Heat capacity of solids
Connect solid heat capacity to microscopic vibrations: derive the Dulong-Petit high-temperature limit and explain low-temperature quantum behavior.
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ContinueA crystal stores thermal energy in collective vibrations, rather than in freely translating gas molecules. At sufficiently high temperature, classical vibration counting predicts the Dulong–Petit limit. At low temperature, quantum excitation makes the lattice capacity fall. The task is to choose a capacity model for the actual material and interval before calculating heat.
Build on heat capacity and the idea of a normal mode. We will distinguish total sample capacity from capacity per mole of atoms.
Count the two energy terms in each mode
A macroscopic crystal of N atoms has approximately 3N vibrational normal modes; removing the few rigid-body motions does not affect its bulk limit. Each harmonic mode has a kinetic term and a restoring-potential term. In its classical regime, each quadratic term has mean energy kBT/2, so a mode contributes kBT to the temperature-dependent energy.
Thus Uₜₕₑᵣₘₐₗ ≈ 3NkBT and CV ≈ 3NkB. Per mole of atoms, this gives C_(V,m) ≈ 3R ≈ 24.9 J/(mol K). For a compound with s atoms per formula unit, the fully excited classical vibrational limit per mole of formula units is 3sR, not 3R. This assumes the relevant modes are all classical and the solid remains in that phase; real materials may not reach that regime before another change occurs.
Ask whether the modes can be excited
A mode of angular frequency ω has a quantum spacing ħω. When kBT is much smaller than that spacing, raising temperature hardly changes its population and it contributes little capacity. When the spacing is small compared with kBT, its contribution approaches the classical value. A high temperature in everyday language need not be high compared with a material’s vibrational scale.
The Debye model accounts for a distribution of lattice frequencies and predicts a lattice capacity proportional to T³ well below the Debye temperature. Electrons in a normal metal can supply an additional term proportional to T; magnetic, structural or superconducting changes can require further terms. Do not call every measured solid capacity a T³ law. The MIT notes on Einstein and Debye models develop the vibrational model and its limits.
A capacity tending to zero means that the derivative of energy with temperature tends to zero. It does not mean all internal energy, binding energy or zero-point energy vanishes. At fixed volume, a temperature-independent energy offset contributes no heat capacity.
Integrate the model over the actual interval
For constant-volume heating with no other work, Q = n∫ C_(V,m)(T) dT. A room-temperature constant cannot be carried into a cryogenic interval without evidence. Likewise, a capacity ratio C(2T)/C(T) is not the heat needed to warm from T to 2T.
Constant-pressure and constant-volume capacities are often close for a solid with modest expansion, but they are not identical. In a stable single phase their difference is related to expansion and compressibility; near a transition the small-difference approximation can fail. Match the supplied capacity to the process before using it.
Guided practice 1
Integrate lattice and electronic contributions
Problem
For a supplied normal-metal model valid throughout 2.0–4.0 K, the molar constant-volume heat capacity is C_(V,m) = γ T + β T³, where γ = 0.010 J/(mol K²) and β = 2.0 × 10⁻⁴ J/(mol K⁴). Heat 0.50 mol from 2.0 K to 4.0 K at constant volume, with no other work. Find the electronic heat contribution, the lattice contribution and their sum. Would multiplying the final heat capacity by the interval give the same result?
Work through the model, then compare your reasoning
Hints
Hint 1
Check the model
Capacity varies throughout the interval. Integrate each temperature power before multiplying by the amount.Show solution step by step
Accumulate each contribution
Method
Use integrals of γ T and β T³.Reason
Capacity is energy change per kelvin, so its area over temperature gives energy.Working
Answer
Qₑ = nγ(Tf²-Tᵢ²)/2 = 0.50(0.010)(16-4)/2 = 0.030 J. Q_lattice = nβ(Tf⁴-Tᵢ⁴)/4 = 0.50(2.0 × 10⁻⁴)(256-16)/4 = 0.0060 J.
Compare with an endpoint shortcut
Method
Add the contributions and evaluate the proposed approximation.Reason
A final capacity does not represent the average of an increasing capacity.Working
Answer
Q = 0.036 J. At 4 K, C_(V,m) = 0.040 + 0.0128 = 0.0528 J/(mol K). Using that value across 2 K gives 0.50(0.0528)(2) = 0.0528 J, about 47% too high. The supplied low-temperature model, not 3R, defines this example’s range.
Heating a solid in the classical regime
Worked example 2
Dulong–Petit heat estimate
Problem
Show full solution
Use the molar heat capacity
Method
Multiply C_(V,m) by the amount and temperature rise.Reason
The high-temperature approximation makes the molar capacity nearly constant, so no temperature integral is needed.Working
Answer
Q ≈ nC_(V,m)Δ T = (2.0)(24.9)(50) = 2.49 × 10³ J.State the estimate
Working
Answer
Q ≈ 2.49 × 10³ J, about 2.5 kJ. The units mol times J mol⁻¹K⁻¹ times K give J. Doubling the rise would double the heat only while the same capacity approximation remains valid.
Check what one mole counts
Check your understanding 1
An idealised crystal contains three atoms per formula unit. Assume every vibrational mode is in its classical regime over a 10 K warming interval. For 0.20 mol of formula units at constant volume, estimate the heat input using R = 8.314 J/(mol K). A colleague uses 3R per mole of formula units: identify the counting error and why this estimate is not automatically valid for a real crystal at 10 K.
Show answer
The classical capacity per mole of formula units is 3(3)R = 9R. Therefore Q ≈ 0.20(9)(8.314)(10) = 149.652 J, about 150 J. Using 3R counts one mole of atoms as if it were one mole of three-atom formula units and underestimates by a factor of three. The 10 K is an interval, not an absolute starting temperature. Classical validity depends on both endpoint temperatures compared with the material’s vibrational scales, not the interval alone.
Practice set
Check your understanding 2
Estimate the heat to warm 0.80 mol of a monatomic solid by 30 K at constant volume in the Dulong–Petit regime. Use 3R = 24.9 J mol⁻¹ K⁻¹.
Show hint
Show answer
Check your understanding 3
Why are the molar heat capacities at constant pressure and constant volume usually close for a solid with small thermal expansion?
Show hint
Consider the extra energy needed when the solid is allowed to expand.
Show answer
Small thermal expansion makes the extra constant-pressure energy requirement small. Thus C_((p, m)) and C_((V, m)) are usually close, though not identical; a gas typically expands much more.
Check your understanding 4
For the same solid in a regime where its lattice heat capacity obeys C ∝ T³, find C(2T)/C(T), with both temperatures in that regime.
Show hint
Show answer
C(2T)/C(T) = (2T)³/T³ = 8. This is a ratio of capacities, not a claim that the heat required for an interval is eight times larger.
What to keep
Count atoms and the kinetic-plus-potential energy of each vibrational mode. The 3R limit is per mole of atoms and needs a classical vibrational regime. At low temperatures, integrate an appropriate quantum and material model; a capacity tending to zero does not imply vanishing stored energy.
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Syllabus and review details
No official syllabus alignment is listed for this lesson.