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The Clausius statement of the second law forbids a cyclic device whose sole effect is transfer of heat from a colder thermal reservoir to a hotter one. The Kelvin-Planck statement of the second law forbids a cyclic device whose sole effect is complete conversion of heat drawn from a single thermal reservoir into work. These are equivalent formulations of the Second law of thermodynamics; an ordinary refrigerator is allowed because supplied work is an additional effect.

Take any fixed material subvolume $B_0$ of the reference configuration, with outward normal vector $N$, current particle velocity $v_i={\dot{x}}_i$, and positive temperature $\theta$. The first law of thermodynamics, including kinetic energy, isThe heat-flux sign is negative because $N_I{q}_I^0$ is outward. The Second law of thermodynamics supplies the corresponding integral Lagrangian entropy inequality:The temperature in each supply or flux term is the local temperature, not a common temperature pulled outside the integrals.

Apply the divergence theorem and use the reference momentum conservation equation ${\rho }_0{\dot{v}}_i=P_{Ii,I}+{\rho }_0{g}_i$. In the total-energy equation, the kinetic derivative cancels the body-force work and the $v_i{P}_{Ii,I}$ term. Since $v_{i,I}={\dot{F}}_{iI}$, the remaining local Lagrangian internal energy balance and Lagrangian entropy inequality areThey hold pointwise because $B_0$ was arbitrary. In particular,

The specific Helmholtz free energy is $\psi =u-\theta \eta$. Substitution yields its balanceand elimination of the heat supply yields the Clausius-Duhem inequalityFor $\psi =\psi (F,\theta ,{\xi }_r)$, expand the derivative to obtainThe usual Coleman-Noll procedure admits independent reversible choices of $\dot{F}$ and $\dot{\theta }$ at a fixed state. Their coefficients must vanish, since either sign and arbitrary magnitude would otherwise violate the inequality. ThusInternal-variable evolution and heat conduction must satisfy this remaining inequality. Substituting the constitutive relations back into the energy balance givesThe last printed equation is missing a factor ${\rho }_0$ on its internal-variable term. The PDF defines $f_r=-{\psi }_{,{\xi }_r}$ with $\psi$ per unit mass, so the expression above is the dimensionally consistent consequence. Equivalently define the volumetric thermodynamic force conjugate to an internal variable ${\hat{f}}_r=-{\rho }_0{\psi }_{,{\xi }_r}$; the term is then ${\hat{f}}_r{\dot{\xi }}_r$ with no extra density. The two conventions must not be mixed.

For thermoelasticity with a temperature-dependent constraint, admissible rates obeyThe reversible coefficient functional need now vanish only on this tangent hyperplane. Hence it can be a scalar multiple $q$ of the constraint normal:ThereforeThe reaction terms cancel on admissible rates, leaving the same internal-variable/conduction inequality and entropy heating equation. The multiplier $q$ is determined by equilibrium and boundary conditions rather than by the free energy alone.

For linear thermoelasticity with prescribed thermal volume, write $F=I+H$, $H_{ij}=u_{i,j}$, and $\vartheta =\theta -{\theta }_0$. Hold the internal variables fixed in this reversible linearization, and evaluate all derivatives below at $(I,{\theta }_0)$ and zero reaction. The stress-free reference state has ${\psi }_{,F}=0$. Since $\partial \det F/\partial F=I$ there, first-order nominal and Cauchy stress tensors agree. Their constitutive laws becomewhere equality of mixed free-energy derivatives givesThe linearized constraint is $u_{i,i}=h^{\prime }({\theta }_0)\vartheta$. Thus volume is thermally prescribed; strict temperature-independent incompressibility corresponds to $h^{\prime }=0$. The coefficient $C_e$ is an entropy slope; the corresponding volumetric heat capacity at fixed deformation is ${\theta }_0{C}_e$.

Together with ${\rho }_0{\ddot{u}}_i={\sigma }_{ji,j}+{\rho }_0{g}_i$, the first-order thermal balance is ${\theta }_0(C_e\dot{\vartheta }-{\beta }_{ji}{\dot{u}}_{i,j}+h^{\prime }({\theta }_0)\dot{q})={\rho }_{0}r-q_{I,I}^0$. A heat-flux constitutive law completes the evolution problem. The paired signs of the coupling tensor follow from the same free energy, rather than from separately postulated mechanical and thermal equations.

The instantaneous-decay approximation keeps the scale factor and physical volume fixed across the decay. Energy conservation and rapid thermalization giveMeanwhile the abundance relation gives $s_{-}={\rho }_{\varphi ,-}/(Ym)\simeq 3M_{\mathrm{Pl}}^2/(Ym{t}_d^2)$. Dividing yields the entropy-injection factorThere is a numerical error in the printed coefficient: part (c) omits the factor $1/3$. The denominator $s_{-}=3M_{\mathrm{Pl}}^2/(Ym{t}_d^2)$ supplies that factor, while the radiation energy contributes $3^{3/4}$ to the numerator. Thus the consistent coefficient is $c_1{3}^{-1/4}{c}_2^{-3/4}$, rather than $c_1{3}^{3/4}{c}_2^{-3/4}$. Both cannot follow from the stated part (b) using the same decay-time approximation.

Its increase is especially transparent without dropping the original radiation. Let $R={\rho }_{\varphi ,-}/{\rho }_{r,-}$. At unchanged $g_{\star }$, $T_{+}^4/T_{-}^4=1+R$, soRelic domination means $R\gg 1$, so the leading estimate is much larger than one. This is entropy production by decay of a dominant relic.

There is no conflict with part (a). The equilibrium gas before the decay was a separately conserved, reversibly expanding system. During decay it receives energy from the nonthermal relic, and the decay products thermalize irreversibly. Its cosmological perfect-fluid continuity equation now has an energy-injection source, so $d({\rho }_{r}V)+P_{r}dV$ does not vanish. The assumptions behind cosmological entropy conservation therefore fail during the event. Moreover the zero-chemical-potential equilibrium expression in part (a) cannot be imposed on the decoupled relic as though it were already part of that same equilibrium radiation bath. The Second law of thermodynamics permits, and here requires, the resulting entropy increase.

Let the closed atmospheric parcel exchange heat and volume work with reservoirs at fixed temperature $T$ and pressure $P$, while reactions rearrange its conserved elemental inventory. For a reversible differential between equilibrium states, the first law of thermodynamics, with chemical work included, isThe Gibbs free energy $G=U+PV-TS$ therefore obeysThe first law of thermodynamics supplies this differential but does not alone choose the direction of a reaction. The Second law of thermodynamics does: for the parcel plus its reservoirs, at fixed $T,P$ and with no non-volume work,Thus spontaneous reactions decrease Gibbs free energy, and stable thermochemical equilibrium is its constrained minimum. For an allowed reaction of extent $d\xi$, let $d{N}_i={\nu }_{i}d\xi$, with positive stoichiometric vector entries for products and negative entries for reactants. At an interior minimum,The condition must hold for every independent allowed reaction, with fixed total atoms of each element and any relevant charge constraint. At a boundary with absent species, only feasible one-sided variations are permitted. These are the hypotheses of constrained Gibbs minimization for chemical equilibrium.

For an ideal gas mixture, ${\mu }_i={\mu }_i^{\circ }(T)+\mathcal{R}T\log (x_{i}P/P^{\circ })$, where $x_i$ is the volume mixing ratio and $\mathcal{R}$ the molar gas constant. ThereforeThis is a dimensionless activity-based equilibrium constant, not a bare product of dimensional pressures.

For a perfect gas with constant specific heat capacity $c_v$, the first law of thermodynamics givesUsing the pressure-density Hugoniot relation for a perfect gas, the entropy production in a perfect-gas shock is thereforeThis is positive for $P>1$, as required by the Second law of thermodynamics for the physical compressive normal shock wave.

Let two systems exchange energy inside an otherwise isolated composite system, so $E_1+E_2=E$ is fixed. By the Second law of thermodynamics, equilibrium maximizes the total entropyAt an interior maximum,Thus $T_1=T_2$, which is entropy maximization under thermal contact and the condition for thermal equilibrium.

Stability requires this entropy maximum to be locally strict, so each ordinary subsystem has a concave entropy-energy relation. Sinceone has $\partial T/\partial E>0$, or equivalentlyThis is the entropy concavity and positive heat capacity criterion.

Now let $N_{\uparrow }=\alpha N$ and $N_{\downarrow }=(1-\alpha )N$. Because the spins are distinguishable, the number of microstates is the binomial coefficientThe Boltzmann entropy $S=k_B\log \Omega$ and the Stirling formula give, to leading order for large $N$,The total energy isThereforeSolving for the up-spin fraction gives the independent spin-one-half two-level system resultFor ordinary positive absolute temperature,with $\alpha \to 0$ as $T\to 0^{+}$ and $\alpha \to 1/2$ as $T\to +\infty$. Since the spectrum is bounded above, the population-inverted range $1/2<\alpha \le 1$ is also mathematically possible and corresponds to negative temperature; $\alpha =1/2$ is the infinite-temperature state.

Substitution givesand direct differentiation yieldsThus the energy increases with temperature throughout either finite-temperature branch, in particular throughout the required positive-temperature range.

The Clausius statement of the second law says that no cyclic device can have as its sole effect the transfer of heat from a colder reservoir to a hotter one. The Kelvin-Planck statement of the second law says that no cyclic device can have as its sole effect the extraction of heat from one reservoir and its complete conversion into work.

Suppose first that the Clausius statement fails. Use the forbidden device to return to a hot reservoir the heat $Q_C$ rejected by an ordinary heat engine. The combined device extracts $Q_H-Q_C$ from the hot reservoir and converts all of it into work, violating the Kelvin--Planck statement.

Conversely, suppose the Kelvin--Planck statement fails. Use the work produced by the forbidden one-reservoir engine to drive an ordinary refrigerator. The engine's heat withdrawal and the refrigerator's heat discharge cancel by the work-energy balance, leaving only a transfer of heat from the cold reservoir to the hot reservoir with no external work. This violates the Clausius statement. Each statement therefore implies the other, so the two forms of the Second law of thermodynamics are equivalent.

The relevant claim concerns a machine that repeats its complete operating cycle while delivering a continuing net output of work without a corresponding supply of energy. A machine cannot supply net work in a cycle without taking in energy or depleting an energy store. This follows from conservation of energy, an experimentally supported physical law rather than a theorem of pure logic.

Let $E$ include the machine's internal energy, kinetic energy and potential energy. With $Q$ the net heat received and $W$ the net work delivered, the first law of thermodynamics givesFor a complete cycle all stores return to their initial values, so $\Delta E=0$ and $W=Q$. If there is neither heat input nor another energy input, $W=0$. A descending mass, unwinding spring, consumed fuel or discharged battery supplies energy, but it cannot be omitted from the accounting while calling the result perpetual motion.

Changing an overbalanced wheel's geometry does not evade the argument. A conservative force gives zero net work around a closed configuration cycle, while friction dissipates usable mechanical energy. Resetting a mechanism or returning a raised weight costs work that must be included.

Indefinitely continuing motion without useful output is a different possibility: an ideal frictionless oscillator or freely moving body need not stop. Connecting a load removes energy and changes that situation. A cyclic machine supplied with heat from only one reservoir and converting all that heat into work is excluded instead by the Kelvin-Planck statement of the Second law of thermodynamics. The first-law and second-law objections address different proposed energy balances.

Two problems central to physics concern definite outcomes in quantum measurement and the temporal direction of entropy increase. They expose different gaps between a dynamical description and the phenomena explained; neither can be resolved merely by saying that physics predicts observations successfully.

For the measurement problem, combine a two-state system with an apparatus initially in a ready state. If an ideal unitary operator correlates basis states with distinct pointer states, its linearity givesThe same dynamics that gives reliable correlations therefore gives an entangled final state, rather than a rule choosing one pointer value. The problem is not that the state has failed to predict possible outcomes, but how definite outcomes and their probabilities are related to that state.

Quantum decoherence explains why interaction with an environment suppresses interference in the apparatus's reduced density matrix: its off-diagonal term is multiplied by the inner product of the corresponding environmental states. Near-orthogonal records make this term very small. Tracing out the environment, however, does not by itself select one term in the global superposition. An objective-collapse theory changes the dynamics, while a many-worlds interpretation changes what is required of an outcome. Distinguishable modifications can be tested; interpretations reproducing the same predictions raise additional explanatory and conceptual questions. Decoherence solves a substantial dynamical part of the problem without automatically settling every interpretation.

The second issue is the thermodynamic arrow of time. In a classical model, time reversal in classical mechanics can pair an entropy-increasing trajectory with a reversed entropy-decreasing one. Liouville theorem in Hamiltonian mechanics preserves phase-space volume, so reversible Hamiltonian dynamics does not imply a universal monotonic increase of fine-grained entropy. The Second law of thermodynamics therefore needs a statistical and boundary-condition account, not a deduction from reversibility alone.

Boltzmann entropy assigns larger entropy to a macrostate occupying more compatible phase-space volume. Starting from a suitable low-entropy condition, overwhelmingly more accessible microstates can evolve toward larger macrostates. But an unrestricted typicality argument applied at an arbitrary low-entropy instant generally points toward increasing entropy in both temporal directions. The past hypothesis supplies an asymmetric low-entropy boundary condition in the past; one must still justify the probability measure and the macroscopic description used.

These are genuine partial resolutions rather than interchangeable cures. Decoherence concerns the accessibility of interference and records; a past boundary condition concerns temporal asymmetry in a statistical explanation. A satisfactory account must say which extra assumptions it introduces and which observable consequences follow. Neither problem licenses dismissing successful physics, but both show why a predictive formalism and a complete interpretation of its success are different achievements.

The zeroth law of thermodynamics states the transitivity of thermal equilibrium: if two systems each equilibrate thermally with a third, they equilibrate with one another. A reproducible thermometric property can therefore label their temperature. Its empirical numerical scale is not itself the complete thermodynamic content; reversible heat-engine relations establish an absolute temperature scale.

The first law of thermodynamics expresses energy conservation. For a closed system, $dU=\delta Q-\delta W$ when $\delta W$ is work done by the system. internal energy is a state function, while heat and work depend on the path. In reversible simple compressible motion, $\delta W=pdV$ and $dU=TdS-pdV$; an open system additionally needs the energies carried by transferred matter. An ideal gas can expand freely with $Q=W=0$ and unchanged internal energy. The first law alone does not decide whether the reverse spontaneous process occurs.

The Second law of thermodynamics introduces entropy and restricts direction. For a reversible path, $dS=\delta Q_{\mathrm{rev}}/T$; for an irreversible process the entropy balance includes nonnegative production. The entropy of an isolated system cannot decrease. In a cycle, the Clausius inequality is $\oint \delta Q/T\le 0$, using the appropriate boundary/reservoir temperatures. The Kelvin–Planck statement forbids a cyclic device whose sole effect is extracting heat from one reservoir and turning all of it into work; the Clausius statement forbids moving heat from cold to hot with no compensating effect. A refrigerator is allowed because it consumes work. A two-reservoir engine has $\eta \le 1-T_c/T_h$, with equality in the reversible limit. Work may be fully dissipated as heat without contradicting this one-way restriction.

The third law of thermodynamics constrains low-temperature entropy. An equilibrium perfect crystal with a nondegenerate ground state approaches zero entropy as $T\to 0$, fixing the conventional absolute entropy reference. Degenerate ground states or frozen nonequilibrium disorder require care when discussing residual entropy. Under regular equilibrium conditions the heat capacity must decrease sufficiently for $\int C/TdT$ to remain finite; a constant positive classical heat capacity cannot be extrapolated to zero temperature. The unattainability formulation states that absolute zero cannot be reached by a finite sequence of admissible cooling operations. These statements concern the limiting thermodynamic regime, rather than predicting that every substance stays in one classical phase during cooling.

The laws work together: conservation sets a device's energy budget, entropy production restricts the attainable budget, and constitutive equations give the properties of its working material. For fixed temperature and volume, Helmholtz free energy decreases toward equilibrium; at fixed temperature and pressure the corresponding potential is Gibbs free energy. Stable equilibrium, phase coexistence and Maxwell relations follow from these potentials with appropriate constraints. Calorimetry, mechanical-work measurements, thermometry and reversible-path comparisons test different parts of this framework. None of the laws by itself supplies a particular gas equation of state or microscopic heat capacity.

Suppose an engine were more efficient than a reversible Carnot engine between the same reservoirs. Couple it to a reversed Carnot engine, scaling the cycles to have equal work. With no net external work, their combined effect would be to transfer heat from the colder reservoir to the hotter one, contradicting the Second law of thermodynamics. No such engine can exceed ${\eta }_C=1-T_c/T_h$. Equivalently, the cyclic entropy inequality gives $-Q_h/T_h+Q_c/T_c\ge 0$, hence the same bound.

The Second law of thermodynamics restricts which energy-conserving processes can occur. In the Kelvin-Planck statement of the second law, no cyclic engine can have as its sole effect the extraction of heat from one reservoir and its complete conversion into work. In the Clausius statement of the second law, no process can have as its sole effect transferring heat from a colder reservoir to a hotter one. An engine or refrigerator combined with the reverse of a reversible device shows that a violation of either statement would violate the other.

A heat engine takes heat $Q_h>0$ from a reservoir at $T_h$, rejects $Q_c>0$ to one at $T_c<T_h$, and returns its working substance to the initial state. Energy conservation gives $W=Q_h-Q_c$ and efficiency $\eta =W/Q_h=1-Q_c/Q_h$. Carnot's theorem states that no engine between these reservoirs is more efficient than a reversible one, and that all reversible engines between them have the same efficiency. Otherwise, drive a reversible engine backwards with the work of a more efficient engine; the composite cycle transfers heat from cold to hot without external work, contradicting the Clausius statement.

A reversible Carnot cycle consists of two isotherms joined by two reversible adiabats. For an ideal gas, $Q_h=N{k}_B{T}_h\log (V_2/V_1)$ and $Q_c=N{k}_B{T}_c\log (V_3/V_4)$. The adiabats give $V_2/V_1=V_3/V_4$, soCarnot's theorem makes this independent of the working substance and defines the thermodynamic absolute-temperature scale consistently with ideal-gas temperature.^[ref]

For any reversible cycle, $\oint \delta Q_{\mathrm{rev}}/T=0$. This makes the integral between two equilibrium states independent of the reversible path, defining the state function entropy:Heat itself is path dependent; entropy is not. An irreversible change can be evaluated using any imagined reversible path between its equilibrium endpoints. For example, a fixed amount of ideal gas with constant $C_V$ has $\Delta S=C_V\log (T_B/T_A)+N{k}_B\log (V_B/V_A)$.

For a general cycle the Clausius inequality is $\oint \delta Q/T_{\mathrm{boundary}}\le 0$. The entropy balance for a process is $\Delta S\ge \int \delta Q/T_{\mathrm{boundary}}$, with equality for reversible heat exchange and no entropy production. The surroundings' entropy must be included, giving $\Delta S_{\mathrm{universe}}\ge 0$. A system's own entropy can decrease by exporting entropy; an isolated system's cannot.

The engine's reservoirs have net entropy change $Q_c/T_c-Q_h/T_h\ge 0$, sowith equality for a reversible engine. Friction, heat flow through a finite temperature difference and unrestrained expansion produce entropy and reduce usable work. Thus energy conservation fixes the balance of heat and work, while the second law fixes their direction and the attainable efficiency.^[ref]

The units supplied for 130 indicate specific heat capacity, so the block's total heat capacity is $C=mc=0.5(130)=65\mathrm{J}{\mathrm{K}}^{-1}$. Treat it as constant, and the lake as a thermal reservoir at $T_0=283.15\mathrm{K}$. The initial block temperature is $T_i=373.15\mathrm{K}$.

Compute the block's entropy change using a reversible comparison path, even though the actual cooling is irreversible:The lake receives $Q=C(T_i-T_0)=5850\mathrm{J}$ at constant temperature, soThereforeThis is entropy generated by cooling a finite body in a reservoir. The positive total is consistent with the Second law of thermodynamics, although the cooling block's own entropy decreases.

Temperature as an equilibrium relation. Classical thermodynamics begins with the zeroth law of thermodynamics: thermal equilibrium is transitive. A thermometer can therefore assign a common empirical temperature to systems in mutual thermal equilibrium. Different thermometer properties initially give different monotone empirical scales; a resistance reading by itself does not define the absolute scale.

The Second law of thermodynamics provides a substance-independent thermodynamic scale. A reversible engine operating between reservoirs has a heat ratio depending only on their temperatures. One chooses absolute temperature so that $Q_h/Q_c=T_h/T_c$, with $Q_h,Q_c$ the positive heat magnitudes. Its reversible efficiency is $1-T_c/T_h$. Equivalently, reversible heat defines entropy through $dS=\delta Q_{\mathrm{rev}}/T$. For an equilibrium simple compressible system,Classical thermodynamics uses this relation without needing a microscopic account of entropy. A dilute-gas thermometer realizes an empirical scale approaching the thermodynamic scale after corrections for nonideal behavior.

Statistical meaning of the same derivative. In the microcanonical description, an equilibrium energy shell containing $\Omega (U,V,N)$ accessible states has $S=k_B\ln \Omega$. Its statistical temperature is consequentlyTo see why this is an equilibrium temperature, let two weakly coupled systems exchange a small energy while their total energy stays fixed. The combined multiplicity is ${\Omega }_A(U_A){\Omega }_B(U_{\mathrm{tot}}-U_A)$. Maximizing its logarithm giveswhich is equality of the two inverse temperatures. Stability requires the total entropy maximum, not only stationarity. The overwhelmingly dominant macrostate in a large system therefore realizes the thermodynamic equilibrium condition.

A large reservoir gives a small subsystem the canonical probabilities $p_i\propto e^{-\beta E_i}$, with $\beta =1/(k_{B}T)$. In that ensemble, $F=-k_{B}T\ln Z$ and $S=-k_B{\sum }_i{p}_i\ln p_i$ agree with macroscopic thermodynamic relations in their usual equilibrium regime. The equipartition identity for a classical dilute monatomic gas, $⟨p^2/(2m)⟩=3k_{B}T/2$, is a useful operational connection between microscopic motion and temperature, but is not a universal definition: quantum modes need not obey equipartition, and a system away from equilibrium need not have one temperature at all.

For finite systems one must specify the entropy convention and energy resolution; surface and fluctuation corrections can distinguish definitions that agree in the ordinary thermodynamic limit. Bounded-energy population-inverted systems can also have a negative entropy derivative and hence a negative statistical temperature. That possibility lies outside an ordinary positive-temperature ideal-gas thermometer; it does not mean temperature is merely an arbitrary average kinetic energy.

Obtaining the energy–temperature conversion. The Boltzmann constant connects entropy measured in $\mathrm{J}{\mathrm{K}}^{-1}$ with the dimensionless logarithm of a state count. Historically it could be determined from the ideal gas law $pV=N{k}_{B}T$: measure the molar gas constant $R$ and independently determine the Avogadro constant $N_A$, then obtain $k_B=R/N_A$. The independent particle-number determination matters; using $R$ alone would not fix energy per particle.

An electrical method uses Johnson-Nyquist noise. A resistor of resistance $R_e$ at known temperature has, in the classical range $hf\ll k_{B}T$, a one-sided open-circuit voltage-noise spectrum $S_V(f)=4k_{B}T{R}_e$. Over a calibrated flat bandwidth $\Delta f$,A practical measurement uses an equilibrium resistor, a calibrated low-noise amplifier and a known transfer function, subtracts or independently characterizes amplifier noise, and replaces a nominal bandwidth by the actual effective noise bandwidth. Varying temperature and fitting the mean-square voltage is more informative than treating one amplified voltage sample as the thermal variance. Other primary methods relate temperature to sound speed in a dilute gas or to calibrated particle diffusion and mobility.

There is a historically relevant unit-definition qualification: by the date of this exam, the revised SI had already taken effect on 20 May 2019. The kelvin is now defined by fixingThe same experiments now realize or compare thermodynamic temperatures using this fixed constant, whereas historically they measured the conversion constant on an independently defined temperature scale. The physical distinction between equilibrium temperature and microscopic state counting is unchanged.^[ref]

(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.

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.

The left panel below shows paddles in a bath connected by an axle to an asymmetric toothed wheel, a spring-loaded pawl and a load. A naive argument lets random paddle kicks turn the wheel in one direction while the pawl blocks reverse motion, apparently extracting work from a single bath. The missing ingredient is the pawl's own thermal fluctuations. At the same temperature, it sometimes lifts and allows reverse slips. Correct forward and reverse transition rates obey detailed balance, so no sustained directed rotation can do work at single-bath thermal equilibrium. The device illustrates, rather than disproves, the Second law of thermodynamics.

A macroscopic shaken ratchet is driven far from thermal equilibrium: imposed shaking supplies organized mechanical energy, and friction or inelastic collisions dissipate it. Its kilogram-scale parts are not moved appreciably by molecular thermal fluctuations. Rectification requires broken spatial symmetry from the teeth or pawl and appropriate dynamical coupling: the shaking must sometimes disengage or drive the pawl over a tooth while contact and dissipation prevent equal reverse motion. The load must remain below the available mean driving torque. Isotropic, aperiodic shaking can satisfy these conditions because the asymmetric mechanism breaks directional symmetry; without the external energy supply, a working rectifier and dissipation, there is no guaranteed directed motion.

Availability and total entropy. The reservoir changes by $d{U}_R=-dU$ and $d{V}_R=-dV$. Its particle number and the system's particle number are fixed. The reservoir's first law of thermodynamics therefore givesAdd the system's entropy change. Because $T_R,p_R$ are constant,Thus define the availabilityAn additive constant in $A$ is immaterial; its value above the equilibrium reference is the exergy. The Second law of thermodynamics gives $\Delta S_{\mathrm{tot}}\ge 0$, hence $\Delta A\le 0$. A stable equilibrium maximizes total entropy over the allowed variations and therefore minimizes $A$. The environmental $T_R,p_R$ enter this definition even when the system has not yet reached their values. Because both thermal and mechanical exchanges are allowed, equilibrium here includes mechanical balance; equality of temperatures alone does not ensure the minimum.

Probability and curvature. For a fixed value of a macroscopic coordinate $x$, let $S(x)$ count the compatible system microstates. The product of the system and reservoir multiplicities isAll factors independent of $x$ enter a normalization constant, givingHere $A(x)$ is the constrained or effective availability for this coordinate, including the multiplicity of any eliminated variables. A nonuniform coordinate measure must likewise be retained or incorporated in that effective availability.

At a stable interior minimum, $A^{\prime }(x_0)=0$ and $A^{\prime \prime }(x_0)>0$. Its Taylor expansion givesKeeping the quadratic term yields a normal distribution,This is availability curvature and equilibrium fluctuations. The quoted fluctuation formula holds at quadratic order in the Gaussian fluctuation approximation; it is not an exact identity for an arbitrary nonlinear $A(x)$. Higher terms can shift the mean away from its most probable value $x_0$.

Bubble equilibrium. Keep the number of air molecules fixed and include the surface energy in the system's total internal energy. For constant surface tension $\Gamma$, this contribution is $\Gamma \Omega =\Gamma \alpha V^{2/3}$, with no surface entropy under the stated approximation. Its differential isHenceIndependent thermal and volume variations give $T=T_R$ andThe air pressure is evaluated at $V_0$; since the bubble is closed it generally depends on volume, so this equation is implicit until an air equation of state is supplied. For a sphere, $\alpha =(36\pi {)}^{1/3}$, $\Omega =4\pi R^2$, and the pressure jump becomes $2\Gamma /R$, the Young–Laplace equation for a single gas–liquid interface. There is no second surface as in a thin soap film.

Isothermal volume fluctuations. With the air maintained at $T_R$, eliminate its entropy in favor of its Helmholtz free energy:The identity $(\partial F_{\mathrm{air}}/\partial V{)}_{T,N}=-p_{\mathrm{air}}$ givesUsing the curvature result,These are the volume fluctuations of a closed gas bubble. Stability requires the denominator to be positive. The negative surface contribution reflects the concavity of $V^{2/3}$; gas compression supplies the positive restoring curvature. For ideal air, ${\kappa }_T^{-1}=p_{\mathrm{air}}$, and mechanical equilibrium reduces the denominator toThe units are consistent: $D$ is pressure, so $k_B{T}_R{V}_0/D$ has units ${\mathrm{m}}^6$, the units of a volume variance. Holding $p_{\mathrm{air}}$ externally fixed while varying $V$ would remove the gas-compression term and change this stability problem; that is not the fixed-particle-number situation here.

Numerical size of the fluctuations. The volume conversion is important:Use the specified $\alpha =4.84$, $\Gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $p_R=10^5\mathrm{Pa}$. ThenWith the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$,Thus the fractional root mean square fluctuation and fractional variance are respectivelyThe absolute root mean square fluctuation is $2.01\times 10^{-19}{\mathrm{m}}^3$. The very small fractional value supports the local Gaussian fluctuation approximation. This estimate neglects exchange of dissolved air, variations of surface tension, shape fluctuations and temperature fluctuations, consistently with the prescribed fixed-number, isothermal bubble model.

The Second law of thermodynamics states that total entropy of an isolated composite system cannot decrease. For a cyclic heat engine exchanging heat only with reservoirs at $T_h>T_c>0$, it gives $-Q_h/T_h+Q_c/T_c\ge 0$. Since $W=Q_h-Q_c$,The bound is attained by a reversible heat engine: no friction, no finite-temperature-difference heat exchange and no other entropy production. A Carnot cycle provides an implementation with two reversible isotherms and two reversible adiabats. All temperatures in these ratios are absolute.

Let $n_{\mathrm{mol}}$ be the fixed gas amount, $r=V_2/V_1>1$, $\Delta T=T_h-T_c$, and let $C_V$ denote the heat capacity of the whole sample. First take the unregenerated Stirling cycle, so that heat rejected during isochoric cooling is not stored and returned during isochoric heating. The reversible quasistatic gas work iswhile the heat inputs are $Q_{12}=n_{\mathrm{mol}}R{T}_h\ln r$ and $Q_{41}=C_V\Delta T$. HenceThe strict inequality follows because $C_V\Delta T>0$ adds heat input without additional work. Quasistatic motion alone does not make heat exchange with a single hot or cold reservoir reversible while the gas temperature changes on an isochore. Indeed, using those two reservoirs gives total external entropy production $C_V\Delta T(1/T_c-1/T_h)>0$ per cycle.