The amount of time within the chamber allows for the cooling.
Somehow the heat has to go somewhere besides up the stem, which is the physics problem I am trying to wrap my head around. I know how the system pipes and the regular gourd calabashes do it (they both precipitate the steam, removing the heat in the water from the smoke) but I don't see the physics principle that makes the reverse calabash work without also condensing the steam. it doesn't matter how much time the smoke spends within the chamber if there is no good route for the heat to escape, and I'm just not seeing one. Wood and lucite are both thermal insulators, and thermal conduction from a gas to a solid (e.g. briar) is notoriously inefficient. so gas to solid conduction doesn't seem like it's a likely candidate (also the high volume to surface area ratio in all the reverse calabash designs are all wrong for conduction anyway).
And contact with cooler air in the chamber and surface of the cooling chamber. Traditional gourd calabashes have a large cooling chamber and the gourd absorbs some of the moisture that might be steam.
The large chamber is helpful if the design is leveraging the ideal gas law PV=nRT, or in this particular application T2 = T1(V2/V1). and system pipes with larger condensation chambers like the gourd calabash and Peterson House pipe all exhibit better cooling properties. The absorbency of the gourd and meerschaum also help .
But if the pipe isn't using condensation as the primary means of cooling the smoke, as is being posited for reverse calabashes, then a chamber with a large internal volume and relatively small surface area and insulating, non-absorbent walls is not going to do a whole lot. Volume goes up roughly by r^3, surface area goes up by r^2 - the larger the chamber the lower the percentage of gas that will contact the walls, or the shorter the contact time. Neither is a good way to cool the smoke.