Specific heat capacity of solids

Pick up a copper coin and a small chunk of aluminium of exactly the same mass, and heat each with the same flame for the same time. They will not warm up by the same amount — copper needs less heat per degree than you might expect, aluminium needs more, because their masses are the same but their atoms are not. Yet weigh out one mole of each — 63.5 g of copper, 27 g of aluminium, wildly different masses — and something remarkable happens: heating either mole by one kelvin costs almost exactly the same amount of energy, about 25 joules. Same for iron, silver, lead, gold, almost any solid element you try, at ordinary room temperature. That is not a coincidence a chemist stumbled onto by trial and error; it is a direct, testable prediction of the same idea you have just used to explain the specific heats of monatomic, diatomic and polyatomic gases — the law of equipartition of energy. In a gas, molecules fly freely and equipartition counts translational and rotational degrees of freedom. In a solid the atoms cannot fly anywhere — they are locked into a crystal lattice, gripped on all sides by their neighbours through strong interatomic bonds. But "locked in place" does not mean "frozen still." Each atom can still vibrate about its fixed mean position, jiggling back and forth in three dimensions like a tiny bead tied to its neighbours by springs. This article treats that vibration with the same equipartition toolkit you already own, counts the degrees of freedom correctly (the trap most students fall into), and lands on a single elegant number: the molar specific heat of essentially any solid element is $C = 3R \approx 25\ \text{J/mol}\,\cdot\,\text{K}$ — the empirical law of Dulong and Petit, explained from first principles almost a century before quantum mechanics existed.

Key takeaways

Atoms in a solid vibrate, they don't translate. Held near fixed lattice sites by strong bonds, each atom oscillates about its mean position along three independent directions.
Each vibrational direction carries TWO quadratic energy terms, not one — kinetic energy $\tfrac12 mv^2$ and potential energy $\tfrac12 kx^2$ of the "spring" — so each direction contributes $2 \times \tfrac12 k_BT = k_BT$ .
Total degrees of freedom per atom = 6 (3 directions $\times$ 2 quadratic terms), giving average energy per atom $E = 6 \times \tfrac12 k_BT = 3k_BT$ .
For one mole: total internal energy $U = 3RT$ , so the molar specific heat is $C = dU/dT = 3R \approx 24.94\ \text{J/mol}\,\cdot\,\text{K}$ — the Dulong–Petit law.
Only one specific heat matters for solids. Because solids barely expand on heating, the work done against external pressure is negligible, so $C_p \approx C_v \approx 3R$ (unlike gases, where $C_p$ and $C_v$ differ by $R$ ).
The law is a room-temperature approximation. At very low temperature real solids show $C$ falling well below $3R$ , eventually toward zero — a breakdown that classical equipartition cannot explain and that needs quantum treatment studied later.
Key idea: modelling each lattice atom as a 3-D oscillator with 6 quadratic degrees of freedom (kinetic + potential in each of 3 directions) and applying equipartition gives U = 3RT and hence C = 3R ≈ 25 J/mol·K for any solid element at ordinary temperature, the Dulong–Petit law.

The core insight

An atom vibrating about a fixed lattice site has 3 directions of motion, each carrying a kinetic and a potential quadratic energy term (6 degrees of freedom total), so equipartition gives average energy $3k_BT$ per atom and molar specific heat $C = 3R \approx 25\ \text{J/mol}\,\cdot\,\text{K}$ for one mole — the Dulong–Petit law, valid at ordinary temperatures for almost any solid element.

Where it fits in the syllabus

This section closes the equipartition sweep of the CBSE Kinetic Theory chapter: after applying the law of equipartition of energy to monatomic, diatomic and polyatomic gases, the same reasoning is carried across to solids. A Class 11 student is expected to model a solid as a lattice of atoms vibrating about fixed mean positions, count degrees of freedom correctly (six per atom, not three), and derive $C = 3R$ , matching the empirically known Dulong–Petit law. The syllabus keeps this section qualitative about why the classical model eventually fails; only a single sentence pointing to Class-12/university-level quantum corrections is appropriate here — a full quantum treatment is out of scope.

The physical picture

Imagine zooming into a copper crystal until individual atoms come into view. Each atom sits at a lattice site, held there not rigidly but elastically — its neighbours behave like a web of tiny springs pulling it back whenever it strays. Thermal energy makes every atom jiggle continuously about its equilibrium position, in all three spatial directions at once: it can be displaced (and restored) along $x$ , along $y$ , and along $z$ , independently. Consider just the one-dimensional motion along a single axis, say $x$ . Like any oscillator, it possesses two distinct energy stores that both depend quadratically on a coordinate: kinetic energy from its instantaneous speed, and potential energy from how far the "spring" is stretched or compressed,

E_x = \frac{1}{2}mv_x^2 + \frac{1}{2}k x^2

where: $m$ is the atom's mass, $v_x$ its instantaneous velocity component along $x$ , $k$ the effective restoring-force constant of the interatomic bond in that direction, and $x$ the instantaneous displacement from the mean lattice position. The physical reading is that this one direction of vibration behaves exactly like a mass on a spring undergoing simple harmonic motion, and — crucially for equipartition — it has two independent quadratic terms in its energy, not one. The law of equipartition assigns average energy $\tfrac12 k_BT$ to every independent quadratic term in the energy expression, so this single direction already contributes $2 \times \tfrac12 k_BT = k_BT$ on average, before the other two directions are even counted.

Figure

A single lattice atom shown vibrating along three perpendicular directions x y and z with springs connecting it to six neighbouring atoms, and a table beside it listing each direction as contributing one kinetic energy term and one potential energy term for a total of six quadratic terms per atom

Extend this to all three directions and every atom in the crystal carries $3 \times 2 = 6$ independent quadratic energy terms — its six degrees of freedom. Equipartition then hands each atom an average total vibrational energy

\overline{E} = 6 \times \frac{1}{2}k_BT = 3k_BT

where: $k_B = 1.38\times10^{-23}\ \text{J/K}$ is Boltzmann's constant and $T$ the absolute temperature. The physical reading is exactly the same logic used for gas molecules — count the independent quadratic terms in the energy, multiply by $\tfrac12k_BT$ each — only here every one of those six terms belongs to a single fixed atom's vibration rather than to translational or rotational motion of a free molecule.

Derivation ADeriving the Dulong-Petit law C equals 3R from equipartition

The goal of this derivation is to turn the per-atom vibrational energy $3k_BT$ into a single testable, element-independent number for the molar specific heat capacity of a solid, using nothing beyond the law of equipartition of energy and the definition $C = dU/dT$ — no calculus beyond a simple derivative, and no assumption about which solid element is being considered, which is exactly why the final answer applies almost universally.

Step 1. Start from the average energy of one vibrating atom found above, $\overline{E} = 3k_BT$ , which follows from six quadratic degrees of freedom (three directions, kinetic plus potential energy in each) and equipartition's rule of $\tfrac12k_BT$ per quadratic term.

Step 2. Scale up to one mole of the solid. One mole contains $N_A = 6.022\times10^{23}$ atoms (Avogadro's number), so the total internal (vibrational) energy of one mole is

U = N_A \times \overline{E} = N_A \times 3k_BT = 3(N_Ak_B)T

Step 3. Recognise that $N_Ak_B = R$ , the universal gas constant — the same substitution used when deriving specific heats of gases from per-molecule energies. This gives the compact result

U = 3RT

which says the total internal energy of one mole of any solid, in this model, is a fixed multiple of temperature, independent of which element it is or what its atomic mass happens to be.

Step 4. Differentiate to get the molar specific heat capacity. Since $C = \dfrac{dU}{dT}$ and $U = 3RT$ is linear in $T$ , the derivative is a constant:

C = \frac{dU}{dT} = 3R

Step 5. Insert the numerical value $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$ to get $C = 3 \times 8.314 = 24.94\ \text{J/mol}\,\cdot\,\text{K} \approx 25\ \text{J/mol}\,\cdot\,\text{K}$ . This is the law of Dulong and Petit: at ordinary temperatures, almost every solid element — copper, iron, gold, lead, aluminium — has this same molar specific heat, because the derivation never used any element-specific property. Only the mass of the atoms differs from element to element, and mass cancelled out completely; it never entered the energy count. Result: $U = 3RT$ and $C = dU/dT = 3R = 24.94\ \text{J/mol}\,\cdot\,\text{K}$ , independent of atomic mass or identity of the solid, matching the empirical Dulong–Petit law.

Worked examples

Board level - Heat needed to warm a known mass of copper

Estimate the heat required to raise the temperature of 250 g of copper (molar mass $63.5\ \text{g/mol}$ ) by 20 K, assuming the Dulong–Petit law applies.

Step 1. Convert the given mass to moles using $n = m/M$ :

n = \frac{250}{63.5} = 3.937\ \text{mol}

Step 2. Take the molar specific heat from the Dulong–Petit law derived above, $C = 3R = 24.94\ \text{J/mol}\,\cdot\,\text{K}$ , and apply $\Delta Q = nC\Delta T$ with $\Delta T = 20\ \text{K}$ :

\Delta Q = 3.937 \times 24.94 \times 20

Step 3. Multiply out: $3.937 \times 24.94 = 98.19\ \text{J/K}$ , so

\Delta Q = 98.19 \times 20 \approx 1963.8\ \text{J}

So roughly 1964 J of heat is needed — this Dulong–Petit estimate is a reasonably good approximation for copper at room temperature, and it required no separate look-up of copper's specific heat: the same $3R$ works for almost any solid element.

Competitive level - Vibrational energy per atom and total internal energy at 400 K

Using the law of equipartition, find the average total vibrational energy of a single atom in a solid at temperature $T = 400\ \text{K}$ , and the corresponding total internal energy of 0.5 mol of that solid.

Step 1. Apply the per-atom result derived above, $\overline{E} = 3k_BT$ , with $k_B = 1.38\times10^{-23}\ \text{J/K}$ :

\overline{E} = 3 \times 1.38\times10^{-23} \times 400 = 1.656\times10^{-20}\ \text{J per atom}

Step 2. Scale to 0.5 mol using the molar form $U = nCT$ with $C=3R$ , i.e. $U = n \times 3RT$ :

U = 0.5 \times 3 \times 8.314 \times 400

Step 3. Multiply out step by step: $3 \times 8.314 = 24.94$ , then $24.94 \times 400 = 9976.8\ \text{J/mol}$ , then $0.5 \times 9976.8 = 4988.4\ \text{J}$ .

So 0.5 mol of this solid has a total vibrational internal energy of about 4988 J at 400 K. Notice the two steps are consistent: multiplying the single-atom answer by $N_A \times 0.5$ atoms would give the same total, confirming that the per-atom and per-mole pictures agree. Timing note: in an exam, do not recompute $3R$ or $N_Ak_B$ from scratch each time — memorise $C = 3R \approx 25\ \text{J/mol}\,\cdot\,\text{K}$ as a ready constant (just like $C_v = \tfrac32R$ for a monatomic gas), and reach straight for $\Delta Q = nC\Delta T$ or $U = nCT$ ; this saves 20-30 seconds per Dulong–Petit question, useful when several numericals share a section.

Common traps

Forgetting that each vibrational direction contributes two degrees of freedom (kinetic + potential energy), not one — this is why solids get $f = 6$ per atom, not 3 as one might guess by wrongly analogising to translational-only motion of a free gas molecule.
Writing $f=3$ and $E = \tfrac32k_BT$ for a solid atom by copying the monatomic-gas result — a solid atom does not translate freely; it oscillates, so both KE and PE terms exist in every direction, doubling the count compared to a free gas molecule.
Assuming $C_p$ and $C_v$ differ significantly for solids the way they do for gases — for solids, thermal expansion is small, so the work done against external pressure on heating is negligible and $C_p \approx C_v \approx 3R$ is a fair approximation at this level.
Applying the Dulong–Petit law ( $C = 3R$ ) at very low temperatures — the law is only accurate at ordinary/room temperature and above; it visibly fails as $T \to 0$ , a limitation that is fine to mention in an answer but not to derive here.
Confusing "specific heat capacity" (per unit mass, $\text{J/kg}\,\cdot\,\text{K}$ ) with "molar specific heat capacity" (per mole, $\text{J/mol}\,\cdot\,\text{K}$ ) — the Dulong–Petit result $3R$ is a molar quantity; converting to a per-kilogram value requires dividing by the molar mass, which is exactly why different solids have different specific heats per kilogram even though their molar specific heats are all close to $3R$ .
Treating $3R$ as an exact universal constant rather than a room-temperature approximation of a classical model — real measured values for different solids cluster near $25\ \text{J/mol}\,\cdot\,\text{K}$ but are not perfectly identical, and light, strongly-bonded solids like diamond deviate noticeably even near room temperature.

You should be able to…

Explain why an atom vibrating in a solid lattice has 6 degrees of freedom (3 directions x 2 quadratic terms each), not 3.
Derive the molar specific heat of a solid, $C = 3R$ , starting from the per-atom energy $\overline{E} = 3k_BT$ using the law of equipartition of energy.
State the law of Dulong and Petit and its approximate numerical value ( $\approx 25\ \text{J/mol}\,\cdot\,\text{K}$ ), and apply $\Delta Q = nC\Delta T$ or $U = nCT$ in numericals.
Explain why $C_p \approx C_v$ for solids, unlike for gases, in terms of negligible thermal expansion.
State (without deriving) that the Dulong–Petit law breaks down at low temperature, requiring the quantum treatment studied in higher classes.
Distinguish molar specific heat (J/mol·K) from specific heat per unit mass (J/kg·K) and convert between them using the molar mass.

Ask the tutor: Want a clearer picture of why each vibration direction counts twice (kinetic and potential energy), or curious why the Dulong–Petit law quietly fails at low temperature — ask the AI tutor for an intuitive walkthrough or more practice numericals.

Why does an atom in a solid get 6 degrees of freedom but a monatomic gas molecule only gets 3?

A free monatomic gas molecule can only translate — move bodily through space along x, y and z — so it has 3 quadratic kinetic-energy terms and nothing else, since it is not bound to any fixed position. An atom in a solid, by contrast, is tied to a lattice site by strong bonds and cannot translate freely; instead it oscillates about that fixed point in all three directions. Each oscillation direction behaves like a spring, storing both kinetic energy (from its speed) and potential energy (from its displacement), so each of the 3 directions contributes 2 quadratic terms instead of 1, giving 6 in total.

Does the Dulong-Petit law mean literally every solid has the exact same specific heat per kilogram?

No — it predicts the same molar specific heat (per mole of atoms), not the same specific heat per kilogram. Since different elements have very different atomic masses, one mole of lead has far more mass than one mole of aluminium, so their specific heats per kilogram differ substantially even though their specific heats per mole are both close to 25 J/mol·K. This is exactly why engineers still look up specific heats in J/kg·K tables for practical heat calculations, even though the underlying molar value is nearly universal.

Why does the Dulong-Petit law fail at low temperature if the derivation looks so solid?

The derivation assumes each vibrational degree of freedom can absorb energy in any tiny, continuous amount, which is the classical picture equipartition relies on. At low temperatures, quantum mechanics takes over: vibrational energy can only be absorbed in discrete packets (quanta), and when the thermal energy k_BT becomes smaller than the spacing between these packets, many vibrational modes effectively "freeze out" and stop contributing to the specific heat. This is why C drops well below 3R and tends toward zero as T approaches absolute zero — a result correctly explained only by the quantum treatments (Einstein's and Debye's models) you will meet in higher studies.

Is this the same equipartition idea used for diatomic and polyatomic gas specific heats, just applied differently?

Yes, exactly the same rule — every independent quadratic term in the total energy expression gets an average energy of half k_BT — is applied to a different physical system. For gases you counted translational, rotational, and sometimes vibrational quadratic terms per molecule; for a solid you count only vibrational terms, but there you get two per direction (kinetic plus potential) instead of the usual translational count of one. The underlying law of equipartition of energy does not change; only what is being counted changes with the physical model of the system.