Absolute entropy: is it valid to write $Q=ST$?

Question

All the basic texts say $dQ = T dS$.  Some say it is possible to define absolute entropy.  That source gives an equation integrating C dT which should reduce to $Q=ST$, if $S(0) = 0$ and $Q$ is defined as all the heat that can possibly be extracted from the object.  But it seems like sources overwhelmingly avoid writing $Q=ST$.  If I could use that equation it should be simple to visualize the nature of entropy... (see below)
This would also seem to make entropy the number of different variables that have been able to take up thermal energy.  But when taking actual entropy values, for example, 131 J/mol K for H2, 192 J/(mol K) for N2, and 193 J/(mol K) for NH3 at STP - the values are many times larger than the gas constant.  Is there a way to conceptualize entropy to break down these numbers into 46 different degrees of freedom, or is this comparison intrinsically flawed?  Why?
EDIT: in light on ongoing discussion, I'd like to quote a very basic exercise from LibreTexts that made me start thinking about this issue:
Calculate entropy change when 36.0 g of ice melts at 273 K and 1 atm. [enthalpy of fusion 6.01 kJ/mol].
Their solution is $Delta S$ = (6.01 kJ mol-1)/272 K * (36 g)/(18 g mol-1) = 1.22 kJ / K
This illustrates we can have more than one entropy value for the same temperature, and also something of a surprising result, namely that the higher the melting point of the substance, the less melting it affects the entropy.  Of course, this is countered by likely requiring more energy to do the actual melting as reflected in the enthalpy value, but still, it made me wonder what this denominator means.  The interpretation I want to give is that there is a pool of heat energy which is larger at higher temperature, and the total entropy has been increased in proportion to the expanded capacity to store heat at the same temperature.  But this would imply thinking of a relationship between these total quantities.

Andrew · Answer

Heat is not a thermodynamic state variable, while entropy is.
In other words, heat is a word that describes the spontaneous transfer of energy between a system and its environment (which we denote $dQ$), which depends on the path, or the way that you apply changes to the system's macroscopic variables. It is meaningless to talk about heat as a property of the system itself (there is no heat function $Q$ that is a function of the system's state).
Entropy, on the other hand, is a state variable, and a system in equilibrium does have a well defined value $S$.
Therefore, $Q=TS$ is meaningless, as $Q$ is meaningless.

Chemomechanics · Answer

It's fine to wish to integrate $T,dS$ and to plan for a lower limit of $S=0$ under the Third Law. It's not fine to bring $T$ outside the integral to obtain $TDelta S=TS$ as if $T$ is constant, because it's not (unless you're integrating from 0 to 0, in which case no net energy is transferred).

Answered by Chemomechanics on May 20, 2021

Wolphram jonny · Answer

In an isothermal process, in which T is a constant, you can integrate and obtain $Q=TDelta S$, in which Q is the heat exchanged to change the entropy by $Delta S$. For non-isothermal processes you will obtain a different relationship though.

Answered by Wolphram jonny on May 20, 2021

Andrew Steane · Answer

The absolute entropy of a system in a given equilibrium state at temperature $T$ and other parameters specified is given by
$$
S = S(0) + int_0^T frac{1}{T} dQ_{rm rev}
$$
where it is understood that the other parameters are being held constant and $dQ_{rm rev}$ refers to heat entering the system by a reversible process, and the limits on the integral are temperatures. Taking $S(0) = 0$, a useful way to write the result is
$$
S(V,T_f) = int_0^{T_f} frac{C_V(V,T)}{T} dT + sum_i frac{L_i}{T_i}
$$
where $C_V$ is the heat capacity at constant volume (in the case of a simple mechanical system) and the sum gives the contribution from first order phase transitions. For the $i$th transition the heat entering is the latent heat $L_i$ and during the transition the temperature ($T_i$) does not change.
The integrals above are not the only way to obtain the absolute entropy, but other ways must be consistent with these. Sometimes it can be obtained by calculation from first principles, and sometimes by a judiciously chosen measurement combined with some thermodynamic reasoning. Here is a paper by myself on that:
https://iopscience.iop.org/article/10.1088/1367-2630/18/4/043022
The total heat that enters the system during such a warming is $Q = int dQ_{rm rev}$ and clearly one will not get $Q = ST$. You could if you like define a quantity $ST$ and denote it by the letter $Q$, but this $Q$ would not be the total heat that can be extracted from the system under some given constraint or path, unless by chance it might be for some carefully chosen path.

Absolute entropy: is it valid to write $Q=ST$?

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