Does entropy always increase with temperature? For any system can we always say that entropy increases with temperature. In other words:
$$\left(\frac{\partial S}{\partial T} \right)_{\{\alpha\}}\ge0$$
where $\{\alpha\}$ is the set of parameters held constant. Either way  can it be proved?
 A: The quantity $\left(\frac{\partial S}{\partial T} \right)_{\{\alpha\}} = TC_v$ is essentially proportional to the heat capacity of the thermodynamic system under study.
As far as I know, there is no principle of thermodynamics that forbids such a quantity to be negative. Considerations such as "yes otherwise matter would not be stable" lie outside the standard axioms of thermodynamics and assumes many additional things in a very vague way.
To try to understand better what is going on we can turn to equilibrium statistical mechanics. In the canonical ensemble (classical will be enough for this discussion) one can write the partition function:
\begin{equation}
Q(\beta,N,V) \equiv \sum_{\{states\}} e^{-\beta E_{state}}
\end{equation}
Where $\beta = 1/(k_BT)$ is often considered as an inverse temperature. Now, for this quantity to make sense it is required that the energy spectrum is bounded from below otherwise it diverges eventually if all states of the spectrum can be sampled by the ensemble.
Moreover, two things can happen depending on the spectrum on the high energy end:


*

*If the spectrum is unbounded from above, which is the case for most physical systems, then a necessary condition for the existence of the partition function is that $\beta$ be positive to more or less ensure convergence. This is the formal reason for the "stability condition" mentioned above.

*If the spectrum is bounded from above, then it is basically defined on a compact and the parameter $\beta$ can be either positive or negative
From the equation of the partition we can get the following identities:
\begin{equation}
\frac{\partial^2 \ln Q }{\partial \beta^2} = Var(E) > 0; \:{\rm and} \:\: C_v = \frac{1}{k_BT^2}\frac{\partial^2 \ln Q }{\partial \beta^2} \: > \: 0
\end{equation}
It follows from it that in the canonical ensemble we have:
\begin{equation}
\left(\frac{\partial S}{\partial T} \right)_{N,V} = \beta \cdot Var(E)
\end{equation}
Since $Var(E) > 0$, it follows that the sign of the quantity you are considering depends on the sign of $\beta$.
In particular, if the system considered is bounded from below and above, $\beta$ can be negative and its is possible for this quantity to be negative.
Finally, there is another case that can be problematic which is that of gravitational systems which have positive temperatures but negative heat capacities (to get an equilibrium statistical mechanics agreement with such an observation one needs to look into the micro canonical ensemble as it is impossible to obtain in the canonical one as we have seen). In that case again the quantity you are looking at will be negative.
A: No, it doesn't. For example, consider a two-state system with ground state $| 0 \rangle$ and excited state $| 1 \rangle$. At zero temperature, the entropy is zero, since only state $| 0 \rangle$ is occupied. As the temperature goes to infinity, the occupancies of $|0 \rangle$ and $|1 \rangle$ become closer and closer to equal. When they are equal, we have the maximum possible amount of entropy, since there's a full missing bit of information per system.
However, continuing past infinite temperature to negative temperature, the occupancy of $|1 \rangle$ becomes greater than that of $|0 \rangle$, so the entropy goes down. As the temperature increases to $-0$, only the $|1 \rangle$ state is occupied, so the entropy returns to zero.
A: No. Under negative thermal temperatures the entropy of a system increases as T is lowered. For example, if the magnetic field is reversed quickly enough around a ferromagnetic material, then the system is in negative thermal temperature. 
