An intuitive reason for the definition of temperature via entropy and internal energy?

[Ax_xiom]

So, temperature is formally defined as the inverse of the rate of change of entropy in respect to internal energy. I'm fine working with this mathematically but I don't get it intuitively. Does anyone have some intuition as to why temperature is defined this way?

 

[anuttarasammyak]

$$\frac{1}{T}=\beta=\frac{dS}{dE}=S'(E)$$
$\beta$ is derivative of entropy increase per internal energy increase.
Large $\beta$, small T, implies that the system is hungry for energy to satisfy its desire for entropy.
$\beta$ is degree of energy avarice of the system to satisfy its desire for entropy.
T is degree of energy satisfaction of the system with its entropy asset.
When two systems of different T's contact, they agree that they should share the same degree of satisfaction to achieve maximum entropy benefit of the whole system and after energy exchange they share same new T.

[EDIT]
Velocity $\frac{dx}{dt}$, Specific heat $\frac{dE}{dT}$ are similar in the sense that derivative matters.

In a closed system with subsystems 1 and 2
$$E=E_1+E_2$$ E is constant.
$$S=S_1+S_2$$
From the second law of thermodynamics
$$\frac{dS}{dt}=\frac{dS_1}{dE_1}\frac{dE_1}{dt}+\frac{dS_2}{dE_2}\frac{dE_2}{dt}
$$ $$=(\frac{dS_1}{dE_1}-\frac{dS_2}{dE_2})\frac{dE_1}{dt}=(\frac{1}{T_1}-\frac{1}{T_2})\frac{dE_1}{dt} \geq 0$$
$T_1<T_2$ :energy flows from 2 to 1.
$T_1>T_2$ :energy flows from 1 to 2.

The tendency of a system to gain energy cannot be considered in isolation: in a closed composite system, that energy must come from another subsystem.

 

[marcusl]

dS=dE/T. At low temperature, systems are highly ordered and a small energy input results in a relatively large increase in disorder. At high T systems are already disordered, so a small energy increase doesn’t change the entropy much.

 

[Chestermiller]

So, temperature is formally defined as the inverse of the rate of change of entropy in respect to internal energy. I'm fine working with this mathematically but I don't get it intuitively. Does anyone have some intuition as to why temperature is defined this way?

when you use a partial derivative, you must specify what its being held constant. The thing that is being held constant is volume.

The equation you present is originally derived from : $$dE=TdS-PdV$$so $$dS=\frac{1}{T}dE+\frac{1}{T}PdV$$so $$\left(\frac{\partial S}{\partial E}\right)_V=\frac{1}{T}$$So the constraint on your equation is that the volume V is constant.

The original equation you wrote is correct for an ideal gas (for which the internal energy is constant), but not for a real gas, where internal energy depends on volume.

 

[Ax_xiom]

dS=dE/T. At low temperature, systems are highly ordered and a small energy input results in a relatively large increase in disorder. At high T systems are already disordered, so a small energy increase doesn’t change the entropy much.

Thank you, I think this gives me more intuition on the definition of temperature.

The equation you present is originally derived from : dE=TdS−PdV

Out of curiosity, is it possible to see how this original expression is derived? And is the $TdS$ term energy input into the system, and the $-PdV$ term work done by the system?

 

[Matterwave]

Out of curiosity, is it possible to see how this original expression is derived?

There's a couple of different ways to help build intuition.

We can start by using the first law of thermodynamics $dE=\delta Q - \delta W$ where $\delta Q$ is heat entering the system and $\delta W$ is work done by the system. We have the Claussius definition of entropy $dS=\delta Q_{rev}/T$. The subscript means equality holds for reversible processes.

Work is force times displacement. If you think of your volume like a piston with volume $V$ at pressure $p$ (which is force per unit area), then you can easily convince yourself the work must be $pdV$ (you can also analyze this in terms of units to help build intuition).

Combining these gets you $dE=TdS-pdV$.

If you are worried about the "reversible" there, the real reason/derivation comes from doing a little bit of statistical mechanics, noting that $S=S(E, V)$ is a proper, monotonically increasing (w.r.t. $E$) state function and therefore can be inverted to get $E=E(S,V)$ from which the rest of the formulae follow.

And is the $TdS$ term energy input into the system, and the $-PdV$ term work done by the system?

Heat input into the system (see above), and if you keep that minus sign then that's work done on the system.

 

[Chestermiller]

Thank you, I think this gives me more intuition on the definition of temperature.

Out of curiosity, is it possible to see how this original expression is derived? And is the $TdS$ term energy input into the system, and the $-PdV$ term work done by the system?

Yes. This is essentially a differential version of the 1st law of thermodynamics along a reversible path.