Q: How can distance and temperature be commensurable? I am sorry to intrude.  The discipline I know something about is metaphysics rather than physics itself.  And on Stack Exchange I spend my time on English Language and Usage.  
I am struggling through Roger Penrose's fascinating book Fashion, Faith and Fantasy in the new physics of the universe (Princeton 2016).  I am in the middle of his unravelling of what he calls the Big Bang paradox.
At 3.6 he provides the following "approximate" equivalents to Planck or absolute units.
"metre= $6.3\times10^{34}$
second=$1.9\times10^{43}$
gram.   =$4.7\times10^{4}$
kelvin.  =$7.1\times 10^{-33}$
C.         =$5.6\times10^{-122}$"  
I can see how distance and time can be commensurable because of the speed of light.  But how can this commensurability apply between distance and (say) temperature?  I have tried the internet, but so far I have not found anything.  If the answer is that I have to go away and do a course on advanced physics - or maths, I should understand.  But if there is a simple explanation, I should be really grateful.

A: Roughly speaking, one can associate a temperature with the energy scale of each particle at that temperature, which for simple situations is 
$$E \sim k_B T.$$
Next, one can associate an energy scale with a frequency scale using the de Broglie relation,
$$E = \hbar \omega.$$
The physical interpretation of such a frequency scale for particles at temperature $T$ would be their typical frequency as waves. We don't need general relativity or even special relativity to make this statement, because the only constants that come in are $\hbar$ and $k_B$. Finally, a frequency/time scale can be associated with a distance scale using the speed of light, as you said.
