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?
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:
- You agitate the air above the liquid, thus facilitating the diffusion of vapor away from the surface; this increases the rate of evaporation (similar to blowing on the surface)
- You increase the surface area - again, more evaporation
- You ensure that the surface temperature is close to the bulk temperature: evaporation causes a liquid to lose heat, and the cooler layer of liquid near the surface will evaporate more slowly. By stirring, you keep replenishing the surface layer with warm liquid.
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".