Thermocouple and cooling

The best possible thermodynamic efficiency (Carnot Engine ) is
= (temp difference in Kelvin) / (High temp in Kelvin)

= 80 / (273 + 100) = 80/373 ~ 20%

So that supports your 10% guesstimate

The, presumably empirical, figure given in the article is “The device can provide a maximum boost of 0.36 Wh,”

Which is 1.3 kJ, an order of magnitude less than @peccavi 's estimate.

Yeah, I knew I was overestimating the thermoelectric efficiency. I finally looked up the numbers for commonly available TEGs and it is about 5-8% (best case for several hundred degree temperature differences at over a thousand degree Thot).: At a Th of 100 C, and a Tc of 20 C, it’s probably more like 3%. Add in the fact that you’re never going to get an 80 C temperature differential (more like 60 C) and the heat losses elsewhere in the cup are probably much bigger contributors than I was estimating, and ~1-2% is probably a better guess than I made.