Specific Heat equations - paradox! We know that enthalpy is $h = u + pv$. Where $u, p, v, h$ are internal energy, pressure, volume, and enthalpy respectively. Now specific heat at constant volume is calculated as $c_v = \frac{\partial u}{\partial T}$, and specific heat at constant pressure as $c_p = \frac{\partial h}{\partial T}$.
I am pretty confused at this point, in volume case it is internal energy, and in pressure case it is enthalpy. Because if $pv$ is a form of energy independent of internal energy, then it should figure in $c_v$ as well, because at constant volume when I am putting energy in the system, the pressure can not stay constant.
What is wrong in the above argument? Appreciate your thoughts!
 A: $pv$ is not "a form of energy". It is just product of pressure and volume. Specific heat is the amount of heat needed to increase temperature by one unit. This heat can be expressed as $Tds$ for quasi-static processes. In constant volume case this can be expressed as $du$, because $du = Tds -pdv$ generally and $dv=0$. In constant pressure case the heat is $dh$ because $dh = Tds+Vdp$ generally and $dp=0$.
A: I know that an answer has been accepted already, but I don't think the current answers are didactic enough (even though they're technically good). It seems to me that the author is still confused, so I'll give it a try myself.
The confusion (aka "paradox") stems from the student forgetting what specific heat really means, in the sea of lowercase letters and partial derivatives that thermodynamic textbooks tend to be. Please allow me to remember: specific heat is merely the average amount of heat needed to increase each unit of temperature in a process (actually that is heat capacity; specific heat is simply that divided by mass).
So, if $2 \text{ kg}$ of something took $250 \text{ kJ}$ to heat up $5 \text{ °C}$, the specific heat in that process was $c = 25 \frac{\text{kJ}}{\text{kg} \cdot \text{K}}$. In general: $$c = \frac{Q}{m \Delta T}$$
Ok, that was easy. Let's just use $q = \frac{Q}{m}$ and note that, in this case, we're usually interested in the instantaneous rate of change, so we'll make $\Delta T \rightarrow 0$: $$c = \frac{\delta q}{\delta T} \tag1$$
$\delta$ means that the infinitesimal changes are dependent on the path the process takes. Now, it's only a matter of knowing a bit about the process. Mainly, there are two special cases:
Constant volume (isometric process)
Let's recap the First Law: $$Q = \Delta U + W \tag2$$
In an isometric process, our substance never expands, so no work can be done ($\delta W = p dV$, remember?). So we have $Q = \Delta U$. Let's plug that into $(1)$, and find our equation specifically for the constant volume process: $$c_v = \left( \frac{\partial u}{\partial T} \right)_v \tag3$$
(don't be scared of the $(\cdots)_v$ thing; it only enforces that the $u$ in question be along an isometric line — simple, but very important!)
Back to the matter in hand, that equation merely reflects the fact that, in an isometric process, all the heat goes directly into the internal energy, so they're the same. It's exactly the same as $\frac{\delta q}{\delta T}$.
The same thinking applies to the constant pressure process, which we'll see now.
Constant pressure (isobaric process)
With constant pressure, $\delta W = p dV$ can be integrated to give $W = p \Delta V$, so $(2)$ becomes: $$Q = \Delta U + p \Delta V = \Delta H$$
Similarly to constant volume, here we find that, in a constant pressure process, all the heat goes directly into enthalpy. So we're quick to turn $(1)$ into: $$c_p = \left( \frac{\partial h}{\partial T} \right)_p \tag4$$
Additional example: generic reversible process
Really, anything you know about $Q$ can be used to define $c$. For instance, in any reversible process, the Second Law gives us: $$Q = T \Delta S$$
Why, we could use that to find that, at any point of any reversible process: $$c_\text{rev} = T \left( \frac{\partial s}{\partial T} \right)_\text{rev} \tag5$$
I mean, I never saw anyone write that (at least not exactly that), but why couldn't we?
A: $\Delta U = Q-W$, according to the first law of thermodynamics.
If $\Delta V = 0$, then $W = 0$
So if $\Delta V = 0$, then $\Delta U = Q$
Therefore, if $\Delta V = 0$, then $C_V \equiv (\frac{\partial U}{\partial T})_V$ is the appropriate heat capacity.
However, if $\Delta P = 0$ and $V$ is free to change, then work must be consider and therefore $C_P \equiv (\frac{\partial H}{\partial T})_P$ is the appropriate heat capacity.
