Entropy Change on Heating

Heat a kettle of water from room temperature to boiling, and its entropy rises in a way you can calculate exactly. The trick is that entropy depends on temperature through the heat capacity, not through heat itself. This connection turns an abstract state function into a quantity you compute from measurable calorimetric data. This matters because entropy changes on heating are the workhorse calculation behind absolute entropy tables, phase-transition entropies, and third-law verifications. Every $S^\circ$ value in a thermodynamic table was built by integrating $C_p/T$ over temperature ranges just like the ones derived here. This article develops the integral $\Delta S = \int C/T\,dT$ from the thermodynamic definition of entropy. You will see the constant-pressure and constant-volume versions. You will see how constant $C$ gives simple logarithms. Then you will work two full numeric examples, one foundational and one exam-level.

Key takeaways

Entropy change on heating comes from integrating $dS = C\,dT/T$ , using the appropriate heat capacity for the constraint (constant $p$ or constant $V$ ).
For temperature-independent $C$ , the integral collapses to $\Delta S = C\ln(T_2/T_1)$ .
Constant-pressure heating uses $C_p$ ; constant-volume heating uses $C_V$ ; the two give different numerical answers for the same gas and same temperature change.
When $C$ varies with $T$ (typical for real substances, especially at low $T$ ), you must integrate the empirical $C(T)$ function or a fitted polynomial.
This calculation is the direct ancestor of absolute entropy determination via calorimetry and of entropy changes across phase transitions.
Key idea: Entropy rises with temperature because $C/T$ , not $C$ itself, is what integrates to give a state function, and the path (constant $p$ or constant $V$ ) fixes which heat capacity applies.

The core insight

Because entropy is defined through reversible heat divided by temperature, $dS = dq_{rev}/T$ , and because $dq_{rev} = C\,dT$ along a constant-pressure or constant-volume path, the entropy change on heating is always the integral of $C/T\,dT$ between the two temperatures—an integral that simplifies to a clean logarithm whenever $C$ can be treated as constant.

The physical picture

Entropy measures the number of ways energy and matter can be arranged, and raising the temperature always increases this multiplicity because more energy levels become thermally accessible. The thermodynamic definition of entropy states that for any reversible process,

dS = \frac{dq_{rev}}{T}

This is exact but not yet useful for computation, because we need to know $dq_{rev}$ as a function of $T$ . Heat capacity supplies exactly that link: by definition $C = dq_{rev}/dT$ along a specified path, so substituting gives $dS = C\,dT/T$ . The heat capacity itself depends on which variable is held fixed while heating occurs—pressure or volume—so the entropy formula naturally splits into two versions, $C_p$ for constant-pressure heating (the common laboratory case, since a beaker open to the atmosphere is at constant $p$ ) and $C_V$ for constant-volume heating (relevant for a sealed rigid container). Crucially, $C_p > C_V$ for any substance that expands on heating, because some of the heat supplied at constant pressure does $p\Delta V$ work on the surroundings rather than only raising the internal energy. That extra heat still counts toward increasing the entropy of the system, so heating a gas from $T_1$ to $T_2$ at constant pressure produces a strictly larger entropy increase than heating it through the same temperature interval at constant volume. At low temperatures heat capacities are not constant—they fall toward zero as $T \to 0$ (a hallmark of the third law)—so the simple logarithmic formula is an approximation valid over ranges where $C$ does not change much, typically room-temperature-and-above ranges for many solids, liquids, and gases.

Figure

Entropy accumulates as the shaded area under C/T versus T

Figure

Same ΔT, two paths: constant-pressure heating gives more entropy than constant-volume heating

The core relations

dS = \frac{C\,dT}{T}

where: $C$ is the heat capacity along the heating path (either $C_p$ or $C_V$ ).

\Delta S = \int_{T_1}^{T_2} \frac{C_p}{T}\,dT \quad \text{(constant pressure)}

where: $C_p$ may itself depend on $T$ .

\Delta S = \int_{T_1}^{T_2} \frac{C_V}{T}\,dT \quad \text{(constant volume)}

where: use $C_V$ for heating in a fixed, rigid volume.

\Delta S = C\ln\!\left(\frac{T_2}{T_1}\right)

where: valid when $C$ (either $C_p$ or $C_V$ ) is constant over the interval $T_1 \to T_2$ .

C_p(T) = a + bT + \frac{c}{T^2}

where: a common empirical fit for solids/gases used when $C$ is not constant, with tabulated $a$ , $b$ , $c$ .

The core derivation

Derivation AFrom dS = dqrev/T to ΔS = C ln(T2/T1)

Goal. Derive the entropy change when a closed system of fixed composition is heated reversibly from $T_1$ to $T_2$ , first for constant pressure, then reduce it to the constant-heat-capacity result.

Step 1 — Start from the definition of entropy. For any reversible process the second law gives

dS = \frac{dq_{rev}}{T}

This holds regardless of what else is happening to the system; we only need $dq_{rev}$ expressed in terms of $T$ .

Step 2 — Express reversible heat via heat capacity. At constant pressure, with no phase change and no non-expansion work, the heat capacity is defined as

C_p = \left(\frac{\partial H}{\partial T}\right)_p = \frac{dq_{rev,p}}{dT}

so that $dq_{rev,p} = C_p\,dT$ . This is exact because at constant $p$ , $dH = dq_p$ for a closed system doing only $pV$ work.

Step 3 — Substitute into the entropy definition. Combining Steps 1 and 2,

dS = \frac{C_p\,dT}{T}

This is the differential form of the target relation. It already shows that entropy and heat capacity share the same sign: since $C_p > 0$ for ordinary substances away from phase transitions, $S$ increases monotonically with $T$ .

Step 4 — Integrate between the two temperatures. Integrating both sides from the initial state $(T_1)$ to the final state $(T_2)$ ,

\int_{S_1}^{S_2} dS = \int_{T_1}^{T_2} \frac{C_p(T)}{T}\,dT

\Delta S = \int_{T_1}^{T_2} \frac{C_p(T)}{T}\,dT

This is the general result: valid for any $C_p(T)$ , including polynomial fits like $C_p = a + bT + c/T^2$ , which simply get integrated term by term.

Step 5 — Specialize to constant heat capacity. Over a temperature range where $C_p$ does not vary appreciably (a good approximation for many liquids and solids well above their Debye temperature, and often adequate for gases over tens to a couple hundred kelvin), $C_p$ can be pulled outside the integral:

\Delta S = C_p \int_{T_1}^{T_2} \frac{dT}{T} = C_p\big[\ln T\big]_{T_1}^{T_2}

Step 6 — Evaluate the logarithm.

\boxed{\Delta S = C_p \ln\!\left(\frac{T_2}{T_1}\right)}

Step 7 — The constant-volume analogue. An identical argument at constant volume, using $C_V = (\partial U/\partial T)_V$ and $dq_{rev,V} = C_V\,dT$ (since $dU = dq_V$ at constant volume with only $pV$ work), gives

\boxed{\Delta S = C_V \ln\!\left(\frac{T_2}{T_1}\right)}

Because $C_p > C_V$ for any substance with a positive thermal expansion coefficient, heating through the same $T_1 \to T_2$ at constant pressure always yields a larger $\Delta S$ than heating at constant volume.

Worked examples

Example 1 — Heating liquid water at constant pressure (foundational). Find $\Delta S$ when 250 g of liquid water is heated reversibly from 298 K to 373 K at 1 bar, given $C_{p,m} = 75.3\ \text{J K}^{-1}\text{mol}^{-1}$ , treated as constant over this range. Moles of water:

n = \frac{250\ \text{g}}{18.02\ \text{g mol}^{-1}} = 13.87\ \text{mol}

Total heat capacity:

C_p = nC_{p,m} = 13.87\ \text{mol} \times 75.3\ \text{J K}^{-1}\text{mol}^{-1} = 1044\ \text{J K}^{-1}

Apply the boxed result:

\Delta S = C_p \ln\!\left(\frac{373}{298}\right) = 1044\ \text{J K}^{-1} \times \ln(1.2517)

\ln(1.2517) = 0.2246

\Delta S = 1044 \times 0.2246 = 234.5\ \text{J K}^{-1}

Sanity check: entropy should increase substantially on heating a mole-scale sample through 75 K, and 234.5 J/K is comparable in magnitude to typical reaction entropies, which is physically reasonable for heating over 3/4 of the water's liquid range. The sign is correctly positive.

Example 2 — Comparing constant-pressure and constant-volume heating of an ideal gas (applied/exam-level). Two moles of a diatomic ideal gas ( $C_{V,m} = \tfrac{5}{2}R$ , $C_{p,m} = \tfrac{7}{2}R$ , both treated as constant) are heated from 300 K to 500 K. Compute $\Delta S$ for (a) constant-pressure heating and (b) constant-volume heating, and confirm $\Delta S_p > \Delta S_V$ . First, the shared logarithmic factor:

\ln\!\left(\frac{500}{300}\right) = \ln(1.6667) = 0.5108

(a) Constant pressure:

C_p = nC_{p,m} = 2\ \text{mol} \times \frac{7}{2}(8.314\ \text{J K}^{-1}\text{mol}^{-1}) = 2\times 29.10 = 58.20\ \text{J K}^{-1}

\Delta S_p = 58.20\ \text{J K}^{-1} \times 0.5108 = 29.73\ \text{J K}^{-1}

(b) Constant volume:

C_V = nC_{V,m} = 2\ \text{mol} \times \frac{5}{2}(8.314\ \text{J K}^{-1}\text{mol}^{-1}) = 2\times 20.785 = 41.57\ \text{J K}^{-1}

\Delta S_V = 41.57\ \text{J K}^{-1} \times 0.5108 = 21.24\ \text{J K}^{-1}

Check: $\Delta S_p - \Delta S_V = 29.73 - 21.24 = 8.49\ \text{J K}^{-1}$ . This should equal $nR\ln(T_2/T_1)$ since $C_p - C_V = nR$ for an ideal gas:

nR\ln(500/300) = 2\times8.314\times0.5108 = 8.49\ \text{J K}^{-1}

The two routes agree exactly, confirming both the arithmetic and the underlying relation $C_p - C_V = nR$ . As expected, constant-pressure heating produces the larger entropy increase because part of the supplied heat performs expansion work, and that heat still carries the $dq_{rev}/T$ entropy contribution.

Limiting cases & common traps

Situation	What to do
$C$ roughly constant over $T_1 \to T_2$	Use $\Delta S = C\ln(T_2/T_1)$ directly
$C(T)$ given as a polynomial (e.g. $a+bT+c/T^2$ )	Integrate term by term: $a\ln(T_2/T_1) + b(T_2-T_1) - \tfrac{c}{2}(T_2^{-2}-T_1^{-2})$
Heating spans a phase transition	Split the integral; add the transition entropy $\Delta H_{trans}/T_{trans}$ separately (see Entropy of Phase Transitions)
Very low temperature range (near 0 K)	$C \to 0$ , so simple constant- $C$ formula fails; use measured $C(T)$ or the Debye $T^3$ law
Gas heated at constant $V$ vs. constant $p$	Use $C_V$ or $C_p$ respectively; never interchange them

The chemist behind it

The entropy-from-heat-capacity integral rests on foundations laid by Rudolf Clausius, who first defined entropy through $dS = dq_{rev}/T$ in the 1850s–60s while formalizing the second law. But it was the American physical chemist Gilbert N. Lewis, along with Merle Randall, who turned this into the practical calorimetric machinery used throughout chemistry: integrating heat-capacity data from near absolute zero up through phase transitions to tabulate absolute entropies. Their 1923 textbook Thermodynamics and the Free Energy of Chemical Substances established the calculation style—heat capacity integrals plus transition entropies—that remains the standard method for building entropy tables today.

Origins · 1 minGilbert N. Lewis and how this was discovered

Gilbert N. Lewis

1875-1946 · Pioneered systematic calorimetric determination of absolute entropies using heat-capacity integration

Ask the tutor: How does this heating integral combine with phase-transition entropies to build a full third-law absolute entropy table from 0 K to 298 K?

Can I use this formula if the substance melts or boils partway through the heating range?

Not directly as a single logarithm. You must split the calculation into three parts: heat the solid (or liquid) up to the transition temperature using its own $C_p$ , add the transition entropy $\Delta H_{trans}/T_{trans}$ at constant temperature, then heat the new phase from the transition temperature to the final temperature using its $C_p$ . Each segment gets its own $C\ln(T_2/T_1)$ term, and the transition entropies are added separately, a topic developed further under Entropy of Phase Transitions.

Is Cp or Cv used when heating a solid or liquid?

For solids and liquids, $C_p$ and $C_v$ are numerically very close because thermal expansion is small, so the volume work term $p\Delta V$ contributes little. Tabulated heat capacities for condensed phases are almost always reported as $C_p$ because constant-pressure (open-vessel) calorimetry is far easier experimentally. For gases, the difference between $C_p$ and $C_V$ is significant and must not be neglected, since $C_p - C_V = nR$ for an ideal gas.

Why does this calculation matter for the third law?

The third law states that a perfect crystal has zero entropy at $T = 0$ K. To get an absolute entropy at any temperature $T$ , you integrate $C_p/T$ from 0 K up to $T$ , adding any phase-transition contributions along the way. This is precisely the entropy-on-heating integral, just extended down to absolute zero using measured low-temperature heat capacities (often fitted to the Debye $T^3$ law near 0 K). The technique covered here is the direct computational core of every third-law entropy table.