Can mass-energy equivalence be used to measure absolute internal energy?

$E = mc^2$ where m is the relativistic mass. $m_0$ is the classical or rest mass.

Consider a closed system at rest with no heat added and no work done, but with an internal chemical or nuclear reaction. At rest means no change in kinetic or gravitational potential energies of the overall system. $\Delta U = 0$ where $U$ is the total internal energy.

Classical thermodynamics considers the energy of the reaction as the "internal energy of formation", and defines $\Delta U$ as $U_{classical \enspace products} - U_{classical \enspace reactants} - U_{formation}. \Delta U = 0$ for this example. $U_{classical}$ is the heat capacity at constant volume for all the moles or nuclei in the system. (See one of the thermodynamics textbooks by Sonntag and van Wylen.) In an engineering thermodynamics context the (classical) internal energy is just the heat capacity.

The energy balance for classical thermodynamics was developed before $E=mc^2$ was understood, hence the need for considering the energy of formation. Using $E=mc^2$ we can express this energy of formation as a change in rest mass.

We can express $U_{formation}$ in terms of the rest masses of the reacting constituents. Consider the constituents in the system; atoms/molecules for a chemical reaction, nuclei for a nuclear reaction. For each constituent $U = U_{classical} + nm_0c^2$ where $m_0$ is the rest mass energy- of an atom/molecule or nucleus- and n the total number of moles or nuclei of the constituent. (See note (a) below.) So for the reaction $a + X ->b + Y$, we have $U_{classical \enspace a} + U_{classical \enspace X} + n_am_{0a}c^2 +n_Xm_{0X}c^2 = U_{classical \enspace b} + U_{classical \enspace Y} + n_bm_{0b}c^2 + n_Ym_{0Y}c^2$

So $U_{formation} = n_am_{0a}c^2 + n_Xm_{0X}c^2 - (n_bm_{0b}c^2 + n_Ym_{0Y}c^2)$; the internal energy of formation is equal to the change in the rest masses. If the product- b and Y- rest masses are less than the reactant- a and X- rest masses, $U_{formation}$ is positive and $U_{classical \enspace products}$ is greater than $U_{classical \enspace reactants}$, due a reduction in rest mass causing an increase in $U_{classical}$.

The change in rest mass is very small for a chemical reaction as contrasted with a nuclear reaction, but the same concept holds. In a chemical reaction the change in rest mass is dictated by the binding energy of the electrons in an atom/molecule. In a nuclear reaction the change in rest mass is dictated by the binding energy of the nucleons in a nucleus.

So far, we have considered the energies of all the reactant and product atoms/molecules in the system. We can also view this entire system, which is at rest, externally as having total internal energy $U_{total} = m_{0 \enspace system} c^2$ that is constant, where $m_0$ considers the overall classical internal energies and rest masses of the constituents in the system.
$m_{0 \enspace system} c^2$ = $U_{classical \enspace a} + U_{classical \enspace X} +n_a m_{0a}c^2 + n_Xm_{0X}c^2 = U_{classical \enspace b} + U_{classical \enspace Y} +n_bm_{0b}c^2 + n_Ym_{0Y}c^2$

So, viewing the system externally, the absolute internal energy is the the total rest mass energy for an isolated system (no heat, work, or mass transfer) that is at rest (no change in overall kinetic or potential energy).

Note (a). For a discussion of the reaction energetics on an atom to atom or nucleus to nucleus basis see my answer at Why is mass defect calculated by the rest mass (energy)? on this exchange.