$$ \begin{align} q_V &= ΔU \tag{1}\\ q_p &= ΔH \tag{2}\\ C_V &= \frac{\mathrm dq_V}{\mathrm dT} = \left(\frac{\partial U}{\partial T}\right)_V \tag{3}\\ C_p &= \frac{\mathrm dq_p}{\mathrm dT} = \left(\frac{\partial H}{\partial T}\right)_p \tag{4}\\ \end{align} $$

Question 1: what is the relation between $q$ and internal energy and enthalpy? Why did I assume that when the volume is constant heat would relate to external energy and when the pressure is constant it would relate to the enthalpy change?

Question 2: why is in the equation of the heat capacity it’s equal to the change of heat over the change of temperature? Why is the change of temperature is in the denominator?

  • 5
    $\begingroup$ Could you explain what you meant in the two questions? Why you assumed something is not the right phrasing of your question, I assume. $\endgroup$ Jul 21, 2020 at 8:31
  • 3
    $\begingroup$ Q2: How do you think the heat capacity is defined ? $\endgroup$
    – Poutnik
    Jul 21, 2020 at 9:26

2 Answers 2


These equations are all based on the definitions of internal energy, enthalpy, and heat capacity.

Internal Energy

The change in internal energy of a system is equal to the sum of the heat transferred $q$ and the work done $W$.

$$\Delta U = q + W$$

In simple systems, most work is pressure-volume work, $W = P\Delta V$. Thus, we can write

$$\Delta U = q - P\Delta V$$

At constant volume, $\Delta V = 0$, so no work is done. Thus we can write

$$\Delta U = q_\mathrm{v}$$

The v subscript in $q_\mathrm{v}$ indicates constant volume.


Constant pressure is a much more common situation. Think of an open reaction vessel. This is constant pressure. The change in internal energy is difficult to calculate here, because we would need to measure the change in volume (might be minuscule).

Enthalpy is a correction to internal energy that takes into account the work needed to get the system to its current pressure and volume.

$$H = U + PV$$

The differential of enthalpy is

$$dH = dU + d(PV) = dq + dW + PdV + VdP$$

Substituting in $dW = -PdV$, we get

$$ dH = dq - PdV + PdV + VdP = dq + VdP$$

At the macro scale, we write

$$\Delta H = q + D\Delta P$$

A constant pressure, $\Delta P = 0$, so

$$\Delta H = q_\mathrm{p}$$

The subscript p in $q_\mathrm{p}$ indicated constant pressure.

Heat capacities

The heat capacity, $C$, of a system is the amount of heat needed to raise the temperature of the system 1 kelvin. Thus, heat capacity is the ratio of the heat exchanged to the change in temperature So:

$$ C = \dfrac{q}{\Delta T}$$

We can rewrite $C$ at the infinitesimal level:

$$C = \dfrac{dq}{dT}$$

Your two equations for heat capacity are just what happens at constant volume and constant pressure.

  • $\begingroup$ As I said in my answer, in thermodynamics, where heat capacity is a physical thermodynamic equilibrium property of the material (rather than a parameter related to the specific process), heat capacity is defined in terms of the internal energy U and the enthalpy H. I stand by this statement. $\endgroup$ Jul 21, 2020 at 18:30

In thermodynamics, the physical property known as heat capacity is not defined in terms of heat. It is defined in terms of internal energy and enthalpy. The change in internal energy and enthalpy between two thermodynamic equilibrium states of a material is a unique property of the material itself, and not of any specific process that brought about the change. On the other hand, there are an infinite number of heat flow and work flow paths from the surroundings to the system that can take the system between these same two thermodynamic equilibrium end states. The only constraint on these, according to the first law of thermodynamics, is that the difference between the heat flow and the work flow must be the same for all the paths (and equal to the change in internal energy). So the reason that the heat flow Q can't be unique (without some constraint on the work) is that the work W can't be unique either.


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