A relationship between entropy and temperature I tried deriving a formula relating entropy (not change in entropy, but entropy itself) to temperature. I’ve only seen two equations really relating to entropy thus far, and only one of them includes the temperature factor. However, that one regards change in entropy, not absolute entropy:
$$dS=\frac{dQ}{T}\tag{1}$$
The other equation doesn’t include temperature:
$$S=k_B\ln{W}\tag{2}$$
So I tried doing my own manipulations:
$$dS=\frac{dQ}{T}$$
$$dQ=mc\:dT$$
$$dS=\frac{mc}{T}\:dT$$
$$\int dS=mc\int\frac{1}{T}\:dT$$
$$S=mc \ln{T} +C$$
I was surprised I managed to get such a result. This seemed to be exactly what I was looking for. However, I had never seen such an equation in my life before. 
My question are: 
Was my derivation correct? If not, what did I do wrong? If it was, then why isn’t this formula more commonly used/taught (assuming $C$ is found)?
And if I’m doing unnecessary steps, is there actually a formula in existence that achieves what I want? A relationship between absolute entropy and temperature?
 A: Your equation is correct only if:$$\mathrm dQ = mc\,\mathrm dT$$ which is not generally true, indeed, common sense tells you that a change in temperature leads to conclusion that an object being heated up. But we do not encounter gases much in our life, which could be regarded as a general case. In reality your assumption is generally false, a good example would be a general gas process: $$\mathrm dQ =\mathrm dW +\mathrm dU$$ Which would take this form for an ideal gas:
 $$P\mathrm dV+\frac{3}{2}NR\,\mathrm dT = P\,\mathrm dV \,\,\,\text{ if } \,\,\,\mathrm dT=0$$
Now you can see that although temperature does not change, heat supply is still possible, at least mathematically. This process is called Isothermal heating, as might have guessed already. This could serve you as an example of non increasing temperature heating.
After that, you should see that $dQ = mc\mathrm dT$ is usually not the case. Because temperature is not the only extensive parameter for a gas, i.e. is not the only thing which determines the energy supply, i.e. is not the only thing which rises when heats comes in. Real equation for entropy of an ideal gas then would look like:
$$\mathrm dS = \frac{1}{T}(P\,\mathrm dV + (3/2)NR\,\mathrm dT) = NR\frac{\mathrm dV}{V} + \frac{3}{2} NR \frac{\mathrm dT}{T}$$
Then $S$ would look like,
$$S = NR \ln{ \left( \left(\frac{V}{V_0} \right) \left( \frac{T}{T_0}  \right)^{3/2}  \right)}$$
for $N = \textrm{constant}$, namely your number of particles does not change, and $V_0$ and $T_0$ play the same role as $C$ in your answer. Bear in mind that this result is not to be used in your studies, since some modifications and generalizations are to be made first, but it is good enough for your understanding.
A: Your derivation is almost correct, and as I'll show below, one does sometimes make use of it. However, it relies on an unwritten assumption that makes it applicable only in a limited range of situations.
That assumption is that the heat capacity $c$ doesn't depend on the temperature. It is this assumption that allowed you to take the factor of $mc$ outside the integral in your second-to-last step. In general, the heat capacity does depend on temperature, so you end up with
$$
\int dS=m\int\frac{c(T)}{T}\:dT,
$$
which would usually have to be integrated numerically, since the function $c(T)$ is some complicated thing that depends on the properties of the material you're working with, and which generally has to be measured rather than being something you can write down an equation for.
The reason your argument does sometimes work is that often the function $c(T)$ is approximately constant over some particular range of temperatures. In that case, to be careful, we should put limits on the integrals, to get
$$
\int_{T_1}^{T_2} dS=m\int_{T_1}^{T_2}\frac{c(T)}{T}\:dT \approx mc\int_{T_1}^{T_2}\frac{1}{T}\:dT,
$$
which gives us
$$
\Delta S = mc(\ln T_2 - \ln T_1) = mc\ln\frac{T_2}{T_1},
$$
which is the formula you'll occasionally see in applications. Note that we're now able to express it as a logarithm of a ratio, so the units are OK.
CuriousOne pointed out a comment that your original formula doesn't work if you plug in $T=0$. The reason for this is the third law of thermodydnamics, which implies that $c$ always goes to zero as $T\to 0$, so the range over which $c$ is approximately constant can't include the zero point on the absolute scale.
A: Here is a link that might be useful
http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/temper2.html#c2
A: Above question poses a new look for entropy and possibily gives solution for entropy with rate of cycles, frequency. As per present understanding of entropy,
$S=k_B\ln W=k_B\ln\frac{N}{N_i}\tag*{}$
is not independent of temperature. The number of microstates reduces as temperature rise because particles go to higher energy, thus increases available energy for work. This is evident from blackbody spectrum, increasing temperature increases intensity manifold.
From the relation of specific heat,
$dQ=m\ c\ dT=N\ c\ dT\Rightarrow S=N\ s\tag*{}$
where $s=Mc$ is specific heat or entropy per cycle or particle and $M$ is mass of a particle, $S=\frac{dQ}{dT}$ is total entropy dependent upon frequency of cycle.
Also, $\ s=\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\ldots =\frac{Q_N}{T_N}$ for $N$ particles
$S=\frac{dQ}{dT}=\frac{Q_1}{T_1}+\frac{Q_2}{T_2}+\ldots +\frac{Q_N}{T_N}=N\ s\tag*{}$
From Maxwell-Boltzmann's distribution the speed of different particles is different and that depends upon temperature. Thus different particles in a system have unequal energy and thus having different temperature or heat level.
