1. I understand that the heat capacity is the gradient of the slope of q against T, which implies that C = $\mathrm{ \frac{dq}{dT}}$, so how do we arrive at C = $\mathrm{\frac{∂q}{∂T}}$?

  2. At constant volume, $\mathrm{C_V}$:

$$\mathrm{dU = dq + dw}$$

$\mathrm{dw = p(dV)}$, but $\mathrm{dv = 0}$, so $\mathrm{dU = dq}$

How do we make the jump to $\mathrm{C_V = \frac{∂U}{∂T}}$?

3. At constant pressure, $\mathrm{C_p}$:

$$\mathrm{H = U + pV}$$ so $$\mathrm{dH = dU + pdV = (dq + dw) - dw = dq}$$ hence $\mathrm{dH = dq}$

How do we make the jump to $\mathrm{C_p = \frac{∂H}{∂T}}$?

  • 2
    $\begingroup$ It is a matter of symbolism. Partial d's are used when a quantity depends on multiple variables and you consider all the variables as constant except one. $\endgroup$
    – ACR
    Commented Mar 9, 2019 at 18:19

2 Answers 2


I think you are starting on the wrong foot.

$U$ and $H$ are state functions. They both depend on several variables. $U$ is a function of entropy $S$ and volume $V$ and composition. For a pure fluid there is no dependence on composition, so we can just say $U = f(S,V)$. Similarly $H$ is a function of entropy and pressure $p$ i.e., $H=f(S,p)$. You go from U to H by using a Legendre Transform (but ignore Legendre Transforms for now... I will give you a good reference to read for thermodynamics at the end, it will make your life better).


Reality has a way of changing things. For instance experiments are done almost always at a constant $T$, and either constant $p$ or $V$. Therefore, we like thinking of things in terms of these variables. Lets face it... entropy is tricky.


Okay, lets now say that $U=f(T,V)$ and $H=f(T,p)$ this is alot easier to handle. In thermodynamics we are always after changes in things. Therefore, the differential forms of $U$ and $H$ are desirable.

Also, this is not a haphazard switch, entropy and temperature are intricately related to each other $T \equiv \left( \frac{\partial U}{\partial S} \right)_{x's}$ where $x's$ are all other independent variables being held constant.

The differentials for both $U$ and $H$ are as follows (with respect to our new independent variables).

$$U(T,V)$$ $$dU = \left( \frac{\partial U}{\partial T} \right)_V dT + \left( \frac{\partial U}{\partial V} \right)_T dV$$ and $$H(T,p)$$ $$dH = \left( \frac{\partial H}{\partial T} \right)_p dT + \left( \frac{\partial H}{\partial p} \right)_T dp$$


It is here that we now define what $C_V$ and $C_p$ are

$$C_V \equiv \left( \frac{\partial U}{\partial T} \right)_V$$

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

These are fundamental definitions and can't be tampered with. In introductory thermodynamics teachers play games and try to frankly dumb things down, and this leads to, in my opinion, improper definitions and understanding. This link provides a definition (without much context) that ties $Q$ into the mix.

If we insert these definitions into our differential equations we have

$$dU = C_V dT + \left( \frac{\partial U}{\partial V} \right)_T dV$$ and $$dH = C_p dT + \left( \frac{\partial H}{\partial p} \right)_T dp$$

I leave it as an exercise, if you wish, to find the definitions for $\left( \frac{\partial U}{\partial V} \right)_T$ and $\left( \frac{\partial H}{\partial P} \right)_T$

Final Resource

Herbert B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd Edition

The man called Callen

The Textbook


In thermodynamics, the definition of heat capacity is somewhat modified and generalized from the version that we learned in freshman physics (i.e., in terms of heat q), which applied only to systems where no work was occurring. This modification is necessary because heat capacity is supposed to be a physical property of the material, independent of process path. However, we know from thermodynamics that q actually is a function of process path. So it is no longer appropriate to use to define heat capacity. Now, in thermodynamics, the heat capacity Cv matches the old definition in terms of q only when volume is constant, and the heat capacity Cp matches the old definition in terms of q only when pressure is constant. Otherwise, no.

The new definitions are much more general than the old definitions and, more importantly, because U and H are functions of state (and not path), so are Cv and Cp.


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