research32 min studyLesson 989 in Physics

Heat capacity and thermal transport diagnostics

Distinguish thermodynamic state counting from heat-current transport and test the assumptions behind low-temperature fits.

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What you’ll understand

  • Extract electronic and phonon coefficients from a conditional low-temperature heat-capacity model.
  • Convert a measured thermal conductance into a conductivity using the correct geometry.
  • Apply a Wiedemann–Franz benchmark without assigning a unique carrier mechanism from one ratio.

Heat capacity asks how much additional energy a sample stores when its temperature rises. Thermal conductivity asks how efficiently excitations carry heat through it. A large density of low-energy states can increase the first while strong scattering suppresses the second. Comparing these observables with electrical transport is therefore informative precisely because they weight states and scattering differently. Neither a fitted linear heat-capacity term nor a residual thermal signal uniquely identifies one exotic excitation without additional controls and a justified extrapolation.

A low-temperature fit is a physical hypothesis

For a conventional three-dimensional metal at temperatures well below relevant electronic and Debye scales, a useful molar model combines a linear electronic heat capacity with a cubic acoustic-phonon term. Plotting heat capacity divided by temperature against temperature squared separates intercept and slope. The electronic coefficient reflects a thermodynamic density of states, including appropriate quasiparticle renormalization in a Fermi-liquid interpretation. It is not the same weighted quantity as a Hall number or one photoemission intensity.

Magnetic excitations, nuclear or impurity Schottky anomalies, low-dimensional modes, glassy defects, and additional transitions can invalidate that two-term fit. Addenda from the platform and grease must be subtracted, the sample amount must be known, and thermal equilibration must be demonstrated. Experiments often measure heat capacity at constant pressure; comparison with a constant-volume model requires that the thermal-expansion correction be negligible or included. Changing the fit window is a physical stability test, not merely a way to improve a regression coefficient.

Cmol(T)≃γT+βT3,CmolT=γ+βT2,ΔSmol=∫T1T2Cmol(T)T dTC_{\rm mol}(T)\simeq\gamma T+\beta T^3,\qquad \frac{C_{\rm mol}}{T}=\gamma+\beta T^2,\qquad \Delta S_{\rm mol}=\int_{T_1}^{T_2}\frac{C_{\rm mol}(T)}{T}\,dT
C_mol is in joules per mole kelvin, gamma in joules per mole kelvin squared, and beta in joules per mole kelvin to the fourth. The entropy integral uses the appropriate equilibrium heat capacity and baseline.↔ Scroll wide equations horizontally.

A thermal current needs a temperature profile

In a steady one-dimensional diffusive bar, heat current density is minus thermal conductivity times temperature gradient. A heater power and a measured temperature difference give conductance, while conductivity additionally requires length and cross section. The power entering the sample can differ from electrical heater power because of heat leaks through wires, radiation, and the mounting structure. A contact temperature drop can also make the apparent sample gradient too large and the inferred conductivity too small.

Use thermometers calibrated in the same field and temperature regime as the measurement, check linearity with heater power, and allow a stationary profile to develop. At very low temperatures the electronic and lattice subsystems or the thermometers can decouple, so the reported temperature may not represent every heat-carrying subsystem. Boundary-limited or ballistic transport can require a description beyond a bulk local conductivity. In a magnetic field, transverse heat flow and thermoelectric corrections can require a tensor treatment rather than a scalar bar formula.

jQ=−κ∇T,Gth=PthroughΔT,κ=GthLA\mathbf j_Q=-\kappa\nabla T,\qquad G_{\rm th}=\frac{P_{\rm through}}{\Delta T},\qquad \kappa=G_{\rm th}\frac{L}{A}
These scalar forms assume a nearly uniform one-dimensional diffusive segment and a small temperature difference. Conductance G_th is watts per kelvin; conductivity kappa is watts per metre kelvin.↔ Scroll wide equations horizontally.

Charge and heat need not be carried in the same way

For degenerate conventional charged quasiparticles dominated by elastic scattering, a low-temperature electronic Wiedemann–Franz benchmark relates thermal and electrical conductivity through the Sommerfeld Lorenz number. The thermal conductivity is the open-circuit electronic value; phonons and neutral excitations add heat without carrying the same electrical current. Inelastic scattering, multiband structure, finite-temperature thermoelectric effects, and nonlocal transport can alter a naive comparison. The benchmark is not an unrestricted identity for all quantum matter.

A residual intercept of thermal conductivity divided by temperature depends on how phonons and other contributions are extrapolated. A fitted intercept can change with boundary scattering or a mistaken temperature calibration, while a genuine electronic intercept need not determine a unique gap structure. Combine temperature and field trends with calorimetric entropy and independent gap probes. Shared geometry errors can partly cancel in a Lorenz ratio when exactly the same segment carries both currents, but mismatched contacts and different heat paths can destroy that cancellation.

κeσT⟶L0=π2kB23e2,κtotal=κe+κph+⋯\frac{\kappa_e}{\sigma T}\longrightarrow L_0=\frac{\pi^2k_B^2}{3e^2},\qquad \kappa_{\rm total}=\kappa_e+\kappa_{\rm ph}+\cdots
The limiting benchmark assumes the stated degenerate elastic-scattering regime. L0 has units of watt ohms per kelvin squared; the additive decomposition assumes independently definable contributions and can need refinement when drag is important.↔ Scroll wide equations horizontally.

Two calorimetric points determine a conditional fit

A hypothetical sample follows the two-term molar model. Its Cmol/TC_{\rm mol}/T is 4 mJ mol−1K−24\,\mathrm{mJ\,mol^{-1}K^{-2}} at 1 K and 7 in the same units at 2 K. Find gamma, beta, and the predicted heat capacity at 3 K.

  1. The horizontal coordinates T squared are 1 and 4 square kelvin. The slope is (7−4)/(4−1)=1 mJ mol−1K−4(7-4)/(4-1)=1\,\mathrm{mJ\,mol^{-1}K^{-4}}, which is beta.

  2. Subtract the phonon contribution at 1 K: γ=4−1=3 mJ mol−1K−2\gamma=4-1=3\,\mathrm{mJ\,mol^{-1}K^{-2}}. Two points determine the model coefficients but cannot test whether extra terms are absent.

  3. At 3 K the electronic contribution is 3(3)=93(3)=9 and the phonon contribution is 1(27)=27 mJ mol−1K−11(27)=27\,\mathrm{mJ\,mol^{-1}K^{-1}}. Their sum is 36 mJ mol−1K−136\,\mathrm{mJ\,mol^{-1}K^{-1}}.

The result

The conditional coefficients are gamma three and beta one in the stated units, predicting 36 millijoules per mole kelvin at 3 kelvin.

Explore the model

Interactive lab

Interactive lab · heat

Diffusion depends on scale as well as diffusivity

For independent runs Reset and set spatial mode n and diffusivity to (1,0.5)(1,0.5) and (2,0.5)(2,0.5). Press Step twice for t=1 second. Predict amplitudes exp(-0.5), about 0.607, and exp(-2), about 0.135. This is one exact periodic heat-equation mode on a dimensionless spatial interval; it does not simulate calorimetry, thermometers, contacts, or a microscopic thermal-conductivity fit.

Diffusion depends on scale as well as diffusivityThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA diffusion decay rate depends on spatial scale as well as the transport coefficient.

Check your understanding

Why can a sample have large electronic heat capacity but modest thermal conductivity?

Try it yourself

For an illustrative conventional elastic metal at 10 K, take ρ=2.0×10−8 Ω m\rho=2.0\times10^{-8}\,\Omega\,\mathrm m and rounded L0=2.44×10−8 WΩK−2L_0=2.44\times10^{-8}\,\mathrm{W\Omega K^{-2}}. If total thermal conductivity is 15 W m−1K−115\,\mathrm{W\,m^{-1}K^{-1}}, find the benchmark electronic contribution and residual.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use kappa_e=L0 T/rho, then subtract it only within the stipulated benchmark assumptions.

Compare with the solution

The benchmark gives κe=12.2 W m−1K−1\kappa_e=12.2\,\mathrm{W\,m^{-1}K^{-1}}, leaving 2.8 W m−1K−12.8\,\mathrm{W\,m^{-1}K^{-1}}. That residual is conditional on the electronic benchmark, matched geometry, and calibration. It could include phonons or other contributions and is not by itself a unique measurement of neutral quasiparticles.

Take it with you

  • A linear-plus-cubic heat-capacity fit assumes a specific low-energy spectrum and reliable addenda subtraction.
  • Thermal conductivity depends on heat paths, equilibration, and geometry as well as excitations.
  • A Lorenz comparison is a conditional electronic benchmark, not a unique mechanism detector.

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