In thermodynamics, entropy is defined as $ d S = \dfrac{\delta q_{\rm }}{T}$. This definition guarantees that heat will transfer from hot to cold, which is the second law of thermodynamics. But, why do we denote entropy as$\dfrac{\delta q_{\rm }}{T}$ other than $\dfrac{\delta q_{\rm }}{T^2}$,$\dfrac{\delta q_{\rm }}{e^T}$,or something else?
Is there an intuitive explanation for this $\dfrac{\delta q_{\rm }}{T}$?
The very short and definitive answer is this is how thermodynamic temperature is defined.
It goes back to very first formulations of the second law of thermodynamics by Carnot and Clausius, to wit, that it is impossible to build a perpetual motion machine of the second kind or "heat can never pass spontaneously from a colder to a warmer body" and the implications of this law to the efficiencies of heat engines. A perpetual motion machine of the second kind is one whose state undergoes a cycle in phase space and, on returning to its beginning state, has pumped heat from a colder to a hotter body without any input of work.
The Wikipedia page on Temperature under the heading "Second law of Thermodynamics" gives a reasonable summary of these ideas; "The Laws of Thermodynamics" (Chapter 44) in the first volume of the Feynman Lectures on Physics is a much fuller exposition.
It all comes down to the efficiencies of reversible heat engines, which, in Carnot's conception, work either by (i) drawing heat from a hotter ("higher temperature", not yet well defined) reservoir and dumping some of it to another cooloer ("lower temperature"", not yet well defined) reservoir whilst outputting the difference as useful work or (ii) work in the inverse way, taking in mechanical work to pump heat from the cooler to the hotter body. A "reservoir" here is a hot body that is so big that any amount of heat added to or taken from it does not appreciably change its macrostate.
By a thought experiment whereby the work output of one reversible heat engine taking heat from hot reservoir and dumping it to the cold reservoir is used to drive another reversible engine taking heat in the opposite direction. After a little work with this idea, it readily follows that the efficiencies of the two reversible heat engines must be the same. Otherwise if one efficiency were greater than the other, we could use the greater efficiency engine as the heat pump and violate the Carnot / Clausius statement of the second law. So we have now Carnot's theorem that:
The efficiencies of all reversible heat engines working between the same two reservoirs must all be the same and depends only on those reservoirs and not on the internal workings of the heat engines
Once you understand this, you now have a way of comparing different reservoirs from the point of view of idea heat engines. Namely, we take a particular reservoir as a standard and call its temperature unity, by definition. Then, if we run a reversible heat engine between a hotter reservoir and this one, and $t$ units of heat is taken from hotter one for each unit of heat dumped to our standard reservoir (thus producing $t-1$ units of work), then we shall call the temperature of the hotter one $t$ units, by definition. Likewise, if we run a reversible heat engine between our standard reservoir and a colder one and we find that our reservoir delivers $t$ units of heat to the engine for every unit dumped to the colder reservoir, then the colder one is by definition at a temperature of $\frac{1}{t}$ units. In general the proportions of heat flowing between reservoirs of temperatures $T_1$ and $T_2$ ($T_1>T_2$) defined in this way in a reversible heat engine (i.e. heat $Q_1$ is drawn from reservoir one and heat $Q_2$ is dumped into reservoir 2, thus producing work $Q_1 - Q_2$) are always in the same proportions and given by:
$$\frac{Q_1}{T_1} = \frac{Q_2}{T_2}$$
From this definition, it then follows that $\frac{\delta Q}{T}$ is an exact differential because $\int_a^b \frac{d\,Q}{T}$ between positions $a$ and $b$ in phase space must be independent of path (otherwise one can violate the second law). This last statement is not altogether obvious: you'll have to look at Feynman for details. So we have this new function of state "entropy" definied to increase by the exact differential $\mathrm{d} S = \delta Q / T$ when the a system reversibly absorbs heat $\delta Q$.
So the entropy expression is the way it is owing to the way we define the thermodynamic temperature, which definition is in turn justified by Carnot's theorem.
Pretty neat, eh?
Anyhow, what happens in practice is the following. Now that we have a definition of the ratio of temperatures in terms of the efficiency $\eta$ of the reversible heat engine running between reservoirs of these temperatures:
$$\frac{T_2}{T_1} = 1-\eta$$
one defines a "standard" unit temperature (e.g. as something like that of the triple point of water), then the full temperature definition follows. This definition can be shown to be equivalent to the definition of temperature for a system:
$$T^{-1} = k\,\partial_U S$$
i.e. the inverse temperature (sometimes quaintly called the "perk") is how much a given system "thermalizes" (increases its entropy) in response to the adding of heat to its internal energy $U$ (how much the system rouses or "perks up"). The Boltzmann constant depends on how one defines one's unit temperature - in natural (Plank) units unity temperature is defined so that $k = 1$.
Now temperature is sometimes said to be proportional to the average energy of a system's thermalized constituent particles. This is true for ideal gasses, but not the general definition. For example, I do the calculation of the temperature of a collection of thermalized quantum harmonic oscillators in this answer here and the thermodynamic temperature is only equal to the mean oscillator energy for temperatures $T$ such that $k\,T\gg \hbar \omega$, where $\hbar \omega$ is the photon / phonon (as appropriate) energy of the oscillator.
$$ \eta = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1} $$
– Papa Smurf Jan 16 '24 at 16:35First off, temperature is an intensive quantity, i.e., not additive. For example, two cups of coffee don't have twice the temperature of one cup. For an extensive (additive) quantity, such as mass, we can't just redefine $m\rightarrow m'=f(m)$, where $f$ is a nonlinear function, because then $m'$ wouldn't be additive. This constraint doesn't apply to temperature, because temperature is intensive.
The relation $dS=dq/T$ is really the definition of temperature, not entropy. (Entropy is really defined as the log of the number of accessible states.) But this still allows us to take $T\rightarrow T'=f(T)$, where $f$ is some nonlinear function, and then we would just define temperature as $T'=f(dq/dS)$. We would need $f$ to be a one-to-one function, because we don't want objects that aren't in equilibrium to have the same temperature.
There is really nothing wrong with this, and for example this question discusses the possibility $f(x)=1/x$. Some equations (e.g., the partition function) actually come out simpler if written using this definition, although others get more complicated (the heat capacity of an ideal gas is no longer constant). However, most possibilities for $f$ result in all equations looking more complicated.
The answer is plain and simple differential calculus $$ds = \frac{\partial s}{\partial e} \large{|}_vde +\normalsize{\frac{\partial s}{\partial v}} \large{|}_e dv$$ What does the differential change $$ds=\frac{\delta q}{T}$$ have to do with the first? For starters, an important question on everyones mind should be what is temperature? Is it a physical quantity that we have intuition about what it represents? In certain circumstances perhaps, but in general we have no intuition about what temperature truly represents. So what is temperature? It is simply defined by $$T=\frac{\partial e}{\partial s}\large{|}_v$$ If you want to go start your own country and define temperature some other way feel free to do so, but no one is going to follow you. The first equation is thus $$ds = \frac{de}{T} +\normalsize{\frac{\partial s}{\partial v}} \large{|}_e dv$$ So how do you get to the second equation? Two simple assumptions: no volume change occurred, in other words no physical work, and the internal energy change was strictly due to heat transfer ($dv=0$ and $de=\delta q$). Voila!
