Thermodynamics of a bar, specific heat I have a problem and I don't understand how the book solves it. It says I have a bar such that $$F=aT^2(L-L_0)$$
where $F$ is the force or tension of the bar, $T$ is the temperature, and $L$ the longitude. $L_0$ is the longitude when $F=0$. It says that $C_{L_0}=bT$ and it asks me about the general function $C_L(L,T)$, being $C_L$ the specific heat at constant $L$.
The book starts solving it with this expression: $$\left(\frac{\partial C_L}{\partial L} \right)_T=-T\left(\frac{\partial^2 F}{\partial T^2}\right)_L$$
From there, I know how to go on, but I have no idea where that comes from. I have tried to get that out of the generalised Mayer relation but if that's the way, I can't do it.
 A: Recall first that the definition of $C_L$ is
$$
  C_L = \left(\frac{\partial U}{\partial T}\right)_L
$$
The first law of thermo for the bar reads
$$
  dU = T dS + F dL 
$$
Taking $S=S(L, T)$ gives
$$
  dS = \left(\frac{\partial S}{\partial T}\right)_L dT + \left(\frac{\partial S}{\partial L}\right)_T dL
$$
Combining these two equations gives
$$
  dU = T\left(\frac{\partial S}{\partial T}\right)_L dT + \left[T\left(\frac{\partial S}{\partial L}\right)_T +F\right]dL
$$
From which we see that
$$
  \left(\frac{\partial U}{\partial L}\right)_T = T\left(\frac{\partial S}{\partial L}\right)_T +F
$$
On the other hand, one has the the following Maxwell relation (I'll leave it to you to derive it, but I can show you how if you want):
$$
  \left(\frac{\partial S}{\partial L}\right)_T = -\left(\frac{\partial F}{\partial T}\right)_L
$$
So plugging this into the last equation gives
$$
  \left(\frac{\partial U}{\partial L}\right)_T = -T\left(\frac{\partial F}{\partial T}\right)_L +F
$$
Taking the partial of both sides with respect to $T$, noting that partial derivatives commute, and then using the definition of $C_L$ gives precisely the equation you want!  Let me know if you want more detail!
Cheers!
