Hottest When Most Probable

$T_{boom}$ being nuclear suggests that energy is generated in excess of energy input driving the matter to higher temperature.  The velocities $v$ of individual particles are, if we admit Kinetic Theory for ideal gas, distributed according to the Maxwell-Boltzmann distribution.  The most probable velocity $v_{p}$ is given by,

$v_{p}=\sqrt{\cfrac{2RT}{M_m}}$  --- (*)

where $T$, is the temperature when most particles attain $v_{p}$

If at $v_{p}$, the particles are in resonance (post "A Small Boom" dated 16 Oct 2017) and is nuclear, then $T_{boom\,p}$ is the temperature when the matter is most nuclear, probablistically.

$v^2_{p}=\cfrac{2RT}{M_m}$

but,

$v^2_{rms}=\cfrac{3RT_{rms}}{M_m}=\cfrac{3}{2}v^2_{p}$

If we set, $v_{p}=v_{boom}$, then

$v^2_{rms}=\cfrac{3}{2}v^2_{p}=\cfrac{3}{2}v^2_{boom}$

and

$T_{boom\,p}=T_{rms}=v^2_{rms}.\cfrac{M_m}{3R}=\cfrac{3}{2}v^2_{boom}.\cfrac{M_m}{3R}=\cfrac{3}{2}T_{boom}$

Boiling or melting may not require the matter to be at its most nuclear.  Sufficient energy to initiate a change in phase will do.

However, the rms value is likely the measured value of $T$, as the average energy (root of the mean square value) content of the whole distribution profile.  However, energy radiating from the matter will peak when $v_{boom}$ is most probable, and so, when $T=T_{boom\,p}$ the matter is most nuclear.  Or simply from (*)

$T_{boom\,p}=v^2_{boom}.\cfrac{M_m}{2R}$

when $v_{boom}=v_{p}$

And we have to adjust all values of $T_{boom}$ calculated previously.