Why does temperature have no uncertainity? Background
My understanding is that temperature is not a quantum mechanical operator. Hence, thereby it should not have any uncertainty. However, any instrument that tries to take the measurement of temperature will have some uncertainty in it for example the mercury postion of the thermometer! Further, we know temperature is 
$$ T = \frac{\partial U}{\partial S}_{(N,V)} $$
But we cannot measure $U$ with arbitrary accuracy so there must be some uncertainty in the slope of $U$ as well!
Questions
Does temperaure have uncertainty? If yes, how is this possible when temperature is not a quantum mechanical observable? If no, how is this possible when it is the derivative of energy with respect to entropy and energy has uncertainty? 
 A: The Heisenberg uncertainty principle applies to dimensions commensurate to $\hbar$, i.e. at the level of particles ( atoms, molecules, elementary particles). Temperature is a classical observable appearing in thermodynamics as a variable, but when analyzed from the emergent statistical ensemble it is not a variable but a statistical average of the kinetic energy of the sample macroscopically displaying temperature $T$.
$$E_k=\frac12 m v^2_\mathrm{rms}=\frac32 kT$$
The formula above is for an ideal gas, but similar formulas exist for all bulk matter. Temperature is called an intensive thermodynamic variable.
As it is not a variable or a function/differential of variables at the quantum mechanical level, it cannot enter into the quantum mechanical equations as an operator. It thus cannot participate in the commutator relations that define whether an observable is limited, when measured, by the uncertainty principle.
A: There are extensions of temperature, I believe, for microscopical objets, but usual temperature is an intensive property of macroscopic objects. The number of degrees of freedom is at least Avogadro Number, and quantum effects tend to cancel. 
You could perhaps as the question for mesoscopic objects. 
