Heat Capacity at Constant Pressure

Heat a metal rod at constant pressure, and its temperature climbs at a rate set by one number: the heat capacity at constant pressure, $C_p$ . This single quantity tells chemists how much energy a system absorbs per degree of warming, and it links directly to enthalpy, the workhorse function of constant-pressure chemistry. Almost every calorimetry experiment, phase diagram, and reaction-enthalpy table in a chemistry course rests on it. This matters because real reactions happen in beakers open to the atmosphere, not in sealed rigid vessels. Constant pressure, not constant volume, is the everyday condition. Knowing $C_p$ lets chemists predict how much heat a process needs, correct enthalpies measured at one temperature to another, and connect molecular structure to macroscopic thermal behavior. This article defines $C_p$ rigorously as a partial derivative of enthalpy. It derives the integrated relation for $\Delta H$ over a temperature range. It works two full numerical examples, then flags common traps before pointing toward $C_p$ versus $C_v$ and Kirchhoff's law.

Key takeaways

$C_p = (\partial H/\partial T)_p$ : the slope of enthalpy versus temperature at constant pressure.
At constant pressure, heat supplied equals enthalpy change, so $q_p = \Delta H = \int C_p\,dT$ .
$C_p$ is always larger than $C_v$ for the same substance, because some heat goes into expansion work.
Over modest temperature ranges $C_p$ is often treated as constant; over wide ranges use $C_p(T) = a + bT + c/T^2$ .
Molar heat capacities near $\frac{7}{2}R$ (diatomic gases) or $3R$ (Dulong–Petit solids) offer quick sanity checks.
Key idea: $C_p$ converts a measured temperature change into an enthalpy change, making it the bridge between calorimetry and thermochemistry.

The core insight

Because enthalpy is the natural energy function at constant pressure, its temperature derivative — $C_p$ — is exactly the quantity that converts an observed heating curve into the enthalpy changes tabulated throughout thermochemistry, including reaction enthalpies and phase-change enthalpies at temperatures other than 298 K.

The physical picture

Consider a beaker of water sitting open on a lab bench, heated slowly by an immersion coil. The pressure stays fixed at atmospheric pressure throughout — the liquid is free to expand negligibly and any gas above it is free to push back the atmosphere. Under this constant-pressure condition, a well-known result from the First Law states that the heat absorbed equals the enthalpy change of the system:

q_p = \Delta H

This is the constant-pressure counterpart of the constant-volume relation $q_V = \Delta U$ , which you have already met. The heat capacity at constant pressure, $C_p$ , simply quantifies how steeply $H$ rises as $T$ rises, at fixed $p$ and fixed composition:

C_p \equiv \left(\frac{\partial H}{\partial T}\right)_p

Physically, $C_p$ answers the question: "how many joules must I supply, per kelvin of temperature rise, to a system held at constant pressure?" Some of that heat raises the internal kinetic and vibrational energy of the molecules (as it would at constant volume), and some does $p\Delta V$ work pushing back the surroundings as the system expands. That extra expansion term is exactly why $C_p > C_v$ for any substance that expands upon heating — gases dramatically, liquids and solids only slightly. Molar heat capacities are tabulated as $C_{p,m}$ (units J K⁻¹ mol⁻¹), and specific heat capacities as $c_p$ (units J K⁻¹ g⁻¹). For an ideal monatomic gas, kinetic theory predicts $C_{V,m} = \frac{3}{2}R$ and $C_{p,m} = \frac{5}{2}R$ . For diatomic gases near room temperature, rotational modes add further contributions, giving $C_{p,m} \approx \frac{7}{2}R$ . For crystalline solids at ordinary temperatures, the Dulong–Petit rule gives $C_{p,m} \approx 3R$ per mole of atoms.

Figure

Enthalpy versus temperature: the slope is Cp

Figure

Why Cp exceeds Cv: extra expansion work

The core relations

C_p = \left(\frac{\partial H}{\partial T}\right)_p

where: $H$ is enthalpy, $T$ absolute temperature, subscript $p$ means pressure and composition held fixed.

q_p = \Delta H = \int_{T_1}^{T_2} C_p \, dT

where: $q_p$ is heat absorbed at constant pressure, equal to the enthalpy change between $T_1$ and $T_2$ .

\Delta H = C_p (T_2 - T_1) \quad \text{(if } C_p \text{ constant over the range)}

where: valid when $C_p$ does not vary appreciably with $T$ across the interval.

C_{p,m}(T) = a + bT + \frac{c}{T^2}

where: $a, b, c$ are empirical constants tabulated per substance, $C_{p,m}$ is molar heat capacity at constant pressure.

C_{p,m} - C_{V,m} = R \quad \text{(ideal gas only)}

where: $R = 8.314$ J K⁻¹ mol⁻¹; this special case is developed fully in the next topic.

Derivation

Derivation AFrom the definition of Cp to ΔH over a temperature range

Goal. Starting from the definition $C_p = (\partial H/\partial T)_p$ , obtain a usable formula for $\Delta H$ when a system is heated from $T_1$ to $T_2$ at constant pressure, and show that this equals the heat absorbed.

Step 1 — Start from the First Law at constant pressure. For a closed system doing only expansion work, the First Law reads

dU = dq - p\,dV

At constant external pressure equal to the system pressure, integrating the work term gives $w = -p\Delta V$ , so

\Delta U = q_p - p\Delta V

Rearranging,

q_p = \Delta U + p\Delta V

Step 2 — Recognize the enthalpy function. Enthalpy is defined as $H \equiv U + pV$ . At constant pressure,

\Delta H = \Delta U + p\Delta V

Comparing this to the boxed expression from Step 1 shows immediately that

q_p = \Delta H

This confirms that at constant pressure, measured heat and enthalpy change are numerically identical — the foundation for using calorimeters (open to the air) to measure enthalpies directly.

Step 3 — Differentiate H with respect to T at constant p. Since $H$ is a state function of $T$ and $p$ (and composition), treat $p$ as fixed and ask how $H$ responds to small changes in $T$ . By definition this response rate is the heat capacity at constant pressure:

C_p \equiv \left(\frac{\partial H}{\partial T}\right)_p

This is an experimentally measurable quantity: supply a known small quantity of heat at constant pressure, record the resulting $dT$ , and $C_p = dq_p/dT$ in the limit of small increments — consistent with Step 2, since $dq_p = dH$ .

Step 4 — Integrate to find the finite enthalpy change. Treating $C_p$ as a function of temperature only (composition and pressure fixed), separate variables:

dH = C_p\, dT

Integrate both sides from $T_1$ to $T_2$ :

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

Step 5 — Apply the constant-Cp approximation. Over a temperature interval narrow enough that $C_p$ does not change appreciably (a good approximation for many liquids and solids over tens of kelvin), pull $C_p$ out of the integral:

\Delta H = C_p \int_{T_1}^{T_2} dT = C_p(T_2 - T_1)

Step 6 — State the general result. Combining Steps 2 and 4, the heat absorbed by a system heated reversibly at constant pressure from $T_1$ to $T_2$ is

\boxed{q_p = \Delta H = \int_{T_1}^{T_2} C_p(T)\, dT \;\approx\; C_p(T_2 - T_1) \text{ for constant } C_p}

This boxed relation is exactly what turns a measured temperature rise in a calorimeter into a tabulated enthalpy change, and it is the starting point for Kirchhoff's law, which asks how a whole reaction enthalpy shifts with temperature.

Worked examples

Example 1 — Heating liquid water (foundational). Calculate the heat required to warm 250 g of liquid water from 20.0 °C to 80.0 °C at constant atmospheric pressure. Take $c_p(\text{H}_2\text{O}, l) = 4.184$ J K⁻¹ g⁻¹, essentially constant over this range. Step 1: convert mass to the heat capacity of the sample.

C_p(\text{sample}) = m \cdot c_p = (250\text{ g})(4.184\text{ J K}^{-1}\text{g}^{-1}) = 1046\text{ J K}^{-1}

Step 2: find $\Delta T$ .

\Delta T = 80.0 - 20.0 = 60.0\text{ K}

Step 3: apply the boxed result.

q_p = \Delta H = C_p \Delta T = (1046\text{ J K}^{-1})(60.0\text{ K}) = 6.28 \times 10^4\text{ J} = 62.8\text{ kJ}

Sanity check: this is a plausible value — heating a quarter-liter of water by 60 °C on a stove takes tens of kilojoules, matching everyday experience (a typical kettle element delivers ~1–2 kW, so this would take roughly 30–60 seconds, which is reasonable).

Example 2 — Enthalpy of nitrogen gas heated over a wide range, with temperature-dependent Cp (applied/harder). Estimate $\Delta H$ for warming 2.00 mol of N₂(g) from 298 K to 900 K at constant pressure, using the empirical expression

C_{p,m} = a + bT + \frac{c}{T^2}

with (for N₂, in J K⁻¹ mol⁻¹, T in K): $a = 28.58$ , $b = 3.77 \times 10^{-3}$ , $c = -0.50 \times 10^{5}$ . Step 1: set up the integral for one mole.

\Delta H_m = \int_{298}^{900} \left(a + bT + \frac{c}{T^2}\right) dT = \left[aT + \frac{b}{2}T^2 - \frac{c}{T}\right]_{298}^{900}

Step 2: evaluate each term at $T_2 = 900$ K and $T_1 = 298$ K. Linear term:

a(T_2 - T_1) = 28.58 \times (900 - 298) = 28.58 \times 602 = 17\,205\text{ J mol}^{-1}

Quadratic term:

\frac{b}{2}(T_2^2 - T_1^2) = \frac{3.77\times10^{-3}}{2}\left(900^2 - 298^2\right) = 1.885\times10^{-3} \times (810000 - 88804)

= 1.885\times10^{-3} \times 721196 = 1359.5\text{ J mol}^{-1}

Inverse term ( $-c/T$ , so subtracting $c/T$ evaluated at limits):

-c\left(\frac{1}{T_2} - \frac{1}{T_1}\right) = -(-0.50\times10^5)\left(\frac{1}{900} - \frac{1}{298}\right)

= 0.50\times10^5 \times (1.111\times10^{-3} - 3.356\times10^{-3}) = 0.50\times10^5 \times (-2.245\times10^{-3})

= -112.3\text{ J mol}^{-1}

Step 3: sum the three contributions for one mole.

\Delta H_m = 17\,205 + 1359.5 - 112.3 = 18\,452\text{ J mol}^{-1} \approx 18.45\text{ kJ mol}^{-1}

Step 4: scale to 2.00 mol.

\Delta H = 2.00 \times 18.45\text{ kJ mol}^{-1} = 36.9\text{ kJ}

Sanity check: a rough constant- $C_p$ estimate using $C_{p,m} \approx \frac{7}{2}R = 29.1$ J K⁻¹ mol⁻¹ gives

\Delta H \approx 2.00 \times 29.1 \times 602 = 35\,000\text{ J} = 35.0\text{ kJ}

This is close to the more careful 36.9 kJ, confirming the calculation — the small discrepancy reflects the mild temperature dependence of $C_p$ captured by the $b$ and $c$ terms.

Limiting cases & common traps

Situation	What to watch for
Ideal gas, small $\Delta T$	Treat $C_p$ as constant; use $\frac{5}{2}R$ (monatomic) or $\frac{7}{2}R$ (diatomic, room T)
Wide temperature range	Use $C_{p,m}(T) = a + bT + c/T^2$ and integrate; do not assume constant $C_p$
Phase change within the interval	Split the integral at the transition temperature and add the latent enthalpy separately; $C_p$ is discontinuous there
Constant-volume process	Use $C_V$ and $\Delta U$ , not $C_p$ and $\Delta H$ — the two are not interchangeable outside ideal-gas contexts
Reaction enthalpy at nonstandard T	Requires $\Delta C_p$ of products minus reactants — this is Kirchhoff's law, covered next

The chemist behind it

The formal relationship between heat capacity and enthalpy, and its application to reacting systems, is inseparable from the work of Germain Henri Hess and later Gustav Kirchhoff, whose temperature-correction law is the direct extension of the ideas in this article. But the very concept of heat capacity as a measurable, tabulatable property of matter owes an enormous debt to Joseph Black, whose 18th-century calorimetric experiments first distinguished "quantity of heat" from "temperature" and introduced the specific-heat concept that $C_p$ formalizes mathematically.

Origins · 1 minJoseph Black and how this was discovered

Joseph Black

1728–1799 · Scottish chemist who discovered latent heat and specific heat, laying the experimental foundation for heat-capacity concepts used throughout thermochemistry

Ask the tutor: How does Kirchhoff's law use ΔCp to correct a reaction enthalpy from 298 K to a different temperature?

Is Cp always constant for a given substance?

No. $C_p$ generally depends on temperature, and sometimes on pressure. Over narrow ranges (tens of kelvin) it is often treated as constant for simplicity, as in Example 1. Over wide ranges, especially for gases used in industrial processes, empirical polynomial fits like $C_{p,m} = a + bT + c/T^2$ are used, as shown in Example 2. Near phase transitions, $C_p$ can even diverge sharply, so tabulated constant values should never be extrapolated across a melting or boiling point without adding the latent enthalpy separately.

How is Cp measured experimentally?

The most common method is calorimetry at constant (usually atmospheric) pressure: a known mass of substance is heated electrically by a measured quantity of energy, and the resulting temperature rise is recorded. Since $q_p = \Delta H = C_p\Delta T$ for small increments, $C_p$ is obtained as the ratio $q_p/\Delta T$ . Differential scanning calorimetry (DSC) extends this to measure $C_p(T)$ continuously across a temperature range, revealing phase transitions as peaks.

Why does Cp matter for thermochemistry if reaction enthalpies are usually looked up at 298 K?

Standard enthalpies of formation and reaction are tabulated at 298 K purely by convention, but real reactions — industrial, biological, geological — often occur at other temperatures. $C_p$ values let chemists correct a known $\Delta H$ at 298 K to the actual temperature of interest using Kirchhoff's law, which is built directly from the integral relation for $\Delta H$ developed here.

Does Cp apply to phase changes, like melting or vaporization?

Not directly across the transition itself. At the melting or boiling point, temperature stays fixed while enthalpy changes by the latent heat, so $dH/dT$ is technically undefined there ( $C_p$ appears to diverge). Away from the transition, within a single phase, $C_p$ behaves normally. Correct calculations spanning a phase change split the integral into three pieces: heating the initial phase, the phase-change enthalpy at constant $T$ , and heating the final phase.

How does Cp relate to Cv, and why does the difference matter?

For any substance, $C_p - C_V$ reflects the extra energy needed at constant pressure to do expansion work as the system heats and grows in volume. For an ideal gas this difference is exactly $R$ per mole, a strikingly simple and universal result derived by combining the ideal gas law with the definitions of $H$ and $U$ . This relationship, developed fully in the next topic, is essential for converting between calorimetric measurements made under different conditions.