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Write $\beta$ for the thermal expansion coefficient, $\nu$ for kinematic viscosity, $\kappa$ for thermal diffusivity, and $k_T=\rho c_p\kappa$ for thermal conductivity. For a temperature difference $\Delta >0$ across depth $h$,The four-thirds convective heat-transfer law posits $\mathrm{Nu}=C{\mathrm{Ra}}^{1/3}$ at high Rayleigh number, and thereforeThe layer depth cancels. A useful physical justification is a mixed interior with thin conductive boundary layers of thickness $\delta$, each maintained near a fixed critical local Rayleigh number: $g\beta {\Delta }_b{\delta }^3/(\nu \kappa )\sim {\mathrm{Ra}}_{\mathrm{crit}}$ and $q\sim k_T{\Delta }_b/\delta$. Eliminating $\delta$ gives the exponent $4/3$. Whether $\Delta$ denotes the whole-layer contrast or the drop across one boundary layer changes the coefficient, not the exponent.

Assume Boussinesq approximation, constant material properties, a laterally extensive statistically turbulent Rayleigh-Bénard convection layer, negligible side-wall heat transfer, no rotation or imposed shear, approximately uniform bulk temperature, and the same specified Prandtl number regime when choosing $C$. The closure is empirical and regime dependent; high Rayleigh number alone does not make it exact or extend it to the impulsive conductive initial transient.

A linear, or “three-thirds”, boundary heat transfer closure needs its own dimensional conductance $L>0$, so $q=L\Delta$. One cannot simply keep the dimensions of the previous nonlinear coefficient $K$. With equal conductances at both boundaries, the mean-temperature equation becomesIntegrating this linear ordinary differential equation givesThe relaxation time is $\rho c_{p}h/(2L)$. If the conductances differed, conservation of energy would instead select their weighted boundary temperature; the arithmetic mean follows from the assumed upper/lower symmetry.

Measure $z$ downward from the iceberg surface. Its mass density is ${\rho }_i(z)=300+12z$ for $0\le z\le 50$ m and $900\mathrm{kg}{\mathrm{m}}^{-3}$ below that. The mass per unit horizontal area isBy Archimedes' principle, the draft is $d=(m/A)/{\rho }_w\simeq 248.78$ m. The iceberg freeboard is therefore $300-d\simeq 51.22$ m. The specified snow-to-ice density is reached atThe more porous snow and firn are consequently above water: atmospheric warming, meltwater infiltration, compaction and refreezing can modify them. At the waterline and below, dense ice is exposed to oceanic heat transfer and wave erosion, which can undercut the berg and cause breakage. If the porous transition were submerged, water infiltration and ocean heat could act directly on that material, changing the erosion and collapse mechanisms.

On the straight sloping leg, $p=p_0(3-V/V_0)$ and $dp=-p_{0}dV/V_0$. Put $v=V/V_0$, which increases from $1$ to $2$. The reversible heat transfer isThus heat is absorbed for $1<v<15/8$ and emitted for $15/8<v<2$, with zero instantaneous heat at $v=15/8$. The switching point has $(p,V)=(9p_0/8,15V_0/8)$. In particular the net heat on this leg, $\frac{3}{2}{p}_0{V}_0$, is not the total positive heat on it.

A thermoelectric generator maintains hot and cold junctions of two materials with different Seebeck coefficients. Mobile charge carriers redistribute in the temperature gradient, producing an open-circuit voltage of magnitude $|V|=|S_p-S_n|\Delta T$ for constant coefficients. A load permits electric current; pairs of p-type and n-type legs are commonly connected electrically in series and thermally in parallel. Maintaining the output requires continuous heat transfer from the hot reservoir and rejection of heat to the cold reservoir. The thermoelectric figure of merit $ZT=S^2\sigma T/\kappa$ expresses the favorable combination of large Seebeck coefficient, large electrical conductivity and small thermal conductivity. Contact losses, Joule heating and parasitic heat leaks reduce practical efficiency.

A pyroelectric generator instead exploits a change of spontaneous electric polarization with temperature. If the relevant component of the pyroelectric coefficient is $p=d{P}_s/dT$, an electrode of area $A$ receives a short-circuit electric current approximately $I=pAdT/dt$. A heating stroke and a cooling stroke produce opposite charge changes. Rectification or a synchronized electrical cycle can transfer energy to storage. A spatial temperature difference drives the thermoelectric device; a temporal temperature change drives the pyroelectric device. Uniform, constant temperature gives no continuing current from pyroelectricity even if the material is hot. Reported nanowire generators deliberately use changing temperature, rather than infer continuous generation from a static hot specimen.^[ref]

Both approaches can be compact, silent and free of macroscopic moving parts. The thermoelectric generator gives continuous output under a sustained gradient, but needs two effective thermal contacts and sufficient heat flow; very small $\Delta T$ gives a small voltage and thermal contacts can consume much of the available gradient. Thin films exhibiting pyroelectricity can respond to intermittent heating without a permanently maintained hot/cold separation across their thickness, but require repeated thermal cycling, low electrical leakage and suitable charge extraction. Their thermal time constant limits useful cycling frequency, and sensible heating/cooling can dominate the heat budget. A steady hot pipe with an available cold sink favors a thermoelectric generator; periodic radiative heating can favor a pyroelectric generator. Neither converts equilibrium thermal motion at a single fixed temperature into sustained useful work.

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

The definition of heat capacity at constant area and the Maxwell relation give the entropy differentialFor a reversible adiabatic process, $dS=0$. Hencefor positive heat capacity $C_A$. Reversible adiabatic stretching cools the film. Reversibility, or negligible entropy production, is essential to this inference. Adiabatic isolation alone only sets the heat transfer to zero; an irreversible stretch can generate entropy, so its temperature change is not fixed by this isentropic derivative.

Replacing the fixed surface temperature by Newton's law of cooling changes the boundary condition but not the radial derivative in steady state. At the surface,HenceHere $\gamma$ is a surface heat transfer coefficient in $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$, unrelated to the surface tension symbol used in C10. An independent energy-balance check uses a length $\ell$: the electrical input $I^2\ell /(\sigma \pi a^2)$ must equal the surface loss $2\pi a\ell \gamma [T(a)-T_{\mathrm{air}}]$, which gives the same answer.

The complete temperature profile isThermal conductivity controls the internal gradient, but cancels from the surface-temperature rise because all generated power must eventually cross the prescribed surface. This assumes positive constant $\gamma$, uniform electrical conductivity and negligible end or additional radiative losses; temperature-dependent material properties could make the heating and boundary problem nonlinear.

(a) Treat the lake as a heat bath at fixed temperature $T_0=288.15\mathrm{K}$ and the block's heat capacity as constant. Its initial temperature is $T_i=473.15\mathrm{K}$, and $C=mc=355\mathrm{J}{\mathrm{K}}^{-1}$. Although the cooling is irreversible, entropy is a state function, so evaluate the block's change along an imagined reversible thermodynamic process:The lake receives heat $Q=C(T_i-T_0)=65675\mathrm{J}$, and thereforeAdding the two contributions givesThe positive result is consistent with the Second law of thermodynamics for heat transfer across a finite temperature difference. The calculation uses $c=710\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ exactly as printed in both the original PDF and the transcription; it does not replace the supplied value by a tabulated value for lead.

(b) The block starts and eventually finishes at the lake's temperature, so its net entropy change is approximately zero. Neglecting buoyancy, air resistance and the final immersion depth, the lost gravitational potential energy $mgh$ becomes internal energy in the lake after the motion and waves have dissipated. The lake remains at $T_0$, givingTemporary heating of the block or temporary storage in kinetic energy does not change this final energy and entropy balance once thermal equilibrium is restored.

Use an ideal gas to illustrate the Carnot cycle, with hot and cold heat baths at $T_h=T_2$ and $T_c=T_1$. The four processes are all reversible thermodynamic processes. Lettered states below belong only to this illustration; the engine later in the question uses numbered states.

Starting at $A$, perform the following four stages:

The clockwise enclosed area in the pressure-volume diagram is the positive net work per cycle, $W=\oint pdV=Q_h-Q_c$.

The complete cycle must be a reversible thermodynamic process: no friction, viscous dissipation, unrestrained expansion, or other source of entropy. Each moving boundary must be arbitrarily close to mechanical equilibrium. Heat transfer to or from a bath must occur across an infinitesimal temperature difference, and the insulated stages must be genuinely adiabatic processes. There must also be no direct heat leak between the heat baths.

For an engine exchanging heat only with these two heat baths, the Second law of thermodynamics givesand thusPractical engines generate entropy through finite temperature differences, friction and other dissipative effects. They therefore fall below the Carnot cycle limit. Approaching reversible heat exchange generally requires vanishingly slow operation for a finite heat-transfer apparatus, limiting useful power.

A persistent barrier to whole-mantle circulation adds an internal thermal bottleneck and permits the deep mantle to retain a larger temperature excess. In a single well-mixed layer there is one interior temperature and one surface heat-loss law. The coupled thermal histories of layered mantle convection arise because separately mixed layers can have different temperatures, exchanging heat through an interfacial boundary layer as well as losing heat at the surface. Their thermal histories then have coupled relaxation modes rather than the single mode in (c).

For example, let ${\theta }_l,{\theta }_u$ be lower- and upper-layer temperature excesses, $H_l,H_u$ their heat capacities per area, andWith no internal heating the budgets are $H_l{\dot{\theta }}_l=F-J$ and $H_u{\dot{\theta }}_u=J-E$. At steady state,Thus a fixed basal flux still leaves the surface eventually with $E=F$, but the lower layer is hotter than it would be in the single-temperature model. For a given deep temperature instead, the extra interfacial drop reduces outward heat transfer. This favours delayed loss of deep primordial heat and distinct upper- and lower-mantle cooling histories. A numerical change in cooling time cannot be obtained from layer count alone: interface transport, layer thicknesses, temperature-dependent viscosity, internal heating, and whether material can cross the interface matter. Merely replacing $d$ by $d/2$ in (c) misses the added interface and is not a model of two-layer cooling.

The estimate assumes the entire $2{\mathrm{m}}^2$ surface remains exposed at $35{}^{\circ }\mathrm{C}$ continuously. Clothing reduces direct exposure and presents a cooler outer emitting surface; the skin temperature also varies across the body and decreases in cooler surroundings through changes in blood flow. Body posture and nearby surfaces reduce the effective exchange area, while blankets and warm surrounding surfaces reduce net radiative heat transfer. These effects make the full-area, constant-temperature daily radiative estimate too large for ordinary clothed conditions.

The unit emissivity approximation is already fairly good for skin in the thermal infrared, so reducing emissivity alone is not the main correction. Also, convection, thermal conduction, and evaporation are additional channels of heat transfer; their existence does not subtract from the calculated radiative term. The computed number is a simplified radiative loss, not a complete metabolic energy budget.

Sadi Carnot's ideal engine was motivated by the practical problem of obtaining mechanical work economically from heat. Existing engines made temperature differences, boilers, condensers and the waste associated with cooling impossible to ignore. His use of a cyclic working substance and two thermal reservoirs therefore retained decisive features of real engine operation. An engine that has to be reset after each stroke cannot be assessed by considering expansion alone.

His crucial step was to strip away friction and uncontrolled heat transfer and ask for a limiting cycle that could be reversed. Combining a putatively better engine with a reversed ideal engine would produce an unacceptable result without a compensating thermal change; the comparison makes the maximum performance depend on reservoir temperatures rather than the identity of the working substance. This is a theoretical limit against which imperfect machines can be judged, not a blueprint that an actual engine can execute at finite power without losses.

There is a historically important separation between this reasoning and its modern reformulation. Carnot worked substantially within a caloric framework in which heat was treated as conserved in transfer. Modern thermodynamics instead distinguishes heat input, work output and heat rejected, with $W=Q_h-Q_c$ and reversible efficiency $1-T_c/T_h$. It would be anachronistic to attribute the complete modern energy-and-entropy analysis to his 1824 text.

Practical experience thus supplied the problem, relevant constraints and examples of irreversibility, while idealization supplied a result that experience alone could not establish by testing finitely many engines. The ideal engine's importance lies precisely in exceeding any particular machine without pretending that its limiting operation is practically attainable at useful power.^[ref]

A state function is determined by the equilibrium state, rather than the path used to reach it. Its differential is exact and its change around a closed thermodynamic cycle is zero. Internal energy, entropy and enthalpy are examples; transferred heat and work are not state functions.

For slow expansion in contact with a large thermal reservoir, the fixed temperature of that reservoir maintains an isothermal process. For a fixed amount of ideal gas, internal energy and enthalpy depend only on temperature, soThe ideal gas law also gives $pV=\mathrm{constant}$. For a reversible expansion the absorbed heat equals the work done by the gas, while its entropy increases by $n_{\mathrm{mol}}R\ln (V_2/V_1)$; constant internal energy does not mean zero heat transfer.

An actual institutional protocol is CTU's Measurement of characteristics of a Peltier cell. Its cooling run uses a water container, an insulating cap and a current-limited supply; both face temperatures are recorded every minute while electric current is monitored. The apparatus includes a water-cooled exchanger, copper blocks and brass containers. The heating run resets the water near room temperature before reversing polarity. The handout specifies no thermometer type and reports no dataset. Its modular junction geometry is shown separately from the external thermal assembly.^[ref]

The following is a proposed quantitative extension. Keep the physical face labels fixed throughout reversal: c denotes the observed object's side and h the exchanger side, even if their actual temperatures change order. Use an independently read thermometer at each thermal contact and log the actual signed electric current and terminal electric potential difference. Maintain reproducible contact pressure and contact layers. A sensor in water measures water temperature, which need not equal the face temperature during a transient; verify mixing and contact equilibration before using it as a face reading. Do not interchange sensor leads when reversing the supply.

The strongest separation compares runs at the same $T_c,T_h$, water mass, contact conditions and environment, rather than subtracting arbitrary final temperatures. One option is short initial-rate runs after resetting the whole assembly to a reproducible state. Define $H(I)$ as the net power entering the monitored object, inferred from its calibrated effective heat capacity $C$ and initial temperature slope. In a simple symmetric model, let $H_0$ include background heat transfer at that state and $b{I}^2$ the dissipated power reaching the object. Thenwhere $a=\alpha T_c$ in the thermoelectric cooling model below. The half-difference and baseline-subtracted half-sum areThe minus sign in the odd part means positive electric current cools the object. The even part has to subtract the zero-current baseline: otherwise heat conduction and ambient exchange remain mixed with Joule heating. Independently measured heat capacity is essential for converting a thermal slope into watts; include the blocks, container and water that actually equilibrate on the measurement timescale.

Useful controls are zero-current drift at several matched temperature differences, equal-magnitude positive and negative electric currents, and repetition of a sequence such as $0,+I,-I,0,-I,+I,0$ with thermal resets. A calibrated small heater on the object can test the heat capacity and lost-heat correction independently. Check Joule heating in supply leads and contacts, bypass heat conduction through supports and wiring, and ambient convection and thermal radiation. Insulation reduces a leak; it does not establish that it is zero. Measure terminal power at the module boundary so lead losses are not mistaken for module heating.

An odd thermal response supports a reversible current-coupled contribution, but does not by itself locate all of it at one junction: the Thomson effect in nonuniform leads can also be odd. Sensor offsets, unequal current magnitudes, drifting thermal contacts and different junction temperatures during the two runs can contaminate both parts. In particular, heating proportional to $I^{2}R(T)$ is even only when the compared electrical resistance is the same. CTU's qualitative reversal illustrates switching heating and cooling; the matched-state controls above would be needed for a coefficient measurement.

The Peltier effect is heat transfer associated with an electric current crossing a junction between materials with different Peltier coefficients. At fixed junction temperature, its contribution changes sign when the conventional current reverses. A junction can absorb heat from its surroundings for one direction and release it for the other. Joule heating, proportional to the square of the electric current, accompanies this transport but does not reverse sign. Consequently, observing a warm device alone does not identify the Peltier effect.

This makes the Peltier effect useful as an electrically controlled heat pump: reversing the supply can switch an object's thermal role without changing any plumbing. In thermoelectric cooling, however, the useful cooling is the junction's absorbed heat minus returning heat conduction and its share of Joule heating. Increasing electric current eventually makes cooling worse. The distinction between transported heat and net cooling is the central practical lesson of the thermoelectric cooler energy balance.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

A radiative-convective climate column applies a radiative-convective equilibrium model to the vertical thermal structure relevant to climate change. Radiative transfer determines heating and cooling; convective adjustment represents unresolved vertical redistribution when the chosen atmospheric lapse rate is too steep. Radiative equilibrium alone need not give a physically admissible lower-atmospheric profile.

In 1967, Syukuro Manabe and Richard T. Wetherald computed equilibrium with prescribed relative humidity, allowing atmospheric water vapor to change with temperature. Their average-cloud case warmed by $2.36\mathrm{K}$ for carbon dioxide increasing from 300 to 600 parts per million, versus $1.33\mathrm{K}$ with fixed absolute humidity. These are model outputs, not measured warming. The comparison isolates an important water-vapor feedback.^[ref]

An original two-temperature calculation exposes the bookkeeping. Let an opaque surface absorb solar irradiance $Q$ and emit blackbody thermal radiation $\sigma T_s^4$. A single isothermal atmospheric layer transmits all sunlight and absorbs a fraction $ϵ$ of surface infrared emission, emitting $ϵ\sigma T_a^4$ both upward and downward. This is a grey thermal-band emissivity, with $0<ϵ\le 1$. Let $H$ be upward nonradiative heat flux, and let $C_s,C_a>0$ be heat capacities per unit area. Then conservation of energy givesThe two equations sum toHere $L$ is outgoing longwave radiation. Both $H$ and downward atmospheric emission cancel internally: they redistribute energy without creating it. The factor two in atmospheric emission counts two faces. Dropping the downward face would spoil the surface budget. Each right-hand term has units $\mathrm{W}{\mathrm{m}}^{-2}$; each $C$ has units $\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

Choose $Q=240\mathrm{W}{\mathrm{m}}^{-2}$, $H=80\mathrm{W}{\mathrm{m}}^{-2}$ and $ϵ=1$. At equilibrium, $L=\sigma T_a^4=240\mathrm{W}{\mathrm{m}}^{-2}$ and $\sigma T_s^4=400\mathrm{W}{\mathrm{m}}^{-2}$, giving $T_a\simeq 255.06\mathrm{K}$ and $T_s\simeq 289.81\mathrm{K}$. Direct substitution gives surface balance $240+240-400-80=0$ and atmospheric balance $400-480+80=0$. These deliberately chosen fluxes illustrate the model; they are not an observed global budget or a reconstruction of the 1967 calculation. Prescribing $H$ illustrates heat transfer accounting but does not close a realistic convection law.

For fixed $Q,H,ϵ$, the positive-temperature equilibrium is locally stable. Put $a=4\sigma T_s^3>0$ and $b=4ϵ\sigma T_a^3>0$. The Jacobian matrix of the temperature equations has trace $-a/C_s-2b/C_a<0$ and determinant $ab(2-ϵ)/(C_s{C}_a)>0$. Thus both eigenvalues have negative real part. Temperature-dependent climate feedbacks alter this test. A single grey layer also omits the optical spectrum, cloud structure and the independent thermal behavior of the troposphere and stratosphere.

A radiometer thermal link is the intentional path for thermal conduction from receiver to reference block. Near a chosen operating point, write $Q_{\mathrm{link}}\simeq G(T-T_b)$ with local conductance $G$ in $\mathrm{W}{\mathrm{K}}^{-1}$. A weaker link increases the steady temperature rise per added power, $\delta T\simeq \delta P/G$, but also increases the uncontrolled relaxation time $\tau =C/G$ when the link dominates other heat transfer. These are local linear approximations: actual thermal conductivity depends on temperature, and active feedback control introduces its own response. NIST explicitly describes the tradeoff between weaker links and smaller receiver heat capacity.^[ref]