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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?

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  • $\begingroup$ Is the beaker in a room, or is it in a chamber containing no air? Can heat be exchanged between the surroundings and the liquid (say, at the liquid surface)? $\endgroup$ – Chet Miller Jun 1 '16 at 11:22
  • $\begingroup$ Although not stated in the question (which is therefore poorly constructed in my opinion) the assumption would be that the beaker is in a room and that heat exchange could take place. This question is to be asked of 15 year olds and I am rapidly coming to the conclusion that they are not equipped by high-school physics to be able to answer it - but that is not a discussion for these boards. I wanted to try and get the physics straight before taking up that point with the exam board. It turns out that I can't answer it too well either hence my post! $\endgroup$ – Dleaf12 Jun 1 '16 at 22:21
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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.

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  • $\begingroup$ Ok I think I understand that - but surely the partial vapour pressure will develop similarly in both experiments? I'm trying to find any explanation that could account for a difference in measured temperature. It is my contention that, if anything, there would be a greater initial temperature drop in the experiment with less volume, but the markscheme I have to use (I am preparing to mark a public exam containing a similar question) claims the opposite, and it has made me doubt my own understanding. Other than the temperature/time cooling curve no other numerical data is given. $\endgroup$ – Dleaf12 Jun 1 '16 at 7:36
  • $\begingroup$ I haven't done the test so can only think of other factors. If the beaker wall is adiabatic (for example), for initial temperature drop, I would agree with you that the temperature decreases faster for the less volume liquid if it has the same surface area. If beaker wall transfers heat, a temperature gradient will form from wall to liquid surface. The sensor location, wall temperature, liquid amount, air temperature all play roles. $\endgroup$ – user115350 Jun 1 '16 at 13:47
  • $\begingroup$ Thanks - so without more information there is no sure way of saying what the effect on the results of the volume change will be - that is, depending on the assumptions made one could argue that a greater OR a smaller temperature difference would obtain. $\endgroup$ – Dleaf12 Jun 1 '16 at 22:28

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