Imagine two evaporative cooling experiments: In the first, a beaker contains a known volume of liquid. The liquid starts at ambient temperature. The temperatue of the body of liquid remaining in the beaker is measured as the liquid is allowed to evaporate. The experiment concludes before the liquid in the beaker has completely evaporated.
The second experiment repeats the first exactly except that the volume of liquid used is less.
Over a period of time the liquid shows a drop in temperature, with a rate of cooling that is greatest at the start but the rate of decrease becomes less.
What effect would the reduction in volume have on the measured results?
I've tried to think about this in small incremental time periods. My understanding is that the initial energy loss due to evaporation would be unaffected as same initial temp and surface area (same beaker), however losing this energy would represent a larger proportion of the internal energy of the remaining liquid (less liquid to start with) so the temperature drop during that first delta(t) time period would be greater in the second experiment.
Would this effect continue or would the now colder liquid lose LESS energy in the next delta(t) period? Clearly the liquid is approaching an equilibrium temperature with its surroundings where evaporative heat loss = heat gained from the now warmer ambient, but will the smaller volume get there quicker?
Is it possible with just the information given to assert that the temperature change measured would be bigger, smaller or unaffected?
What could we say about the temperature measured in each experiment at a set time T from the start BEFORE equilibrium was reached?
Best Answer
At the start when there is no water vapor above water, the evaporation rate is the fastest considering the partial vapor pressure is zero. After a while, the vapor accumulates and partial vapor pressure approaches the saturated point. They then change the mode by diffusing out driven by water vapor concentration gradient. This slows down the process. At the same time, you cannot ignore the thermal process, which also plays a role here.