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The three-dimensional Debye model normalizes its phonon density of states to $3N$ modes: $g(\omega )=9N{\omega }^2/{\omega }_D^3$ below the Debye frequency ${\omega }_D$. The thermal energy is $U={\int }_0^{{\omega }_D}g(\omega )\hslash \omega /[e^{\hslash \omega /(k_{B}T)}-1]d\omega$. Differentiate at fixed volume and substitute $x=\hslash \omega /(k_{B}T)$ to obtain the displayed heat capacity at constant volume.

At $T\ll {\Theta }_D$, integration by parts gives ${\int }_0^{\infty }{x}^4{e}^x/(e^x-1{)}^{2}dx=4{\int }_0^{\infty }{x}^3/(e^x-1)dx=4{\pi }^4/15$. Therefore $C_V\sim (12{\pi }^4/5)N{k}_B(T/{\Theta }_D{)}^3$. At $T\gg {\Theta }_D$, each mode approaches the equipartition theorem value $k_B$ of heat capacity, so $C_V\to 3N{k}_B$, the Dulong-Petit law. Fixed harmonic zero-point energy does not contribute to either limit.

The effective isotropic acoustic speed used to replace separate branches in the Debye model. With atomic number density $n$, counting three branches gives $k_D=(6{\pi }^{2}n{)}^{1/3}$ and ${\omega }_D=v_D{k}_D=k_B{\theta }_D/\hslash$. Matching the low-frequency density of states for one longitudinal and two transverse branches gives the displayed inverse-cube average. A Debye temperature alone does not determine both individual sound speeds.

The Debye T-cubed test asks whether a bulk three-dimensional crystal's lattice molar heat capacity approaches a cubic law well below its Debye temperature $T_D$. It tests the low-frequency acoustic phonon branches in the Debye model, rather than asserting that every contribution to a real sample's heat capacity is cubic. Peter Debye's continuum description makes a striking connection: a thermal measurement can reveal the low-frequency vibrational density of states without resolving individual vibrations.^[ref]

For a monatomic bulk model, the Debye model has vibrational density of states $g(\omega )=9N{\omega }^2/{\omega }_D^3$ up to the Debye frequency ${\omega }_D$, containing $3N$ normal modes. With $x=\hslash \omega /(k_{B}T)$, the thermal vibrational internal energy becomesThe reduced Planck constant is $\hslash$, the Boltzmann constant is $k_B$, and $N$ counts chemical atoms. For $T\ll T_D$, replace the upper limit by infinity. The integral is ${\pi }^4/15$, as in the Stefan–Boltzmann law derivation. Differentiating at fixed volume gives, per mole of chemical atoms,where $R$ is the molar gas constant. The cubic exponent comes from three-dimensional acoustic phonon branches with linear low-wavevector dispersion relations. The coefficient encodes their propagation speeds. Formula-unit normalization in a compound requires its number of chemical atoms per formula unit; do not silently substitute moles of formula units for moles of chemical atoms.

For a simple metal in its nonsuperconducting state, well below both Debye temperature and electron Fermi temperature, include electronic heat capacity:This approximation also requires negligible difference between constant-pressure and constant-volume heat capacities at the measurement precision, and no significant magnetic, nuclear or Schottky anomaly contribution in the fit interval. Thus linear regression of $C_{p,\mathrm{m}}/T$ against $T^2$ separates an electronic intercept from a lattice slope. For an insulating crystal with no relevant low-energy magnetic excitations, the electronic term can instead be negligible; a superconductivity transition changes the electronic response.^[ref]

After subtracting calorimeter addenda, Brando's second gold sample, $72.6\mathrm{mg}$, gave a linear $C_{p,\mathrm{m}}/T$ versus $T^2$ plot in the nonsuperconducting state. The reported electronic intercept was $7.0\times 10^{-4}\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-2}$; deviations from reference heat capacities were below $2\%$.^[ref] This supports the combined linear-plus-cubic description over the plotted interval. Because the platform subtraction used reference gold, it is also an instrument validation with a shared material standard, rather than a wholly independent determination of the Debye model from an uncalibrated instrument.

For a worked original inference, stipulate values of $y=C_{p,\mathrm{m}}/T$ equal to $(0.001100,0.002300,0.004300)\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-2}$ at $T=(1,2,3)\mathrm{K}$. Assign independent $u(y)=0.000020\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-2}$ and negligible temperature error. Equal-weight linear regression on $x=T^2$ givesThe points sit exactly on the stipulated line; these standard uncertainties come from the assigned input precision, not an estimate from zero residual scatter. With $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, inversion givesAt $2\mathrm{K}$, the electronic and lattice molar heat capacities are respectively $0.0014$ and $0.0032\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. Their equality occurs at $\sqrt{\gamma /\beta }=1.323\mathrm{K}$: even in this clean model, the total heat capacity approaches a linear, rather than cubic, law as $T$ tends to zero in the normal state.

A straight line over a narrow interval is a useful diagnostic, not a universal proof. Inspect residuals and repeat the fit over several upper and lower temperature cutoffs; curvature can signal higher lattice terms, an unremoved Schottky anomaly, or a faulty calorimeter addenda subtraction. Correlated thermometer calibration and sample-amount errors affect both axes and can dominate the illustrative linear regression uncertainty. Low-dimensional samples, additional low-energy excitations and proximity to a phase transition need other models. The test is valuable precisely because its failures can reveal physics beyond the bulk acoustic approximation.

The spherical cutoff in the three-dimensional Debye model preserves three vibrational modes per atom. For atom density $n_a$, $q_D=(6{\pi }^2{n}_a{)}^{1/3}$ and ${\omega }_D=c_s{q}_D$ in a single-speed approximation. A monovalent free-electron metal has $k_F=(3{\pi }^2{n}_a{)}^{1/3}$, giving $q_D=2^{1/3}{k}_F$. This cutoff is a mode-counting approximation, not an exact spherical boundary of the crystal's Brillouin zone.

The Einstein model of a solid represents the vibrations of $N$ chemical atoms by $3N$ independent one-dimensional quantum harmonic oscillators, all with the same positive angular frequency $\omega$. Equivalently, each chemical atom has three independent oscillator directions. The common angular frequency is a simplifying model assumption, not a claim about every vibration of a real solid. Hold the volume, $N$ and $\omega$ fixed. Define the characteristic temperature ${\theta }_E=\hslash \omega /k_B$ and the dimensionless ratio $x={\theta }_E/T$. The vibrational internal energy isThe first term is the zero-point energy $3N\hslash \omega /2$. It is constant at fixed $\omega$, so it contributes no heat capacity at constant volume. The temperature-dependent term givesThis is the vibrational heat capacity at constant volume of the model, with no expansion or other contributions included. ^[ref]

For a worked comparison, choose the model parameter ${\theta }_E=300\mathrm{K}$. The normalized heat capacity at constant volume depends only on $T/{\theta }_E$:

At $T={\theta }_E$, one mole of chemical atoms therefore has vibrational heat capacity at constant volume $3R(0.92067)\simeq 22.965\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, using the unrounded model value. Even at twice the characteristic temperature, the model remains about $2.06\%$ below the Dulong-Petit law limit. These are calculations for the ideal Einstein solid, not material measurements.

A short derivation uses the thermal partition function of a quantum harmonic oscillator. The energy eigenvalues $E_n=\hslash \omega (n+1/2)$ and their Boltzmann factors form a geometric series, giving the one-oscillator canonical partition function $z=e^{-x/2}/(1-e^{-x})$. The mean occupation is $1/(e^x-1)$, hence the displayed $U$ after summing over $3N$ independent quantum harmonic oscillators. Taking its derivative with $dx/dT=-x/T$ gives the displayed heat capacity at constant volume. ^[ref]

The two limits explain the shape. At high temperature, a Taylor series in small $x$ givesAt low temperature, $1-e^{-x}\to 1$, soThe zero-point energy remains even when the thermal response is tiny: freeze-out means that changing temperature scarcely changes the occupations, not that the chemical atoms have zero energy. This is the useful lesson of the Einstein model of a solid: the classical counting rule reappears when many levels are accessible, while discrete excitation spacings explain its failure upon cooling.

The exponential low-temperature decay is also the model's limitation. Real three-dimensional crystalline solids have acoustic phonon branches with arbitrarily small angular frequencies in the bulk limit. The Debye model includes these collective normal modes and predicts a low-temperature vibrational contribution $C_V\propto T^3$, which the single-frequency Einstein solid cannot reproduce. This qualification concerns the vibrational contribution; additional excitations can change a material's total heat capacity. ^[ref]

Einstein introduced the quantum treatment of solid vibrations in his 1907 paper Die Plancksche Theorie der Strahlung und die Theorie der spezifischen Wärme. Its historical importance is that quantum excitation energies explained why the classical Dulong-Petit law could fail; the modern quantum harmonic oscillator formulation above includes the zero-point energy explicitly. ^[ref]

Choose a positive constant $\Sigma$ and the illustrative interface lawHere $T_h$ and $T_c$ are the hot and cold interface-side temperatures, not distant thermometer readings. This is a low-temperature phonon model under restrictive assumptions, not a universal fit to Kapitza's measured exponent or a material-property table. Its coefficient $\Sigma$ is not the Stefan-Boltzmann constant: the carriers in this model are phonons, rather than photons crossing a vacuum.

The fourth power has a simple counting explanation. In the three-dimensional low-temperature Debye model, a linear phonon dispersion relation gives a density of states proportional to ${\omega }^2$. A phonon has energy $\hslash \omega$ and Bose-Einstein distribution occupation $[\exp (\hslash \omega /k_{B}T)-1{]}^{-1}$. If propagation speeds and the angle-averaged transmission introduce no additional frequency scale over the occupied modes, the outgoing energy flux has the formExtending the upper integration limit assumes occupied modes lie well below the cutoff. Balancing the two outgoing fluxes, with coefficients satisfying zero net heat flux at equal temperatures, gives the chosen difference of fourth powers. Frequency-dependent transmission, different excitations or additional transport channels can change this conclusion.

The finite-drop area-specific thermal resistance follows by factoring a difference of powers:Near thermal equilibrium, write $T_h=T+\delta T$ and $T_c=T$. Expansion gives $q=4\Sigma T^3\delta T+O((\delta T{)}^2)$, so the differential area-specific Kapitza resistance isAt a finite bias, varying only the hot side gives $(\partial q/\partial T_h{)}^{-1}=1/(4\Sigma T_h^3)$; this is generally different from $r_{K,\sec }$. Stating what is held fixed makes a differential resistance reproducible.

For an original normalized example, choose reference scales $T_0>0$ and $A_0>0$, and setFor contact area $A=2A_0$, the total heat transfer rate is $\dot{Q}=2.1472A_0\Sigma T_0^4$. The finite-drop resistance satisfies $\Sigma T_0^3{r}_{K,\sec }=0.2/1.0736=125/671\approx 0.18629$. Using the equilibrium tangent at $T_0$ instead would predict $j=4(0.2)=0.8$, illustrating why a substantial temperature rise requires the nonlinear law. Doubling the common equilibrium temperature reduces $R_K$ by eight in this chosen model.

A series check can be done without guessing any material properties. Stipulate a hot-side slab with dimensionless areal bulk resistance $b_s=\Sigma T_0^3{L}_s/k_s=0.10$, and neglect the cold-side bulk drop. At the same $j=1.0736$, the hot reservoir must have $T_H/T_0=1.2+b_{s}j=1.30736$. The imposed reservoir drop splits into $0.10736$ in the bulk and $0.20$ at the interface. Applying the fourth-power law directly to $T_H$ would assign the bulk drop to the interface and overestimate the heat flux.

Finally, transfer heat at rate $Aq$ between maintained interface-side temperatures $T_h>T_c>0$. The hot side loses entropy at rate $Aq/T_h$, and the cold side gains it at rate $Aq/T_c$. With no interface entropy accumulation, the entropy production per unit area isFor the normalized example it is ${\dot{S}}_{\mathrm{prod}}/(A\Sigma T_0^3)=1.0736(1-1/1.2)=0.178933\dots$. The same energy crosses both sides in a steady state, while total entropy increases, consistently with the Second law of thermodynamics. A finite temperature jump during heat transfer therefore expresses dissipation, not a failure of energy conservation.

For three-dimensional Debye model mode density $D(\omega )=9N{\omega }^2/{\omega }_D^3$, the Bose-Einstein distribution becomes $n_{\omega }\simeq k_{B}T/(\hslash \omega )$ when $T\gg {\Theta }_D$. Integrating the occupations gives $N_{\mathrm{ph}}\simeq (9/2)NT/{\Theta }_D$. The number of phonon quanta is not the number $3N$ of vibrational modes and need not be bounded by it. If a stipulated electron-phonon scattering model makes resistivity proportional to this phonon number with a temperature-independent coefficient, the high-temperature electrical resistivity is proportional to $T$.

Three distinct effects spoil indefinite $T\propto B$ cooling: extra entropy outside the spin subsystem, additional magnetic energy scales, and irreversible heat or work input. They can be distinguished with a thermodynamic derivative. For a smooth equilibrium Helmholtz free energy obeying $dF=-SdT-mdB$, a Maxwell relation givesHere $C_B$ is the total heat capacity participating in the stage's thermal equilibrium. The ideal model has $(\partial m/\partial T{)}_B=-C_{B,\mathrm{mag}}/B$. If its spins share temperature with a field-independent lattice and load of combined heat capacity $C_{\mathrm{other}}>0$, thenThe positive slope still means cooling when $dB<0$, but its smaller magnitude gives less cooling per fractional field reduction. For example, a low-temperature Debye model gives a lattice entropy proportional to $T^3$. As that lattice entropy falls, the spins take up the released entropy; $x$ can no longer remain constant. A weak enough coupling to neglect its equilibrium interaction energy can still be strong enough to exchange heat on the sweep timescale.

Interactions between magnetic moments, including exchange interaction and the fields of neighboring magnetic dipole moments, introduce energy scales that do not vanish with the applied $B$. Once they compete with $k_{B}T$ or $\mu B$, the independent-site partition function fails. Correlations and possible magnetic ordering alter the available magnetic entropy and the cooling curve. Simply setting the external magnetic field to zero does not eliminate those interactions. Different refrigerants therefore need their own measured equilibrium properties; there is no universal final temperature from the applied-field ratio alone.

Finally, a real open heat switch has residual thermal conductance; supports, wires and radiation carry incoming heat. Field changes may also dissipate energy. Locally near thermal equilibrium, with incoming heat rate ${\dot{Q}}_{\mathrm{leak}}$ and internal entropy production rate ${\dot{S}}_{\mathrm{prod}}$, the stage balance becomesThis uses a common stage temperature; a non-equilibrated load requires separate balances. The sweep must allow spin-lattice relaxation while remaining fast enough to limit accumulated parasitic heat. At fixed $B$, the same balance explains warming and finite hold time. Useful refrigeration is the controlled absorption of a load's heat during that finite interval, followed by a renewed isothermal process of magnetization to export its entropy.

Integrating the Bose-Einstein distribution against the three-dimensional Debye model phonon density of states gives $N_{\mathrm{ph}}=9N(T/{\Theta }_D{)}^3{\int }_0^{{\Theta }_D/T}{x}^2/(e^x-1)dx$. For $T\ll {\Theta }_D$, expand $(e^x-1{)}^{-1}={\sum }_{j\ge 1}{e}^{-jx}$ and integrate termwise: the limiting integral is $2{\sum }_{j\ge 1}{j}^{-3}=2\zeta (3)$. This counts thermal quanta, not available modes or half quanta of zero-point energy.

Suppose a three-dimensional Debye model is normalized to $3N_m$ vibrational modes in volume $V$. Its low-temperature heat-capacity coefficient is the one displayed, and its spherical wavevector cutoff obeys $k_D^3=6{\pi }^2{N}_m/V$. The Debye temperature is ${\Theta }_D=\hslash v_D{k}_D/k_B$. At fixed measured $\beta$ and volume, changing the counting convention by $N_m\mapsto c{N}_m$ raises both ${\Theta }_D$ and $k_D$ by $c^{1/3}$; the inferred average sound speed is unchanged.

For a mole of formula units containing $r$ atoms each, a conventional full atomic-mode count uses $N_m=r{N}_A$ and tends to lattice heat capacity $3rR$ at high temperature. A reduced convention using $N_m=N_A$ instead has three effective modes per formula unit. These are different continuum normalizations, not a claim that the real crystal has only three modes per formula unit. A real multi-atom primitive cell has three acoustic branches and additional optical branches; a Debye sphere preserving the full atomic count is an effective representation of that spectrum, rather than its literal first Brillouin zone.

The Debye model treats a three-dimensional solid as an isotropic elastic continuum with three independent acoustic phonon polarizations. It assumes harmonic normal modes, a linear dispersion relation $\omega =vq$, and equilibrium Bose-Einstein statistics with zero phonon chemical potential. Real anisotropy and nonlinear dispersion are replaced by one effective Debye sound speed. A cutoff restores the finite number of atomic vibrational degrees of freedom; the continuum itself would otherwise have infinitely many modes.

For sample volume $V$, the wavevector state density is $V/(2\pi {)}^3$ per polarization. Counting a sphere givesThere are $3N$ modes for $N$ atoms. HenceThe associated phonon density of states isIf longitudinal and transverse sound speeds differ, use $3/v^3=1/v_L^3+2/v_T^3$ for the low-frequency count. The Debye frequency is an effective mode-counting cutoff, not a claim that every actual branch terminates at exactly that frequency.

The FCC conventional unit cell has four atoms, so the cube containsatoms, neglecting surface corrections. A phonon is one excitation quantum of a vibrational normal mode; there can be many quanta per mode, and zero-point motion does not add half a phonon to the thermal population.

For the low-temperature thermal phonon number, integrate the Bose-Einstein distribution against the Debye model phonon density of states:At $10\mathrm{K}$, ${\Theta }_D/T=21.5$, so extending the upper limit to infinity has negligible effect at the precision required. The supplied integral yieldsThis is the thermally excited phonon number, not the number $3N$ of available modes.

The continuum approximation fails when the thermally occupied wavelengths become comparable with the cube size $L$. From the Debye temperature,A lowest nontrivial acoustic frequency is of order $v/L$; a standing-wave estimate ${\omega }_{\min }\sim \pi v/L$ corresponds to $\hslash {\omega }_{\min }/k_B\sim 0.25\mathrm{K}$. Periodic boundary conditions instead give $2\pi v/L$ and $0.50\mathrm{K}$. The precise spectrum depends on mechanical boundary conditions and branch speeds, neither of which is specified.

This finite-crystal phonon freeze-out requires summing discrete normal modes. For a finite positive lowest frequency, $N_{\mathrm{ph}}\sim g_{\min }{e}^{-\hslash {\omega }_{\min }/(k_{B}T)}$, rather than extrapolating the bulk $T^3$ law indefinitely. Rigid translations and rotations of an isolated crystallite are not internal thermal phonons. Surface acoustic modes can modify the intermediate crossover, but do not justify inventing a unique boundary spectrum.

In the spin-degenerate free electron model, a monovalent metal has conduction-electron number density $n$. The density of wave-vector states is $1/(2\pi {)}^3$ per unit real-space volume, with two spin states at each wave vector. Filling a sphere at zero temperature givesFor the Debye model, there are three acoustic polarizations and $3n$ modes per unit volume in total. Replacing their dispersions by $\omega =ck$ with the effective Debye sound speed, mode counting givesThe factor two in the electron count is spin, whereas the factor three in the phonon count is polarization; the two counts are not interchangeable. If branch speeds differ, the Debye effective average obeys $c^{-3}=(c_l^{-3}+2c_t^{-3})/3$ in the isotropic approximation.

Dividing gives $T_F/{\theta }_D=\lambda (3{\pi }^{2}n/2{)}^{1/3}$, where $\lambda =\hslash /(2m_{e}c)$. A monatomic FCC conventional cell contains four atoms, so $n=4/a^3$. ThereforeHere and below $c$ is the speed of sound, not the speed of light. The Debye cutoff counts acoustic modes in a spherical approximation to the Brillouin zone; it is not an electron-state cutoff.

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

The Debye model treats the crystal as an isotropic elastic continuum with independent harmonic vibrational modes. There are three acoustic polarizations, approximated by linear dispersions and a common effective speed. A cutoff is chosen to retain $3N$ modes for $N$ atoms. The model neglects anharmonic interactions and the detailed high-wavevector dispersion. In a multi-atom primitive cell, optical branches require additional treatment; copper has one atom per primitive cell even though its conventional cubic cell contains four atoms.

A frequency interval contains, per volume,Each phonon mode has Bose-Einstein distribution occupation $(e^{\hslash \omega /(k_{B}T)}-1{)}^{-1}$. The fixed zero-point energy contributes no heat capacity. Its thermal energy density isFor $T\ll {\theta }_D$, extending the upper limit to infinity gives $u_{\mathrm{ph}}=(3{\pi }^4/5)n{k}_B{T}^4/{\theta }_D^3$. ThereforeThe cubic law reflects the three-dimensional acoustic mode density proportional to ${\omega }^2$. It is a low-temperature limit, not a valid extrapolation through the classical high-temperature regime.

At fixed chain length, each harmonic mode contributes a Bose–Einstein thermal occupation and a temperature-independent zero-point energy. The latter contributes no heat capacity. With $x=\hslash \omega /(k_{B}T)$,At low temperature only the approximately linear part of the acoustic dispersion is excited, so $g(\omega )\simeq 1/(\pi v_s)$. Using ${\int }_0^{\infty }{x}^2{e}^x/(e^x-1{)}^{2}dx={\pi }^2/3$ givesIt is linear in temperature, rather than the three-dimensional Debye $T^3$ law. At high temperature every scalar oscillator tends to $k_B$, henceThe cutoff temperature is ${\Theta }_m=\hslash {\omega }_m/k_B=305.5$ K. The curve rises smoothly from zero toward this plateau; ${\Theta }_m$ sets a crossover scale, not a sharp transition. In normalized variables its initial slope is $C_V/(N{k}_B)\simeq (2\pi /3)T/{\Theta }_m$. The low-temperature continuum limit also requires temperature above the finite-size level spacing; uniform center-of-mass translation is treated separately. Below the lowest nonzero mode energy a finite chain does not retain a continuum linear law down to arbitrary temperature.

In the harmonic approximation, each in-plane displacement coordinate contributes one quadratic kinetic energy term and one quadratic restoring potential energy term. The equipartition theorem assigns $k_{B}T/2$ to each. There are two such coordinates per chemical atom, henceHere the fixed-size heat capacity is at constant area in this two-dimensional model. On cooling, phonon excitation freezes out and the heat capacity decreases. For each nonzero angular frequency, the quantum harmonic oscillator contributeswhich is below $k_B$ at finite $x$ and approaches zero as $T\to 0$. Two linearly dispersing in-plane acoustic phonon branches give the familiar two-dimensional Debye model scaling $C_v\propto T^2$ at sufficiently low temperature. Out-of-plane modes are excluded by the specified model.

The Debye model fixes its cutoff by requiring the continuum to contain the $2n$ vibrational degrees of freedom:Therefore the Debye frequency and Debye temperature are

For a classical ideal monatomic gas, $C_{V,\mathrm{He}}=3N{k}_B/2$ and $C_{p,\mathrm{He}}=5N{k}_B/2$. Treat copper as having one conduction electron per atom. Its degenerate-electron contribution isConsequently the electron-only estimates at5K areThe constant-pressure gas versus electronic metal heat capacity convention makes the second numerator appropriate to a constant-pressure gas measurement. For the solid, the low-temperature difference between $C_p$ and $C_V$ is usually small, but its phonon contribution is not fixed by $E_F$. In a Debye model,The supplied $E_F$ alone therefore does not determine the ratio to copper's total heat capacity; the electronic-only result is conditional.^[ref] An estimate using additional material data is possible: the specific-heat fit in Material Properties: OFHC Copper by National Institute of Standards and Technology gives $c_{p,\mathrm{Cu}}(5\mathrm{K})\simeq 0.155\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Using copper's rounded molar mass $0.0635\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$ gives $C_{p,\mathrm{Cu}}\simeq 9.85\times 10^{-3}\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, and henceThis supplementary estimate includes copper's lattice contribution and retains the ideal-gas approximation for helium. The NIST fit reports a 10% deviation from its source data below 15 K; neither that fit nor the stated pressure alone supplies helium nonideality corrections.^[ref]

Lattice energy is transported by phonons; long-wavelength acoustic modes usually dominate low-temperature conduction. Their speed is set by elastic wave propagation. In the Debye model their volumetric heat capacity grows as $T^3$ at low temperature and approaches the classical lattice limit at high temperature. With boundary-limited mean free path, this gives a low-temperature $T^3$ conductivity; increasing phonon scattering can instead produce a peak and a declining high-temperature trend.

In metals, electrons near the Fermi surface carry both heat and charge. Their heat capacity is linear in T and their typical transport speed is the Fermi velocity, not a classical gas thermal speed. Degeneracy restricts the active thermal fraction but the large speed and long mean free path can make electronic heat conduction dominant. Insulators have few mobile electrons, so phonon conduction usually dominates. The approximate parallel-channel sum is $\kappa ={\kappa }_e+{\kappa }_{\mathrm{ph}}$; which channel dominates is material- and temperature-dependent. ^[ref]