Lesson 3 · 35 min

Heat capacity of solids and third-law entropy

Why does a mole of metal atoms hold about 25 J/K while diamond holds a quarter of that? Start from the Dulong–Petit law and atomic masses by the method of mixtures, fit Einstein’s quantised vibrations to the diamond data, then integrate copper’s low-temperature heat capacity from 0 K to its standard entropy.

In this lesson

Explain the Dulong–Petit law Cv ≈ 3R by equipartition, and use it with chemical analysis to fix an atomic mass
Use Einstein’s model to explain why heat capacities fall at low temperature and in light, hard solids, and fit its characteristic temperature to measured data
Obtain third-law entropies by integrating Cp/T from low-temperature data, and separate electronic and lattice contributions with a plot of Cp/T against T²

In 1819 Pierre Louis Dulong and Alexis Thérèse Petit in France measured the specific heats of a dozen solid elements and noticed something odd: specific heat times atomic weight came out nearly the same for all of them. In today’s units a mole of atoms holds about 25 J/K — whether the atoms are light aluminium or heavy lead. A bulk property, the specific heat, was counting atoms.

The rule soon became a working tool for chemists. Chemical analysis gives an element’s equivalent mass (the mass that combines with 8 g of oxygen) but not its valence, so an atomic weight could be off by a factor of two or three; a specific heat tells you which factor. Yet diamond, boron and silicon refused to obey, with room-temperature heat capacities only a fraction of the expected value. The puzzle lasted more than half a century, until in 1907 Einstein solved it with quantised vibrations — the first time quantum theory was applied to solids.

Predict first

Copper has a specific heat of 0.385 J g⁻¹ K⁻¹ and a molar mass of 63.5 g/mol. Aluminium’s molar mass is 27.0 g/mol. Roughly what is its specific heat?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

About 0.9 J g⁻¹ K⁻¹ Also about 0.385 J g⁻¹ K⁻¹ About 0.16 J g⁻¹ K⁻¹ It cannot be estimated from this

Dulong and Petit: 3kT per atom

In a crystal each atom is tied to its place by its neighbours, like a ball that can vibrate in three directions. Classical equipartition gives each direction two squared terms, kinetic and potential, at ½kT each, so each atom stores 3kT on average and a mole stores 3RT. Every kelvin of warming then costs 3R ≈ 24.9 J — the Dulong–Petit law.

For most metals at room temperature it works well. The plotted Cp values sit slightly above 3R, partly because tables give the heat capacity at constant pressure: a solid also expands as it warms, and for metals Cp exceeds Cv by a few per cent.

C_V \approx 3R = 24.9\ \mathrm{J\,K^{-1}\,mol^{-1}},\qquad c \approx \frac{3R}{M}

Specific heat c (per gram) is inversely proportional to the molar mass M, so a specific heat estimates an atomic mass.

$C_V$: molar heat capacity at constant volume
$c$: specific heat capacity
$M$: molar mass

50.0 g of metal from a boiling bath at 100.0 °C → 100.0 g of water at 20.0 °C · cup 10 J/K · room 22 °C

Measure a specific heat by mixtures, then fix an atomic mass

The apparatus (sketch)

Thermometer reading

First heat the metal through in the boiling bath, then drop it into the calorimeter.

Heat, then drop, and read the temperature curve: extrapolating the post-drop line back to the moment of the drop gives T_f and hence the specific heat; Dulong–Petit gives the size of the atomic mass, the mass of oxide gives the exact equivalent mass, and together they fix the valence and the atomic mass.

Einstein: vibrations are quantised

Einstein carried Planck’s idea over to solids: an atom’s vibrational energy can only be a whole multiple of hν. At high temperature kT ≫ hν, many levels are populated and the classical 3R holds; at low temperature kT ≪ hν, almost every oscillator sits in its ground state, can hardly be lifted by even one quantum, and contributes almost nothing — the vibrations are “frozen out”.

The dividing temperature θ_E = hν/k is set by the vibration frequency. Light atoms and stiff bonds vibrate fast and give a high θ_E: diamond’s carbon atoms are light and its C–C bonds stiff, so θ_E exceeds 1000 K, room temperature is still “cold” for diamond, and its heat capacity is only about a quarter of 3R.

C_V = 3R\,\frac{x^2 e^x}{(e^x - 1)^2},\qquad x = \frac{\theta_\mathrm{E}}{T} = \frac{h\nu}{kT}

As x → 0 (high temperature) the factor tends to 1 and 3R returns; for large x it falls to zero like x²e⁻ˣ.

$\theta_\mathrm{E}$: Einstein temperature
$\nu$: atomic vibration frequency

Weber’s specific heats of diamond (the table in Einstein’s 1907 paper) · Einstein and Debye models

Einstein’s diamond

Einstein temperature θ_E = 600 K

Also draw the Debye model (fitted θ_D)

Weber’s measurements
Einstein model

θ_E = 600 K: rms misfit 9.14 J K⁻¹ mol⁻¹ (at the best θ_E = 1296 K it is 0.50). The matching vibration wavelength is λ = hc/(kθ_E) = 24.0 μm.

Drag θ_E until the curve runs through Weber’s points. Einstein fixed θ_E ≈ 1325 K from the single point at 331 K; fitting all 12 points gives about 1296 K with a misfit under 0.5 J K⁻¹ mol⁻¹ — one parameter describes everything from 220 to 1260 K.

Debye, electrons and third-law entropy

Einstein’s model falls too steeply at very low temperatures. Debye (1912) pointed out that a crystal does not vibrate at one frequency but carries a whole spectrum of sound waves up to a maximum; the lowest-frequency waves can be excited however cold it gets, and at low temperature Cv ∝ T³. Metals add one more term: only the few conduction electrons near the Fermi energy can take up heat, and they contribute γT. So at a few kelvin, Cp/T plotted against T² is a straight line whose intercept is γ and whose slope gives the Debye temperature.

Heat capacities have a bigger use too. The definition dS = δq/T says that integrating Cp/T upwards from near 0 K gives a substance’s absolute entropy; the third law guarantees S = 0 for a perfect crystal at 0 K. Many of the standard entropies S° in thermodynamic tables were measured exactly this way, point by point, by low-temperature calorimetry.

S(T) = \int_0^T \frac{C_p}{T^\prime}\,\mathrm{d}T^\prime,\qquad C_p \approx \gamma T + \beta T^3\ (T \ll \theta_\mathrm{D}),\quad \theta_\mathrm{D} = \left(\frac{12\pi^4 R}{5\beta}\right)^{1/3}

The first is the third-law entropy; the second is a metal’s heat capacity at the lowest temperatures: an electronic term plus Debye’s T³ term.

$\gamma$: electronic heat-capacity coefficient
$\beta$: lattice T³ coefficient
$\theta_\mathrm{D}$: Debye temperature

Heat capacity of copper evaluated by NBS (NSRDS-NBS 18) · S(T) = ∫ Cp/T dT

Integrating the entropy of copper from 0 K

Upper limit T = 100 K

The area under Cp/T is the entropy

At the lowest temperatures: Cp/T is linear in T²

S(100 K) = ∫₀ Cp/T dT = 10.03 J K⁻¹ mol⁻¹ (NBS table: 10.03); here Cp = 16.01 J K⁻¹ mol⁻¹ and the Debye model (θ_D = 344 K) gives 14.74.

Low-temperature fit: Cp = γT + βT³, γ = 0.691 mJ K⁻² mol⁻¹, β = 0.0479 mJ K⁻⁴ mol⁻¹ → θ_D = (12π⁴R/5β)^(1/3) = 344 K.

Drag the upper limit to 298.15 K: integrating 49 heat capacities gives S = 33.15 J K⁻¹ mol⁻¹, copper’s standard entropy. On the right, the points from 1 to 8 K lie on a line with intercept γ ≈ 0.69 mJ K⁻² mol⁻¹, and its slope gives θ_D ≈ 344 K.

Worked example

An atomic mass from a specific heat and an oxide

A metal has a specific heat of 0.129 J g⁻¹ K⁻¹; 1.000 g of it burns completely to 1.0772 g of oxide. Find its valence and atomic mass.

Concept check

Why does diamond hold only about 6 J K⁻¹ mol⁻¹ at room temperature, while lead holds nearly 27?

Carbon atoms are light and C–C bonds stiff, so the vibration frequency is high, θ_E lies far above room temperature and most oscillators are still in their ground state Each atom in diamond vibrates in fewer than three directions Diamond does not conduct, so it lacks an electronic heat capacity Diamond is less dense than lead

Try the calculation

Taking θ_E = 1300 K, use Einstein’s formula to calculate Cv of diamond at 298.15 K (J K⁻¹ mol⁻¹).

Key takeaways

Dulong–Petit: a mole of atoms has Cv ≈ 3R = 24.9 J/K and c ≈ 3R/M; together with the chemical equivalent mass it fixes valence and atomic mass.
Einstein: vibrations are quantised and freeze out for T ≪ θ_E; diamond has θ_E ≈ 1300 K and only a quarter of 3R at room temperature.
Debye’s T³ plus the electronic γT describe a metal at low temperature; integrating Cp/T from 0 K gives the third-law entropy, 33.15 J K⁻¹ mol⁻¹ for copper.

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Data and references

P. L. Dulong, A. T. Petit, “Recherches sur quelques points importants de la théorie de la chaleur”, Ann. Chim. Phys. 10, 395–413 (1819).
A. Einstein, “Die Plancksche Theorie der Strahlung und die Theorie der spezifischen Wärme”, Ann. Phys. 22, 180–190 (1907); diamond data of H. F. Weber on p. 190.
P. Debye, “Zur Theorie der spezifischen Wärmen”, Ann. Phys. 39, 789–839 (1912).
G. T. Furukawa, W. G. Saba, M. L. Reilly, Critical Analysis of the Heat-Capacity Data of the Literature and Evaluation of Thermodynamic Properties of Copper, Silver, and Gold from 0 to 300 K, NSRDS-NBS 18 (1968).
M. W. Chase, NIST-JANAF Thermochemical Tables, 4th ed., J. Phys. Chem. Ref. Data Monograph 9 (1998), via the NIST Chemistry WebBook.