Specific heat capacity of solids, water and gases

Heat capacity, specific heat and molar specific heat

C depends on the nature of the substance, its temperature and the conditions under which heat is supplied.

NCERT Physics Part II (Class 11), Section 11.6, page 231

(In the printed NCERT Eq. 11.6, page 231, the mole symbol μ appears as the proportional sign ∝; read it as μ.) These definitions are the same as in the specific heat capacity topic of the previous chapter. What is new here is that thermodynamics can predict some of these values: for solids, and for the difference between the two molar specific heats of a gas.

Molar specific heat of solids

Picture a solid as N atoms, each vibrating about its mean position. NCERT uses the law of equipartition of energy (explained in the next chapter, Kinetic Theory) to give each atom an average energy k_BT for each of the three directions it can vibrate in (each vibration has ½k_BT of kinetic and ½k_BT of potential energy), 3k_BT in all (page 231). In the classical harmonic approximation, for one mole of atoms, with N_A atoms and k_BN_A = R:

The temperature-dependent vibrational contribution is U − U_₀ = 3RT, where U_₀ is a temperature-independent reference contribution.

NCERT neglects thermal expansion and the associated change in lattice energy, so ΔQ ≈ ΔU and C_ₚ ≈ C_ᵥ in this model. Small external PΔV work alone does not prove that C_ₚ and C_ᵥ are equal for every real solid or liquid. In the classical high-temperature limit:

The experimental rule that most solid elements have C ≈ 3R is known as the Dulong–Petit law (a name not used in NCERT); equipartition explains it. NCERT Table 11.1 (page 231) shows how well it works at room temperature: aluminium 24.4, copper 24.5, lead 26.5, silver 25.5 and tungsten 24.9 J mol^⁻¹ K^⁻¹. Carbon, at 6.1, is the exception (page 232).

The agreement is known to break down at low temperatures.

NCERT Physics Part II (Class 11), Section 11.6, page 232

Notice that the specific heats per kilogram in the same table differ a lot (aluminium 900, lead 127.7 J kg^⁻¹ K^⁻¹), while the molar values are nearly equal. A mole of any element has the same number of atoms, N_A, so the heat per mole is nearly the same; a kilogram of a light element such as aluminium has far more atoms than a kilogram of lead, so it needs far more heat per kelvin (Examples 1–2).

Specific heat capacity of water

The calorie was first defined as the heat needed to raise the temperature of 1 g of water by 1 °C. More precise measurements showed that the specific heat of water changes slightly with temperature: NCERT Fig. 11.5 (page 232) shows it falling from about 1.008 cal g^⁻¹ °C^⁻¹ near 0 °C to a minimum near 40 °C, and rising again to about 1.008 at 100 °C. So the temperature interval had to be fixed:

One calorie is defined to be the amount of heat required to raise the temperature of 1g of water from 14.5 °C to 15.5°C.

NCERT Physics Part II (Class 11), Section 11.6, page 232

Near 15 °C, NCERT uses water’s specific heat as approximately 4186 J kg^⁻¹ K^⁻¹, or 4.186 J g^⁻¹ K^⁻¹, giving 1 historical 15 °C cal ≈ 4.186 J. The thermochemical calorie is instead defined as exactly 4.184 J; water’s specific heat is temperature-dependent. The old “mechanical equivalent of heat”, the work needed to produce 1 cal of heat, is just this conversion factor between two units of energy. Since heat, work and every other form of energy are now measured in joules, NCERT notes that the term is no longer needed:

the term mechanical equivalent is now superfluous and need not be used.

NCERT Physics Part II (Class 11), Section 11.6, page 232

Molar specific heats of a gas

For a gas the conditions matter a great deal, so two molar specific heats are defined (page 232):

• C_v, at constant volume;
• C_p, at constant pressure.

For an ideal gas they are linked by a simple relation, often called Mayer’s relation:

Proof (NCERT, page 232). Start from the first law for 1 mole, ΔQ = ΔU + PΔV.

• At constant volume, ΔV = 0, so C_v = (ΔQ/ΔT)_v = (ΔU/ΔT)_v = ΔU/ΔT (NCERT Eq. 11.9). The subscript v can be dropped in the last step because the internal energy of an ideal gas depends only on its temperature.
• At constant pressure, C_p = (ΔQ/ΔT)_p = (ΔU/ΔT)_p + P(ΔV/ΔT)_p (NCERT Eq. 11.10). For the same reason the subscript p can be dropped from the first term, so it equals C_v.
• For a mole of ideal gas PV = RT, so at constant pressure PΔV = RΔT, that is, P(ΔV/ΔT)_p = R (NCERT Eq. 11.11).

So C_p = C_v + R, that is, C_p − C_v = R.

In words: for the same rise in temperature, the gas gains the same internal energy either way, but at constant pressure it also expands and does work RΔT per mole, so it needs that much more heat. The real gases in NCERT Table 10.4 fit this closely: for each, C_p − C_v lies between 8.3 and 8.5 J mol^⁻¹ K^⁻¹, close to R = 8.31 J mol^⁻¹ K^⁻¹ (see molar specific heats of gases). Chemistry reaches the same result through enthalpy (see heat capacity in chemistry).

Solved examples

Take R = 8.31 J mol^⁻¹ K^⁻¹.

Common mistakes to avoid

Quick revision summary

• Heat capacity S = ΔQ/ΔT; specific heat s = S/m; molar specific heat C = S/μ. They depend on the substance, its temperature and the conditions.
• Classical harmonic solids: the vibrational energy varies as 3RT per mole of atoms, giving C_ᵥ ≈ 3R; this often approximates room-temperature elemental solids, with carbon and low temperatures as important exceptions.
• The historical 15 °C calorie is about 4.186 J; the thermochemical calorie is exactly 4.184 J. Water’s specific heat varies with temperature.
• Gases: C_p at constant pressure, C_v at constant volume; for an ideal gas C_p − C_v = R.

Questions students ask

How do you find Cp and Cv from γ?

For an ideal gas, C_p and C_v follow from γ by combining Mayer’s relation, C_p − C_v = R, with γ = C_p/C_v: putting C_p = γC_v gives C_v = R/(γ − 1) and C_p = γR/(γ − 1), per mole. For example, a gas with γ = 1.30 has C_v = 8.31/0.30 ≈ 27.7 J mol^⁻¹ K^⁻¹ and C_p = 1.30 × 27.7 ≈ 36.0 J mol^⁻¹ K^⁻¹.

Is Cp − Cv = R true for solids and liquids?

No, C_p − C_v = R holds per mole for an ideal gas, not for solids and liquids. For many solids and for water near room temperature, C_ₚ and C_ᵥ are fairly close, but the difference depends on thermal expansion and compressibility as well as temperature. It cannot be inferred solely from the small external PΔV work; neither 3R nor Mayer’s relation is universal for liquids. The gas values themselves, such as C_v = (3/2)R for a monatomic gas, come from kinetic theory.

What is the dimensional formula of molar specific heat?

Molar specific heat has the dimensional formula [M L^² T^⁻² K^⁻¹ mol^⁻¹] and the SI unit J mol^⁻¹ K^⁻¹, the same as the gas constant R, which is why C_p − C_v = R is dimensionally consistent. Specific heat capacity per kilogram, by contrast, has the dimensions [L^² T^⁻² K^⁻¹], because dividing by mass removes the M (see the table of dimensional formulae).

What is the molar specific heat of water?

The molar specific heat of liquid water is about 75 J mol^⁻¹ K^⁻¹: multiplying its specific heat, 4186 J kg^⁻¹ K^⁻¹, by its molar mass, 0.018 kg mol^⁻¹, gives C = Ms ≈ 75.3 J mol^⁻¹ K^⁻¹. That is about 9R, far more than the 3R of a solid metal. In older units it is about 18 cal mol^⁻¹ K^⁻¹, since water’s specific heat is close to 1 cal g^⁻¹ K^⁻¹ and its molar mass is 18 g mol^⁻¹.

What is the Dulong–Petit law, and why does it fail at low temperatures?

The Dulong–Petit law says that the molar specific heat of most solid elements is about 3R ≈ 24.9 J mol^⁻¹ K^⁻¹, as equipartition predicts for atoms vibrating in three directions. It fails at low temperatures because the energy of an atom’s vibration can change only in steps (a quantum result beyond Class 11): once k_BT is small compared with a step, the atoms cannot take up their full share of energy, so C falls below 3R and tends to zero as T approaches 0 K. Carbon’s atoms are light and strongly bonded, so its steps are large, which is why its value at room temperature, 6.1 J mol^⁻¹ K^⁻¹ in NCERT Table 11.1, is far below 3R.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Why is C_p greater than C_v for a gas?

   Show answerHide answer

   At constant pressure the gas expands as it warms and does work on its surroundings, so it needs extra heat beyond what raises its internal energy. For an ideal gas the extra is R per mole per kelvin.

2. Q2. Oxygen has C_p = 29.4 J mol^⁻¹ K^⁻¹ (NCERT Table 10.4). Treating it as ideal, find C_v.

   Show answerHide answer

   C_v = C_p − R = 29.4 − 8.31, about 21.1 J mol^⁻¹ K^⁻¹, matching the table.

3. Q3. Why did the calorie need a definition with a particular temperature interval, 14.5 °C to 15.5 °C?

   Show answerHide answer

   The specific heat of water changes slightly with temperature (NCERT Fig. 11.5), so the heat needed for a 1 °C rise depends on where that degree is. Specifying the interval and pressure makes that historical definition reproducible; its conversion to joules is measured. The distinct thermochemical calorie is exactly 4.184 J by definition.