Suppose I cook a stew or soup and then take it off of the stove and leave it uncovered to cool. Of course, it will cool faster if I stir it (more surface area), but it will, eventually, cool to room temperature even if I leave it alone. As it cools, it loses liquid to evaporation — steam rises from it — until is has cooled a bit. Does it lose more liquid, total, if I stir it as it cools, or if I leave it alone — or does it lose the same amount either way?
[Physics] Does stew lose more liquid if stirred as it cools
coolingevaporationfoodthermodynamics
Best Answer
The stew loses heat mostly by two mechanisms: evaporation, and conduction to the environment (which probably means convection - air flowing past the pot). Radiative heat losses are very small for objects below 100 C.
When the air is completely still, vapor will build up above the surface of the stew and this will slow the rate of evaporation (even more so if you cover the stew with a lid). If you stir, you do several things:
All three things will increase evaporation, and will therefore speed up the rate at which the stew cools down and result in higher loss of liquid. But since the heat loss is likely to be mostly by evaporation anyway (for an uncovered stew), the difference is not massive: it's more the time than the amount of liquid lost. But that depends a bit on the shape of the pot, its thermal conductivity, the relative humidity of the air, etcetera.
All of which means the answer is a cautious "yes".