > 0, Fourier’s law is q = −k dT/dx. Combining it with the entropy force gives σ = q d(1/T)/dx = k(dT/dx)²/T² ≥ 0. A temperature gradient produces heat flow; the inverse-temperature gradient is its entropy-conjugate force. The two statements agree because d(1/T)/dx = −(dT/dx)/T².

Our layers have k₁ = 0.50 W/(m K), L₁ = 0.010 m and k₂ = 0.20 W/(m K), L₂ = 0.020 m. The contact has areal resistance R″c = 0.020 m² K/W. The hot and cold boundary exchange coefficients are hhot = 20 and hcold = 10 W/(m² K). There is no internal source, radiation, bulk motion or storage. All these coefficients are constant assigned values in this case.

Per unit area, the resistances add: R″tot = 1/hhot + L₁/k₁ + R″c + L₂/k₂ + 1/hcold = 0.290 m² K/W. Thus q = (330−300)/R″tot = 103.4483 W/m² and Q̇ = 1.03448 W. Each temperature drop is qR″. The lower-conductivity second layer has a steeper temperature slope; equal heat flux does not imply equal gradient.

The convective relations q = hhot(Th−Ts,hot) and q = hcold(Ts,cold−Tc) connect bulk reservoir and solid face temperatures. They aggregate boundary exchange. No physical film thickness is given, so we do not draw invented linear spatial profiles through those exchange regions.

Where the wall produces entropy

In any steady pure-heat segment with no storage, Ṡ_gen = Q̇(1/Tout−1/Tin). In a solid layer this is the volume integral of k|∇T|²/T². The contact instead produces entropy per area q(1/Tcold face−1/Thot face). Both contact faces are at x = 0.010 m, with temperatures 322.758621 K and 320.689655 K. Their finite temperature jump and finite resistance are part of the boundary model.

Choose the two solids plus the contact as the control volume. Heat enters at 324.827586 K and leaves at 310.344828 K. Its entropy inflow is 0.003184713 W/K, outflow is 0.003333333 W/K, and production is their difference, 0.000148620 W/K. Using 330 K and 300 K at these solid boundaries would include two external exchange regions that are outside this control volume.

For the complete path between the reservoirs, the five contributions add to 0.000313480 W/K = Q̇(1/300−1/330). The hot reservoir loses entropy at Q̇/330, and the cold one gains it at Q̇/300. The wall has no entropy accumulation. The increased entropy of the reservoir pair equals the complete intervening production.

Chemical potential and the direction of mass transfer

In an ordinary equilibrium homogeneous mixture, μᵢ = (∂G/∂nᵢ) at fixed T, p and every other species amount, with any additional work coordinates also held fixed. Here nᵢ is the amount in mol and G = U + pV − TS is Gibbs energy, with U internal energy, V volume and S entropy. Chemical potential is therefore a constrained partial molar property, in J/mol. It accounts for composition, temperature, pressure and interactions. A concentration gradient is therefore not the most general driving force for matter.

For a stable ideal binary mixture at fixed T, μ₁−μ₂ = reference term + RT ln[x₁/(1−x₁)], where x₁ is the mole fraction of species 1 and R is the molar gas constant. For the assigned stationary mixture the reference contribution is constant along the sample: there is no additional pressure or stress driving. The logarithm is defined for 0 < x₁ < 1. Increasing x₁ raises this chemical-potential difference. In a nonideal mixture the activities replace mole fractions, and the thermodynamic factor can change the concentration-to-force relation.

Pressure differences, electrical potential differences, stress and thermal coupling can also matter. At fixed temperature an ion has electrochemical potential μ̃ᵢ = μᵢ + zᵢF_Aφ, with valence zᵢ, electrical potential φ and Faraday constant F_A. A model using electrochemical driving must count electrical energy consistently. A measured concentration slope alone does not determine a universal current direction.

Two opposite species currents

Take a rigid stationary ideal binary mixture at 300 K, total concentration c = 1000 mol/m³, length L = 1.00 mm and assigned constant D = 1.00×10⁻⁹ m²/s. Species reservoirs maintain x₁ = 0.40 on the left and 0.60 on the right. We assume equal-site, constant-total-concentration equimolar counterdiffusion, J₁ = −J₂, with no bulk advection, electric driving or reaction. This is a specified model, not a rule for every mixture.

The steady composition profile is linear. Fick’s law gives J₁ = −cD dx₁/dz = −2.00×10⁻⁴ mol/(m² s), while J₂ = +2.00×10⁻⁴ mol/(m² s). Species 1 goes left and species 2 right. Zero net molar current leaves two nonzero species currents; it does not eliminate transport or dissipation.

Write Δμ = μ₁−μ₂ and Xc = −d(Δμ/T)/dz = −R[x₁(1−x₁)]⁻¹ dx₁/dz. Both J₁ and Xc are negative, so σmatter = J₁Xc is positive. At the midpoint it is 1.330314 W/(m³ K); at either end it is 1.385744 W/(m³ K). The flux is constant while the force and production vary with composition.

Equivalently, J₁ = Lchem Xc with Lchem = cD x₁(1−x₁)/R > 0. Constant D does not imply constant Lchem. Integrating gives production per area −J₁[Δμ(right)−Δμ(left)]/T = 0.001348490 W/(m² K), where the chemical-potential difference increases by 2022.734691 J/mol. Because the mixture is isothermal, the entropy term involving the energy current is zero; that does not require the transported energy itself to vanish.

Further: derive the entropy-conjugate currents

To avoid hiding species-carried energy inside an undefined “heat flux”, first choose a rigid continuum with no bulk flow, reaction or mechanical work. Let u and s be energy and entropy per volume and cᵢ the molar densities. Local equilibrium gives ds = du/T − Σᵢ(μᵢ/T) dcᵢ. Use the full nonadvective energy current Jᵤ and species currents Jᵢ in the conservation laws ∂tu = −∇·Jᵤ and ∂tcᵢ = −∇·Jᵢ.

The product rule then gives ∂ts + ∇·Jₛ = σ, with Jₛ = [Jᵤ − Σᵢ μᵢJᵢ]/T and σ = Jᵤ·∇(1/T) − ΣᵢJᵢ·∇(μᵢ/T). The energy current includes energy carried by diffusing species. Fix consistent species-energy and chemical-potential reference levels throughout; changing them transforms the paired currents and forces together. With J₁ = −J₂ = J in one dimension, σ = JᵤXᵤ + JXc, Xᵤ = d(1/T)/dz and Xc = −d[(μ₁−μ₂)/T]/dz. These are the pairs used in our coupled example.

Other conventions are useful, but their forces must change with their currents. For example, define a reduced current Jq′ = Jᵤ − aJ. Then σ = Jq′Xᵤ + J(Xc + aXᵤ), an algebraic identity at each local state. If a is the appropriate partial-molar-enthalpy difference for a particular mixture/frame, this is the familiar sensible-heat convention with a correspondingly transformed chemical force. Calling Jᵤ “conductive heat” while keeping Xc unchanged loses that identity.

The entropy flux Jₛ is not generally conserved: its divergence supplies the local production in steady state. Total energy and conserved species currents may be constant through a one-dimensional steady membrane while reduced heat currents change with local partial molar enthalpies. This is why a membrane interface needs explicit current definitions as well as face states.

Further: linear response and honest units

Near a stable local equilibrium state, a passive material may admit Jα = ΣβLαβXβ to first order in small driving forces. The coefficients depend on the reference state and frame. Cross coefficients allow thermal driving to induce species transfer and chemical driving to induce energy transfer. The cross coefficients used to describe Soret and Dufour effects depend on the chosen current–force convention. A measured temperature or composition response is unchanged by a consistent change of convention. Our illustrative matrix is for full energy current and is not a fitted coefficient set for either effect.

Choose Xᵤ* = 10⁻⁴ /(K m), Xc* = 1 J/(mol K m) and σ* = 10⁻³ W/(m³ K). Define f₁ = Xᵤ/Xᵤ*, f₂ = Xc/Xc*, j₁ = Jᵤ/(10 W/m²), and j₂ = J/(10⁻³ mol/(m² s)). Then σ/σ* = j₁f₁ + j₂f₂. Scaling conjugate currents and forces together preserves this entropy product and the reciprocity relation.

Our assigned dimensionless relation is (j₁,j₂) = ℓ(f₁,f₂), with ℓ = [[2,0.60],[0.60,1]]. For f = (1,−0.20), j = (1.88,0.40). Thus Jᵤ = 18.8 W/m², J = 4.00×10⁻⁴ mol/(m² s), and σ = 0.001800 W/(m³ K). The energy channel contributes +0.001880 and the chemical channel −0.000080 W/(m³ K). One current is driven against its own conjugate force by the coupled channel; the total still satisfies the second law.

This local linear illustration uses small perturbations. Over 1 mm, Xᵤ corresponds to a temperature decrease of about T²XᵤL = 0.009 K near 300 K; |Xc|L = 0.0002 J/(mol K) at the chosen force. These estimates describe driving scales, not a solved spatial boundary-value problem. The response-family figure varies f₂ at f₁ = 1; it is not a time trajectory.

Further: positivity, coupling and stability

Only the symmetric part Ls = (L+Lᵀ)/2 contributes to XᵀLX, because XᵀLaX = 0 for the antisymmetric part La. The second law therefore requires Ls to be positive semidefinite on the allowed force space. For two unrestricted scalar channels, L₁₁ ≥ 0, L₂₂ ≥ 0 and L₁₁L₂₂ − [(L₁₂+L₂₁)/2]² ≥ 0. Strict inequalities give positive definiteness; a zero eigenvalue permits a nondissipative direction and needs separate interpretation.

For the symmetric ℓ here, det ℓ = 1.64 and its eigenvalues are about 0.718975 and 2.281025, so every nonzero force produces positive entropy. Along f₁ = 1, σ/σ* = 2 + 1.2f₂ + f₂². The chemical product f₂(0.6+f₂) is negative only for −0.6 < f₂ < 0, while the total stays positive.

Positive diagonal entries alone are insufficient. The trial matrix [[1,1.2],[1.2,1]] has det = −0.44. Applied to f = (1,−1), it gives j = (−0.2,0.2) and f·j = −0.4. It cannot be a passive linear constitutive law on that full force domain.

Dissipation positivity and thermodynamic stability are different tests. The local equation of state must also describe a stable branch with the appropriate susceptibilities or free-energy curvature; the full evolution and boundary conditions matter for dynamic stability. A positive mobility multiplying a negative thermodynamic factor can signal growth of a concentration fluctuation on a phase-unstable branch, where a gradient-energy/phase-separation model is needed. The neighboring stability and phase-equilibrium subjects develop that mechanism.

Further: what Onsager and Casimir add

At equilibrium, microscopic reversibility can imply reciprocal linear transport coefficients for correctly paired variables. The second law alone does not imply L₁₂ = L₂₁: it restricts the symmetric quadratic form but leaves an antisymmetric part untested. Onsager reciprocity adds microscopic information to thermodynamic positivity.

For variables aα with time-reversal parity εα = ±1 and entropy-conjugate transport laws, the usual Onsager–Casimir relation is Lαβ(B) = εαεβLβα(−B), also reversing any other time-odd bias such as imposed rotation. The parities refer to the underlying variables; using their time derivatives as currents changes both parities together and leaves their product unchanged. One must identify the actual conjugate variables, reference equilibrium and microscopic reversibility, rather than apply a symmetric-matrix slogan to arbitrary fitted coefficients.

Our conserved energy and species densities are both time even, so their parity product is +1. With no time-odd bias, the illustrative coefficients can be symmetric. To show the field-reversal distinction algebraically, let ℓ(b) = [[2,0.60+0.40b],[0.60−0.40b,1]], with b = −1,0,+1 reversed under time reversal. Then ℓ₁₂(b) = ℓ₂₁(−b), although the matrix at b ≠ 0 is not symmetric.

The symmetric part of this family remains [[2,0.6],[0.6,1]]. Its antisymmetric part changes the currents but cancels from f·j. At f = (1,−0.2), the scaled production is 1.8 for all three b values. This is an assigned algebraic illustration, not a measured magnetic-field law or a claim that every driven, active or far-from-equilibrium system obeys the same reciprocity.

A separate experiment: a slab cools to one bath

The steady wall connects two maintained reservoirs. Now take a different homogeneous slab, initially 330 K, cooling to one 300 K bath through both broad faces. Each face area is 0.010 m², full thickness d = 0.010 m, density 1000 kg/m³, specific heat 1000 J/(kg K), conductivity 0.50 W/(m K), and each boundary has h = 5 W/(m² K). There are no edge losses or sources.

The thermal diffusivity is α = k/(ρc) = 5.00×10⁻⁷ m²/s. The characteristic length is Lc = V/Aexposed = d/2 = 0.005 m, since both faces exchange heat. Bi = hLc/k = 0.050 indicates relatively weak internal temperature differences, supporting a lumped-temperature approximation. It does not make the slab exactly uniform at finite h.

In that approximation C dT/dt = −hAexposed(T−T₀), where C = ρcV = 100 J/K. The solution is θ = (T−T₀)/(Ti−T₀) = exp(−t/τ), with τ = C/(hAexposed) = 1000 s. The internal diffusion scale Lc²/α = 50 s is shorter, but is not an exact eigenmode relaxation time. A spatial conduction problem with the actual boundary conditions supplies the correction when the lumped approximation is inadequate.

Initially the slab loses 3.00 W; after 1000 s its approximate temperature is 311.036383 K. As t tends to infinity it approaches the bath and the current tends to zero. This final equilibrium differs from the wall’s driven steady state.

Entropy during finite-rate cooling

Within the lumped model, the outgoing heat rate per planform area is qout,A = 2h(T−T₀). The slab entropy rate is −qout,A/T per area, while the bath gains qout,A/T₀. Their sum is σA = qout,A(1/T₀−1/T) ≥ 0. This is production for slab plus exchange to the bath per planform area; it is not a claimed local volume production inside an exactly uniform slab.

The heat capacity per planform area is C_A = ρcd = 10000 J/(m² K). Integrating gives Sgen/A = C_A[ln(T/Ti)−(T−Ti)/T₀]. For complete cooling, the bath receives 3000 J and the total entropy increases by 0.468982 J/K. The slab’s own entropy decreases by C ln(300/330), while the bath’s increases by 3000/300 = 10 J/K; their sum is positive.

The approximation neglects the small internal gradients and their production. A resolved slab calculation divides production between the interior and boundary exchange using the actual surface temperature. A reversible comparison would require a succession of baths differing infinitesimally in temperature, not this finite-rate cooling to one fixed bath.

Boundary conditions and transport regimes

A transport equation needs boundary and initial data. A prescribed face temperature, a prescribed normal flux, and a Robin exchange law qnormal = h(Tsurface−Tbulk) are different physical controls. An insulated boundary has zero normal heat flux; it need not force every tangential gradient to vanish. At a finite contact, conservation fixes the transmitted steady heat flux and a jump law relates the two face temperatures.

The coefficient h summarizes the boundary flow and geometry. It depends on the flow regime, fluid properties and length scale, and can change with temperature or position. A turbulent heat exchanger needs suitable correlations or a resolved flow model. Radiation adds its own nonlinear boundary transfer; phase change adds latent energy and moving-interface conditions. Neither can be absorbed into our constant h without an explicitly justified approximation.

A continuum Fourier or Fick model can fail in rarefied gases, ballistic nanoscale transport, ultrafast heating or strong nonequilibrium. Use mean-free-path and relaxation scales, not specimen size alone, to assess the local approximation. Anisotropic conduction requires a tensor conductivity: its symmetric dissipative part must be positive semidefinite in ∇T·k∇T/T²; a scalar k cannot describe every directional response.

Further: flow, reaction and applications

Uniform translation does not by itself create an intrinsic entropy-production term. In a fluid, advection transports energy, species and entropy through the control volume. Velocity gradients can produce viscous dissipation. In a consistent fluid frame, the relevant term is τ:∇v/T for the viscous stress τ, with dissipative constitutive restrictions. Reaction adds Σr A_r r_r/T, where r_r is the extent rate per volume and A_r = −Σᵢνᵢr μᵢ for stoichiometric coefficients positive for products. Each added term needs its own units, forces and balance convention.

Membrane separation, drying and thermal diffusion illustrate why energy and mass transfer often need to be solved together. A thermal gradient can sustain a species current against its chemical driving; the heat source and cold sink then supply the overall entropy budget. Electrical or chemical work can likewise support uphill transfer. An individual negative channel product does not describe an isolated spontaneous reversal of the total second law.

The opposite streams in a counterflow heat exchanger require local balances along both flow directions. At every passive heat-transfer element, heat still goes from the locally hotter stream to the locally colder one. A global outlet-temperature comparison does not replace those local temperatures. The two-species counterdiffusion example here has no bulk streams and should not be used as the exchanger’s flow model.

These applications connect back to entropy and useful-work loss and to potentials under specified reservoir constraints. The fluid conservation and viscous-flow subjects supply momentum and transport mechanisms; “Microfluidic, reacting, biological, and geophysical flows” develops applications involving advection and diffusion. Chemical potential and phase equilibrium supplies activities and phase constraints. Microscopic statistical subjects explain fluctuations and transport coefficients. Those links deepen this lesson without changing its local definitions.

What the results do and do not imply

A stationary state can continuously produce entropy; a constant heat flux can coexist with a nonconstant entropy-production density; zero net molar flux can hide nonzero counterdiffusion; and positive diagonal coefficients can coexist with an inadmissible coupled matrix. Each conclusion follows from a balance or a complete quadratic form, rather than from the sign of one plotted arrow.

There is no unrestricted rule that a driven system chooses the state of minimum entropy production. Restricted near-equilibrium theorems require stated linear coefficients, boundary constraints and an allowed class of variations. Far-from-equilibrium pattern selection and dynamic stability require the governing equations and their perturbations. Minimizing an unqualified entropy-production formula is not a substitute for solving the transport problem.

A reliable calculation therefore names the system and frame, defines the currents, writes the balances, identifies the conjugate forces, chooses constitutive laws in their regime, imposes the actual boundary data, and then checks total entropy production. The wall and mixture examples show that these checks remain useful even when the final arithmetic is simple.

Worked examples

Try it yourself

1. If the wall area doubles while all areal resistances and reservoir temperatures stay fixed, what happens to q and Q̇?

   Show the answer and reasoning

   q stays 103.448276 W/m²; Q̇ doubles to 2.068966 W.

   The areal resistance law fixes flux. Multiplying that flux by area fixes total rate; this assumes the same one-dimensional boundary coefficients and negligible edge effects.

2. The wall has dU/dt = 0. Must its entropy production vanish?

   Show the answer and reasoning

   No. Its steady maintained heat current produces entropy continuously.

   Energy inflow and outflow balance, while entropy outflow exceeds inflow. There is no need for entropy storage in a steady dissipative state.

3. Compare the magnitude of the steady temperature slopes in the two layers.

   Show the answer and reasoning

   Layer 1: 206.896552 K/m; layer 2: 517.241379 K/m.

   The common flux obeys |dT/dx| = q/k. The smaller conductivity requires the larger gradient; the thickness sets the resulting total drop.

4. What happens to the contact jump and its entropy production as R″c tends to zero, with the other resistances fixed?

   Show the answer and reasoning

   Both tend to zero, while the full heat flux increases to 30/0.270 = 111.111111 W/m².

   The contact drop is qR″c and the two temperatures coincide in the perfect-contact limit. Production elsewhere remains because the reservoirs still differ in temperature.

5. Which temperatures belong in the steady entropy balance for only the two solid layers and contact?

   Show the answer and reasoning

   324.827586 K at entry and 310.344828 K at exit.

   These are the actual outer solid faces. The reservoir temperatures include external exchange regions and therefore apply to a larger selected path.

6. In a constant-k layer, can σ vary even though q and dT/dx are constant?

   Show the answer and reasoning

   Yes: σ = q²/(kT²), so it is larger at the colder end.

   The inverse-temperature gradient varies with T even for a linear temperature field. Entropy is not conserved like the steady energy current.

7. Does J₁+J₂ = 0 imply that neither species crosses the mixture?

   Show the answer and reasoning

   No. Here J₁ = −0.0002 and J₂ = +0.0002 mol/(m² s).

   The signed sum cancels. Each current remains nonzero and their combined chemical entropy production is positive.

8. Why is the chemical mobility smaller at x₁ = 0.40 than at x₁ = 0.50 if D is constant?

   Show the answer and reasoning

   Lchem is proportional to x₁(1−x₁): the factors are 0.24 and 0.25.

   The ideal-mixture chemical susceptibility is composition dependent. Fick diffusivity and entropy-conjugate mobility are related coefficients, not the same constant.

9. At fixed uniform T, does the vanishing Jᵤ∇(1/T) term imply Jᵤ = 0?

   Show the answer and reasoning

   No. The force vanishes, so the product vanishes for any finite energy current.

   Diffusing species can carry energy. The isothermal entropy calculation does not erase that energy from the conservation law.

10. If Jq′ = Jᵤ − aJ, what chemical force keeps the same entropy production?

    Show the answer and reasoning

    Xc′ = Xc + aXᵤ.

    Substitution gives JᵤXᵤ+JXc = Jq′Xᵤ+J(Xc+aXᵤ). Changing only the current would change the entropy budget.

11. Along f₁ = 1, for which f₂ does the assigned chemical channel have a negative entropy product?

    Show the answer and reasoning

    −0.60 < f₂ < 0.

    Its product is f₂(0.60+f₂). The current and its own force have opposite signs only between the two roots, while total production remains positive.

12. For a symmetric matrix [[2,a],[a,1]], what restriction follows from entropy positivity for all forces?

    Show the answer and reasoning

    |a| ≤ √2.

    The diagonal entries are positive and the determinant condition is 2−a² ≥ 0. Strict inequality gives positive definiteness; equality leaves a zero-production direction.

13. Can the second law determine the sign of an antisymmetric off-diagonal part by itself?

    Show the answer and reasoning

    No. It cancels from XᵀLX.

    The second law tests the symmetric part. Microscopic symmetry and other physical information can restrict the antisymmetric response, but positivity alone cannot.

14. For energy and species densities, why is the parity product in Onsager–Casimir reciprocity +1?

    Show the answer and reasoning

    Both underlying densities are unchanged by microscopic time reversal.

    Their transport currents are both time odd, giving the same product of parities. Time-odd biases must also be reversed; an arbitrary matrix at fixed bias need not be symmetric.

15. Does a positive mobility alone prove a concentration field is dynamically stable?

    Show the answer and reasoning

    No. The thermodynamic susceptibility, gradient terms and boundary conditions also matter.

    A phase-unstable free-energy curvature can turn down-chemical-potential motion into growing composition fluctuations. Dissipation positivity and phase stability test different parts of the model.

16. Why use d/2 rather than d for Lc in the lumped slab’s Biot number?

    Show the answer and reasoning

    Both broad faces exchange heat, so V/Aexposed = Ad/(2A) = d/2.

    The characteristic length depends on the exposed geometry. Using one face would represent a different boundary-value problem and change the cooling time.

17. At what approximate time does the slab lose half its initial temperature excess over the bath?

    Show the answer and reasoning

    t = τ ln 2 = 693.147181 s.

    The lumped model gives θ = exp(−t/τ); setting θ = 1/2 yields this result. It is a half-excess time, not the time to reach half the absolute temperature.

18. Can the cooling slab’s own entropy decrease while total entropy increases?

    Show the answer and reasoning

    Yes. Its decrease is 100 ln(300/330) = −9.531018 J/K, while the bath gains 10 J/K.

    The slab is not isolated. Heat exports entropy to the bath; their complete entropy increase is +0.468982 J/K.

19. What does an insulated face impose locally?

    Show the answer and reasoning

    Zero normal heat flux across that face.

    For scalar Fourier conduction this imposes a zero normal temperature derivative there. It does not require a uniform temperature along the face or throughout the solid.

20. May an arbitrary far-from-equilibrium steady state be selected by minimizing σ?

    Show the answer and reasoning

    No. No unrestricted minimum-production principle applies.

    A valid variational theorem needs its own linear-response, coefficient, boundary and variation constraints. Conservation and constitutive evolution still determine the physical state.

What to carry forward

• A maintained steady current can produce entropy without accumulating energy or entropy in the selected material.
• Entropy production is a sum of correctly conjugate flux–force products; total production must be nonnegative, while an individual coupled contribution may be negative.
• A contact jump, an external exchange region and a solid gradient belong to different pieces of the model and use their own actual boundary states.
• Chemical or electrochemical potentials give the driving for matter; concentration-only diffusion laws need composition, frame and regime assumptions.
• The symmetric transport matrix must be positive semidefinite. Reciprocity additionally needs microscopic reversibility, conjugate variables and the proper time-reversal conditions.
• A useful transport prediction needs conservation, constitutive laws, initial/boundary data and scale checks; positivity alone supplies neither a rate nor a universal stability or minimum-production principle.