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In the Carnot cycle, let $1\to 2$ be reversible isothermal expansion at $T_h$, $2\to 3$ reversible adiabatic expansion, $3\to 4$ isothermal compression at $T_c$, and $4\to 1$ adiabatic compression. The pressure–volume curve is clockwise. The entropy–temperature figure below uses temperature on the horizontal axis and entropy on the vertical axis, literally an $S$–$T$ plot: isotherms are vertical and adiabats horizontal. The hot isotherm increases entropy by $\Delta S$ and absorbs $Q_h=T_h\Delta S$; the cold one rejects $Q_c=T_c\Delta S$. Thus $W=(T_h-T_c)\Delta S$ and ${\eta }_C=1-T_c/T_h$.

For the second engine, choose the working direction with heat addition during $A\to B$ at $p_1$ and rejection during $C\to D$ at $p_2$. A reversible ideal-gas adiabat obeys $T\propto p^{(\gamma -1)/\gamma }$. DefineThe two constant-pressure heat transfers are $Q_{\mathrm{in}}=c_p(T_B-T_A)$ and $Q_{\mathrm{out}}=c_p(T_C-T_D)=Q_{\mathrm{in}}/r$, where rejected heat is quoted positive. Therefore the Brayton cycle efficiency isfor $\gamma =5/3$. All heat-capacity amount conventions cancel in the ratio.

Since $T_B>T_A$ and $T_C=T_B/r$, $T_D=T_A/r$, the maximum is at B and the minimum at D. The Brayton-Carnot efficiency comparison isThe inequality is strict for a nondegenerate cycle with positive heat input. Although each process can be reversible, the isobaric heat transfer spans a range of temperatures; making the whole cycle reversible requires a succession of matched-temperature reservoirs, rather than just the hottest and coldest reservoirs.

For an ideal diatomic gas with active rotation and frozen vibration, $\gamma =7/5$ andHere $0<p_2/p_1<1$. This comparison holds at the same pressure ratio and assumes temperature-independent heat capacities in the stated mode regime. Frozen rotation or excited vibration would require their actual heat capacities.

A microstate specifies the microscopic state: classically the positions and momenta of all particles, or quantum mechanically an allowed many-body state. A macrostate specifies a few coarse variables such as energy, volume and particle number. Many microstates can give the same macrostate. A statistical ensemble is a probability assignment over hypothetical copies prepared with the same macroscopic constraints; it does not require a sample to be literally divided into many physical copies. An observable's ensemble mean is $⟨A⟩={\sum }_i{p}_i{A}_i$, with an integral replacing the sum in classical phase space.

For an isolated system at fixed $U,V,N$, the microcanonical ensemble assigns equal probability to accessible microstates in a narrow energy shell. If their number is $\Omega (U,V,N)$, the Boltzmann entropy is $S=k_B\ln \Omega$. Thermal contact between two initially isolated systems allows energy to be shared while their sum stays fixed. The most probable division maximizes $S_1(U_1)+S_2(U-U_1)$, so equilibrium requires equal $\partial S/\partial U$. This gives the statistical temperature definitionFor positive temperature, energy tends to flow from lower entropy slope to higher entropy slope until those slopes agree. The narrow energy-shell width represents macroscopic resolution; it must not exclude the physically accessible states.

For a small subsystem exchanging energy with a much larger bath, the bath has energy $U_{\mathrm{tot}}-E_i$ when subsystem microstate $i$ has energy $E_i$. Expanding its entropy gives $S_b(U_{\mathrm{tot}}-E_i)=S_b(U_{\mathrm{tot}})-E_i/T+\cdots$. Counting compatible bath states therefore yields the Boltzmann distribution and canonical ensemble:The exponential weights individual microstates. An energy level of degeneracy $g_j$ has total probability $g_j{e}^{-\beta E_j}/Z$; a more degenerate excited level can outweigh one ground-state level even though each of its microstates is less probable. The Gibbs entropy $-k_B{\sum }_i{p}_i\ln p_i$ gives $S=k_B\ln Z+U/T$. The equilibrium energy variance is $⟨(\Delta E{)}^2⟩=k_B{T}^2{c}_V$, linking measurable fluctuations to a macroscopic response.

If particle exchange is also allowed, a reservoir fixes $T$ and chemical potential $\mu$. The grand canonical ensemble weights a state by $e^{-\beta (E_i-\mu N_i)}/\Xi$, with $\Xi ={\sum }_{N,i}{e}^{-\beta (E_i-\mu N)}$. Mean particle number is $k_{B}T\partial \ln \Xi /\partial \mu$, while volume remains a constraint in this version. Thus the ensemble follows the physical exchange conditions: fixed energy, thermal contact, or thermal and particle contact.

In ordinary large short-range systems, relative fluctuations often decrease as $N^{-1/2}$ and the ensembles give the same bulk thermodynamics away from singular limits. Finite systems, critical fluctuations and long-range interactions require care. Interpreting a time average as an ensemble average additionally requires appropriate sampling or mixing dynamics; it is not guaranteed solely by naming an ensemble. thermal equilibrium probability assignments do not describe every driven or relaxing state.

For the entropy balance of a thermostatted resistor, electrical work becomes heat $Q=I^{2}Rt=2.0\times 10^7\mathrm{J}$. At fixed temperature the resistor has no net change of thermodynamic state; it exports this heat to its thermostat. Treating the electrical supply as a source of work,Entropy is produced by Joule heating even though it does not accumulate in the resistor. Additional entropy production inside a nonideal power supply is outside the specified model.

For a monatomic ideal gas, $\gamma =C_P/C_V=5/3$. If the expansion is reversible adiabatic, $T{V}^{\gamma -1}$ is constant, soAdiabatic isolation alone does not imply this temperature law. For example, free expansion into vacuum has $Q=W=0$, so an ideal gas remains at $300\mathrm{K}$. Thus $75\mathrm{K}$ is the usual intended answer with the additional isentropic assumption; the literal adjective “adiabatic” does not uniquely select it.^[ref]

Use absolute thermodynamic temperatures and assume a constant-parameter Arrhenius equation, with composition held fixed so that reaction rate is proportional to the rate constant. Write $R(T)=A{e}^{-\beta /T}$, where $\beta =E_a/\mathcal{R}$. two-temperature Arrhenius extrapolation eliminates the unknown Arrhenius prefactor and givesConsequentlyA tenfold increase for one $20\mathrm{K}$ interval does not imply another tenfold increase for the next interval: the Arrhenius plot is linear in inverse temperature. Two measured rates alone do not specify a unique extrapolation without a rate law for its temperature dependence; the result above states that modelling assumption explicitly.

The Stefan–Boltzmann law gives solar luminosity $L_{\odot }=4\pi R_{\odot }^2\sigma T_{\odot }^4$. At orbital distance $r$, the incident irradiance is $\sigma T_{\odot }^4{R}_{\odot }^2/r^2$. A spherical absorber intercepts a disc of area $\pi R_M^2$ but radiates over $4\pi R_M^2$. Thus the uniform-temperature radiative equilibrium estimate isThe planetary radius cancels. This is a black-body radiation estimate with zero reflected fraction and complete redistribution of absorbed heat. More generally the balance fixes $⟨T^4{⟩}^{1/4}$, rather than the arithmetic mean of a nonuniform surface temperature. It excludes atmospheric heating and thermal lag; no measured surface temperature is inferred.

Polarization density $\mathbf{P}$ is electric dipole moment per unit volume. Displacing positive and negative bound charges leaves the bulk nearly neutral locally but can produce uncompensated charge where polarization varies or terminates. For a dielectric occupying a volume with outward unit normal $\mathbf{n}$,The divergence theorem makes their total sum zero for a body described entirely by bound dipoles. Uniform polarization gives no interior volume charge but generally gives opposite surface charges on different faces.

In a linear isotropic dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ and $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}=ϵ\mathbf{E}$. Gauss's law becomes $\nabla \cdot \mathbf{D}={\rho }_f$, separating free from bound charge. Across a static interface, tangential electric field is continuous, while the normal displacement jumps by the free surface charge. Polarization charges alter the internal field, as in the dielectric cylinder calculation. Molecular distortion and orientation contribute differently, and response can depend on temperature and frequency. Bound charge is a redistribution of material charge, not an additional free-charge reservoir.

Write magnetization as magnetic dipole moment per volume. In weak-field linear response, $\mathbf{M}=\chi \mathbf{H}$ and $\mathbf{B}={\mu }_0(\mathbf{H}+\mathbf{M})$.

Paramagnetism has positive susceptibility: permanent spin or orbital moments tend to align with a field against thermal disorder. Independent localized moments in the classical weak-field regime give $\chi \propto 1/T$, the Curie law. Conduction-electron Pauli paramagnetism has a different origin and need not follow that temperature law.

Diamagnetism is the induced magnetic response opposing the applied field, giving negative susceptibility. Changes in orbital motion underlie this response; it exists even without permanent moments. It is often relatively weak and only weakly temperature dependent, although particular quantum systems have stronger or more complicated behaviour.

Ferromagnetism arises from interactions favouring parallel magnetic order. Below the Curie temperature a domain can have spontaneous magnetization even at zero applied field. A sample can nevertheless have nearly zero net moment if differently oriented domains cancel. Domain rearrangement and rotation produce a nonlinear magnetization curve, saturation, remanence and coercivity; hysteresis dissipates energy in a field cycle. Above the ordering temperature the spontaneous order disappears, and a Curie-Weiss susceptibility is a useful mean-field approximation. The signs of weak-field susceptibility distinguish para- and diamagnetism; spontaneous order and domain dynamics distinguish ferromagnetism from independent-moment alignment.

At coexistence, the two phases have equal chemical potential. For a single component, $d\mu =-sdT+vdp$, with entropy and volume measured on the same molar or mass basis. Equating the phase derivatives along coexistence gives $\Delta vdp=\Delta sdT$. Reversible latent heat satisfies $L=T\Delta s$, henceHere $p$ and $T$ are coexistence pressure and absolute temperature, $L$ is latent heat for transforming phase 1 to phase 2, and $\Delta v=v_2-v_1$ uses the same amount of material as $L$.

For vaporization, neglect liquid molar volume compared with ideal-vapour volume $RT/p$. The ideal-vapour Clausius-Clapeyron approximation becomesUsing the normal boiling reference $T_b=373.15\mathrm{K}$, $p_b=1.01325\times 10^5\mathrm{Pa}$ and the supplied constant $L_e$ givesThis deliberately constant-latent-heat extrapolation does not reproduce the supplied triple-point pressure exactly. It predicts the liquid-vapour branch, metastable against ice below the triple-point temperature; it is not a precision sublimation-pressure prediction.

The phase diagram has solid, liquid and vapour regions meeting at the triple point. The sublimation and vaporization boundaries have positive slopes, whereas the ordinary ice–water boundary slopes negatively because ice has larger specific volume. The liquid–vapour line terminates at a critical point; the original schematic below emphasizes topology and marks the supplied triple point.

For the melting-pressure estimate, Archimedes' principle gives ${\rho }_i=0.90{\rho }_w=900\mathrm{kg}{\mathrm{m}}^{-3}$, soConvert the molar evaporation heat using water molar mass $M\simeq 0.0180\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, rounded from the tabulated value $18.0153\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref] An enthalpy cycle through vapour then givesThis uses the question's temperature-independent evaporation heat all the way to the triple point; no replacement experimental fusion heat is inserted. The melting pressure inferred from flotation and latent heats, on holding $L_f$ and $\Delta v$ constant, isusing $T_t=273.16\mathrm{K}$ and $T=270.15\mathrm{K}$. Linearizing the logarithm about the triple point gives $53.4\mathrm{MPa}$ instead of $53.7\mathrm{MPa}$. Both are estimates within the stated constant-property approximation; the negative melting slope makes a positive pressure necessary to lower the melting temperature.

Capacity convention. The printed equation requires $c$ to be volumetric heat capacity, in $\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. If “specific heat capacity” means per unit mass instead, the storage term is ${\rho }_{m}c{\partial }_{t}T$ and both denominators in the printed equation must contain ${\rho }_{m}c$. This distinction is fixed by dimensions.

For a fixed material region $V$, conservation of thermal energy gives the heat balance with volumetric heat generation,Use the divergence theorem, $\mathbf{J}=-\kappa \nabla T$ and constant coefficients. Since the region is arbitrary,The heat source is power per volume, while $\kappa$ is thermal conductivity rather than thermal diffusivity.

For uniform electrical conductivity and imposed current, current density is $j=I/(\pi a^2)$ and Joule heating is $H=j^2/\sigma$. A long, steady, cylindrically symmetric wire therefore satisfiesIntegration gives $T^{\prime }=-Hs/(2\kappa )+C/s$. Regularity at $s=0$ makes $C=0$, and the surface condition fixes the remaining constant. Thus the Joule-heated cylindrical wire hasHere $s$ denotes the printed radial coordinate $\rho$ and avoids confusing it with mass density. The centre is hotter by $I^2/(4{\pi }^2{a}^2\kappa \sigma )$.

For the air boundary, the internal flux at $s=a$ is $-\kappa T^{\prime }(a)=Ha/2$. Equate this to $\alpha [T(a)-T_{\mathrm{air}}]$. The Newton cooling of a Joule-heated wire givesMultiplication by circumference $2\pi a$ verifies that surface loss per length equals $I^2/(\pi a^2\sigma )$, the electrical power per length. The full air-cooled profile uses this surface temperature plus the same parabolic interior excess, assuming constant material properties and negligible end losses.

A heat engine's thermal efficiency is net work out divided by heat absorbed,where both heat quantities are positive magnitudes. The last equality follows from the first law over a complete cycle.

Use $n$ moles of an ideal gas with constant molar $C_V,C_p$ and $\gamma =C_p/C_V>1$. For the Otto cycle, reversible adiabats give $T_2=T_1{r}^{\gamma -1}$ and $T_3/T_4=r^{\gamma -1}$. Isochoric heat input and rejection are $Q_{\mathrm{in}}=n{C}_V(T_3-T_2)$ and $Q_{\mathrm{out}}=n{C}_V(T_4-T_1)$. ThereforeThe denominator's temperature difference is $r^{\gamma -1}$ times the numerator's difference.

For the ideal Atkinson cycle, heating at fixed volume gives $T_3/T_2=r_p$. The longer expansion ends when $p_{4^{\prime }}=p_1$. Combining the adiabatic pressure–temperature relation with $p_2/p_1=r^{\gamma }$ yieldsAlso $V_{4^{\prime }}=V_1{r}_p^{1/\gamma }>V_1$ for $r_p>1$. The PDF asks explicitly for $T_3,T_{4^{\prime }}$; the transcription additionally names $T_2$, which is supplied above too.

Heat input remains $n{C}_V{T}_1{r}^{\gamma -1}(r_p-1)$, while the isobaric return rejects $n{C}_p{T}_1(r_p^{1/\gamma }-1)$. Hence the Atkinson efficiency advantage from extended expansion is quantified byCompare cycles with the same initial state, compression ratio and heat input. The extra expansion and constant-pressure return enclose positive extra work,It uses heat that the Otto cycle would reject, while $Q_{\mathrm{in}}$ is unchanged. Equivalently, concavity gives $\gamma (r_p^{1/\gamma }-1)<r_p-1$, so ${\eta }_A>{\eta }_O$ for $r_p>1$ under this matched comparison. This is an ideal-cycle result; the question's named real-engine application is not needed to establish it.

With the standard liquid-water approximation $\rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, each portion has mass $100\mathrm{kg}$. The equal heat capacities give final temperature $T_f=310\mathrm{K}$ by energy conservation. For an insulated mixing process with no other changes,The reversible integrals $mc\int dT/T$ evaluate entropy differences between equilibrium states; the actual heat transfer is irreversible. These are two portions of the same substance, so there is no additional ideal-mixture compositional entropy.

Let absorbed solar flux per unit surface area be $F_{\odot }=\sigma T_0^4$. The single-layer greenhouse model emits infrared upward and downward. Top-of-atmosphere balance gives $\sigma T_a^4=F_{\odot }$, hence $T_a=T_0$. Surface balance is $\sigma T_s^4=F_{\odot }+\sigma T_a^4$, soThe layer simultaneously satisfies $\sigma T_s^4=2\sigma T_a^4$. This ideal radiative model excludes convection, spectral windows and atmospheric temperature structure; it is not a quantitative prediction for the real atmosphere.

Use the rotating reference frame equilibrium model, in which the centrifugal potential energy of a particle of mass $m_i$ is $U_i(r)=-m_i{\omega }^2{r}^2/2$. The Boltzmann distribution then givesDefine the concentration ratio as $c=n_1/n_2$. At the axis,Thus the axis favours the lighter species. The expression assumes a dilute classical mixture at one temperature, equilibrium in the rotating reference frame, and sampling sufficiently close to $r=0$.

Magnetization $\mathbf{M}$ is magnetic dipole moment per unit volume. In SI units,in a linear isotropic response. The internal field matters: for a uniformly magnetized ellipsoid it is ${\mathbf{H}}_{\mathrm{int}}={\mathbf{H}}_{\mathrm{applied}}-{\mathbf{N}}_d\mathbf{M}$. Sample shape therefore affects an apparent magnetic susceptibility.

Diamagnetism opposes the applied field, giving small negative magnetic susceptibility from the induced orbital response. It is often only weakly temperature dependent. Paramagnetism aligns pre-existing moments and gives positive magnetic susceptibility; independent localized moments obey approximately $\chi =C/T$ in weak fields before saturating. Pauli paramagnetism of conduction electrons is a different, weakly temperature-dependent mechanism.

Ferromagnetism involves collective exchange-driven alignment and spontaneous magnetization below the Curie temperature. A specimen can nevertheless have nearly zero total moment because magnetic domains point in different directions. Field-driven domain-wall motion and rotation change the net magnetization. Antiferromagnets have compensating ordered sublattices below a Néel temperature; ferrimagnets have oppositely ordered but unequal sublattice moments and retain a net moment. Ferrites are useful examples of electrically resistive ferrimagnetic core materials.

Ferromagnetic response is nonlinear and history dependent. A hysteresis loop displays remanent magnetization, magnetic coercivity and saturation. A single constant magnetic permeability cannot describe the entire loop: differential magnetic permeability describes its local slope. The quasistatic energy loss per unit volume and cycle is the loop area $\oint HdB$. Soft magnetic materials have low magnetic coercivity and usually high useful magnetic permeability; hard magnetic materials retain magnetization and suit permanent magnets.

An electromagnet normally uses a soft, high-permeability core to guide flux generated by a coil. For a uniform-section core with length ${\ell }_c$ and air gap $g$, neglecting leakage and fringing,An air gap can dominate the magnetic reluctance even when the core magnetic permeability is large. Core saturation limits the improvement with current, while winding resistance causes heating. For alternating operation, laminations suppress eddy-current loops in metallic cores; high-resistivity ferrites also reduce those losses. Low hysteresis loss is important in repeatedly cycled cores, whereas high magnetic coercivity is valuable in a permanent magnet. Useful design therefore depends on the full magnetic response, electrical losses and operating frequency, rather than the largest quoted magnetic permeability alone.

For speed distribution $f(v)$ normalized to one and isotropic directions, the incident number flux isThe positive normal velocity weights the population reaching the wall. Consequently its mean translational energy isThe bulk value $3k_{B}T/2$ would omit the arrival weighting.

For the two vessels, let $r=\sqrt{T_1/T_0}$, with subscripts h and c for hot and cold. Equal opposing effusion rates give $p_h/\sqrt{T_1}=p_c/\sqrt{T_0}$. Number conservation in equal, fixed volumes adds $p_h/T_1+p_c/T_0=2p_0/T_0$. HenceBoth final pressures must be determined; holding the cold temperature fixed does not hold its particle number fixed.

For aperture area $A_h=\pi d^2/4$, the common one-way flux is $J=p_c/\sqrt{2\pi m{k}_B{T}_0}$. The translational heat transfer from hot to cold is thereforeThis is the total heat transfer for a monatomic gas with no active internal modes, neglecting radiation and conduction through vessel walls. For a molecular gas, add $A_{h}J[u_{\mathrm{int}}(T_1)-u_{\mathrm{int}}(T_0)]$ for its internal energy per molecule. The maintained unequal temperatures describe a stationary nonequilibrium state even though the net particle current is zero.

A thermodynamic potential packages the equilibrium state in variables suited to the imposed constraints. For a simple fluid, $U(S,V,N)$ has differential $dU=TdS-pdV+\mu dN$. Legendre transforms give $H=U+pV$, $F=U-TS$ and $G=U+pV-TS$, with natural variables of a thermodynamic potential $(S,p,N)$, $(T,V,N)$ and $(T,p,N)$ respectively. At fixed particle number, Helmholtz free energy is minimized at fixed $T,V$; Gibbs free energy at fixed $T,p$.

A flow vessel is an open control volume, even if its walls are enclosed. For each inlet, pressure forces carry flow work $pdV$ in addition to internal energy. The steady first-law balance isWith no heat exchange, shaft work, or significant kinetic and gravitational energy differences, the enthalpy fluxes are equal. Mass flow must also balance. This equality concerns mass-flow-weighted specific enthalpies, not an unweighted sum over pipes.

Use the liquefaction yield with separate warm-gas and liquid outlets. The wording specifies that the outgoing gas is at 14 K; it does not require the outgoing liquid to have that temperature. The two outlet streams need not be in mutual thermal equilibrium. Taking $H$ as the printed specific enthalpy, the steady-flow energy balance givesFor fixed outlet enthalpies the greatest yield occurs at the smallest inlet enthalpy. Among the tabulated pressures this is 30 atm. Approximating the atmospheric-pressure gas enthalpy by the tabulated zero-pressure limiting value givesThis uses a low-pressure approximation: $p_0=1$ atm is not literally the zero-pressure table entry. Linear interpolation to 1 atm would instead give $H(p_0)\simeq 8.651\times 10^4\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$ and $\alpha \simeq 0.193$. The sparse table does not determine an exact continuous optimum between entries.

Atmospheric-pressure liquid helium must be much colder than the 14 K gas outlet: its normal boiling point is about 4.224 K.^[ref] A counterflow heat exchanger can return cold gas through the incoming stream before that gas leaves the overall control volume, while the collected liquid leaves separately at the cold end. This is a possible interpretation of the whole apparatus, rather than an extra diagram specified by the question. The enthalpy calculation needs only the stated inlet and outlet properties; it does not assume one equilibrium vessel containing both phases at 14 K. In particular, $H(p_0)-H_{\ell }$ is an outlet-stream enthalpy difference, not the latent heat of vaporization at one common temperature.

The coldest day has outside temperature $-5^{\circ }\mathrm{C}$, so the indoor–outdoor temperature difference is 25 K. If $K$ is the building's heat-loss coefficient, the required heating rate is ${\dot{Q}}_H=25K$. For direct electrical heating,This sets the reference electrical cost for the same delivered heat and tariff.

A heat pump delivers $Q_H=Q_S+W$: electrical work moves heat from a source as well as contributing energy itself. Its heating coefficient of performance can therefore exceed one, whereas direct resistance heating supplies at most its electrical input as useful room heat. For an ideal reversible pump,with absolute temperatures and $T_S<T_H$. This is an ideal limit; actual costs also depend on compressor performance and tariffs.

For the annual ground-temperature wave, model the soil as a homogeneous semi-infinite conductor, with depth positive downward and a prescribed surface temperature. Here $\kappa$ is thermal conductivity and $C$ is heat capacity per unit volume, so the thermal diffusivity is $\kappa /C$. Conservation of energy gives $C{\partial }_{t}T=\kappa {\partial }_z^{2}T$. After transients decay, put $T-T_0=\mathrm{Re}[A{e}^{i\omega t-qz}]$. Substitution gives $q^2=i\omega C/\kappa$; the solution bounded at depth uses $q=(1+i)/\delta$. Matching the surface then givesThis approximation identifies the ground surface with air temperature and neglects the pump's perturbation of the ground, as required by the model. The attenuation and phase delay arise from heat diffusion, not propagation at one fixed wave speed.

The graph below covers a full annual cycle. At $z/\delta =\pi /2$, the amplitude is $20e^{-\pi /2}=4.16$ K and the minimum lags January by a quarter-year. At $z/\delta =\pi$, the amplitude is only $0.864$ K and the minimum lags by half a year: the ground is slightly warmer than its annual mean in January.

The seasonal source depth for a reversible heat pump has a subtle limit: a deep source approaches $T_0$ and avoids severe winter cooling, but infinite depth is not the thermal optimum of this single-frequency model. In January, with $x=z/\delta$, maximize $T_S=T_0-T_A{e}^{-x}\cos x$. The first and largest maximum is at $x=3\pi /4$, where $\cos x+\sin x=0$, giving $z=3\pi \delta /4$ and $T_S\simeq 16.34^{\circ }\mathrm{C}$. It is also the optimum of the ideal annually integrated heating work: the heating load is even about January, so its weighted sine term integrates to zero and the depth dependence is again proportional to $e^{-x}\cos x$. Installation cost, finite tube spacing, heat extraction, weather and real pump performance alter this result. The supplied model therefore supports a finite seasonal thermal optimum, while not fixing a practical burial depth in metres.

Phase equilibrium minimizes Gibbs free energy at specified temperature, pressure and composition. Coexisting phases have equal temperature, pressure and chemical potential. For a pure substance, a first-order transition has continuous equilibrium Gibbs free energy but a jump in its first derivatives, such as entropy or volume; the latent heat is $T\Delta S$. Continuous transitions lack that latent heat and can instead have singular or discontinuous higher responses.

The van der Waals molar equationcombines an excluded-volume correction with attractive interactions. Its inflection conditions giveBelow $T_c$, the oscillatory isotherm contains a mechanically unstable segment with $(\partial p/\partial v{)}_T>0$. Equilibrium instead replaces the loop by a constant-pressure coexistence line. Equality of chemical potentials gives the Maxwell construction,On isothermal compression the gas reaches saturation, then converts progressively into liquid at fixed pressure. The mean molar volume obeys $v=(1-x)v_{\ell }+x{v}_g$, where $x$ is the gas fraction. Metastable superheated or supercooled states can persist when nucleation is delayed; they are not the global equilibrium selected by this construction.

A pressure–temperature phase diagram contains coexistence curves separating solid, liquid and gas. Three phases meet at a triple point; the liquid–gas coexistence curve ends at a thermodynamic critical point, above which liquid and gas can be connected without crossing a first-order phase boundary. A generic solid–liquid line continues beyond the liquid–gas critical temperature. For one component the Gibbs phase rule is $F=3-P$, so a three-phase equilibrium fixes both temperature and pressure.

Along a two-phase coexistence curve, $d\mu =-sdT+vdp$ for each phase givesThis is the Clapeyron equation, valid with consistent molar or specific quantities. For vaporization, neglecting the liquid volume and treating the vapour as ideal gives $\Delta v\simeq RT/p$, hence $d\ln p/dT=L/(R{T}^2)$. With approximately constant molar latent heat,This Clausius–Clapeyron approximation explains the rapid rise of vapour pressure with temperature. It fails close to the thermodynamic critical point, where phase volumes converge, and is not a universal expression for melting. The sign of a melting-line slope depends on its volume change; water's denser liquid gives a familiar negative slope.