Is our definition of entropy unique? Suppose we don't know anything about statistical mechanics, not even the existence of atoms.
Why is entropy defined as 
$$\delta S=\frac{\delta Q}{T}$$
instead of, say, 
$$\delta S=\frac{\delta Q}{T^2}$$
or any other function that will conserve the 2nd Law of thermodynamics? In a nutshell, is our entropy definition unique?
 A: In thermodynamics, the definition of entropy is unique up to redefinitions of temperature. 
The Zeroth Law of thermodynamics tells us that a temperature scale exists, but it doesn't specify anything more than that. Therefore, we are free to replace temperature $T$ with any monotonic function $f(T)$, in which case the definition of entropy becomes
$$\Delta S = \frac{\Delta Q}{f(T)}.$$
As you've seen, this doesn't upset the Second Law. It does change what "temperature" means, though. The Carnot efficiency, the form of the ideal gas law, etc. all have to be changed.
This might sound puzzling, because nobody ever seems to mention this. The reason is that there are many ways of sneaking the choice $f(T) = T$ in. For example, a standard thermodynamics book could begin with the ideal gas law, which defines temperature by $T = pV/nR$. Then it can be used to derive the efficiency for an ideal gas Carnot engine. Comparing other cycles with this one in turn leads to the Clausius inequality, and hence the usual definition of entropy $\Delta S = \Delta Q / T$. 
On the other hand, if one starts with the Carnot engine without specifying the working fluid, then the most one can conclude is that a Carnot engine running between reservoirs of temperature $T_1$ and $T_2$ has an efficiency $\eta$ obeying
$$1 - \eta(T_1, T_2) = \frac{g(T_2)}{g(T_1)}$$
as can be shown by considering the composition of two Carnot engines in series. If the book is sloppy, then at some point in this analysis it will implicitly take $g(T) = T$, thereby fixing a temperature scale. This, of course, agrees with the "ideal gas" temperature $T = pV/nR$. But we're also free to take any function $g$, and choosing a nontrivial $g$ is equivalent to choosing a nontrivial $f = g^{-1}$ above.
Luckily, all of this discussion is moot, because there really is a sense that $f(T) = T$ is the best choice. That's because in statistical mechanics, we have a more fundamental definition of entropy,
$$S = k_B \log \Omega.$$
This definition is unique, and it forces the choice $f(T) = T$.
A: It can be proved that, if the definition is $\delta S=\delta Q_\text{rev}/T$, the entropy is a state function for an ideal gas (a different definition would not ensure "statefunctionness"). That was the reason why Clausius chose $\delta S=\delta Q_\text{rev}/T$, but then he decided that the entropy so defined is a state function for ANY system. The proof he offered, however, was invalid. Since then the statement "entropy is a state function" has been repeated so many times that nowadays nobody would question it. Still it is just as unjustified as in the time of Clausius. 
