Temperature dependence of entropy $$\text{Entropy}=\frac{\text{Heat absorbed}}{Temperature}$$
$$\Rightarrow S=\frac{Q}{T}$$
$$[S]=[ML^2 T^{-2} K^{-1}]$$
If entropy increases with increase in temperature of the system, then it suggests that entropy is directly proportional to the temperature of the system. But the equation of entropy from second law of thermodynamics shows temperature being in denominator, i.e., S(entropy) = Q(heat) / T(temperature). Why is it so? Being in denominator suggest inversely proportional, but we know entropy is directly proportional to temperature.
 A: As already noted, the proper definition of entropy is
$$dS=\frac{\delta Q_{rev}}{T}$$
$$\Delta S=S_{2}-S_{1}=\int_1^2 \frac{dQ_{rev}}{T}$$
The application of this definition does not conflict with your statement ..."entropy increases with increase in temperature of the system". 
For example, we know that heat transfer only occurs spontaneously from a higher temperature body to a lower temperature body. Since $\delta Q$ is positive for the lower temperature body there is in an increase in entropy of the lower temperature body. Generally (see exception below) it also results in an increase in the temperature of the lower temperature body. So here we have an increase in entropy associated with an increase in temperature.
It should be noted that although an increase in temperature means an increase in entropy, an increase in entropy does not necessarily mean an increase in temperature. An example is a reversible isothermal expansion of an ideal gas in which entropy increases but temperature does not. 
Hope this helps.
A: The first thing to note is that the statement that $S=Q/T$ simply isn’t true. In particular, there is no such thing as the “heat of a system.” The correct statement is that an infinitesimal change in entropy is an infinitesimal change in heat, divided by the temperature, or $\mathrm{d}S=\delta Q/T$.
The change in heat of a system, $\delta Q$, may in general have nontrivial dependence on the temperature. For instance, in an ideal gas, we have $U=c_VT$, and so the first law tells us that
$$c_V\mathrm{d}T=\delta Q -p\mathrm{d}V,$$
Which means that the change in heat is directly proportional to a change in temperature (at constant volume). This leads to the entropy being logarithmic in temperature, simply by integrating $\delta Q/T$.
Basically, dimensional analysis is useful, but in general we may have constants (in this case $c_V$) that can convert units in a nontrivial way.
A: The conventional natural variables of entropy are $S(E,V,N)$ (see Callen's Thermodynamics book). From this you get -
$$\frac{\partial S}{\partial E}=\frac{1}{T}$$ 
So this fundamental definition of temperature can already suggest that $S$ has units of energy over Kelvin (this is the SI definition), even though its dependence on temperature is implicit. Only through Lagrange transform this relation become explicit, as you redefine entropy as function not of its natural variables,but its derivative with respect to them. 
It's Illuminating to look at Boltzmann's definition of entropy - $$S=k_B\log\Omega$$ entropy has the units of $k_B$, whatever they are. They are set to be consistent with thermodynamic definitions.
So the conclusion is that units of entropy and temperature are interlinked, and defined to be consistent with thermodynamic definitions. If you insist that temperature would have units of its own, change in units of temperature will result in change in units of entropy. For example, using the equipartition relation as our intuition, you can set $k_B$ to be dimensionless, and temperature have units of energy. Thus entropy will be dimensionless as well.
