Assume that the water is 2 cm deep in a flat pan (if that is relevant).

EDIT: As a fun experiment, I'm trying to evaporate a mixture of powdered root and water in my kitchen. I've heated the mixture to 45 °C, and I'm wondering how long it might take to completely evaporate and leave behind a dry powder.

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    $\begingroup$ My guess is that we are speaking about days. Just pour the solution into something shallow and leave for several days in place with strong ventilation, I used this scheme for evaporation of solutions of some salts. $\endgroup$ – permeakra Sep 8 '15 at 4:41
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    $\begingroup$ Might also depend on ambient humidity. If the relative humidity in the air is 100%, then the air is already saturated with water vapor and the water will never evaporate. $\endgroup$ – chipbuster Sep 8 '15 at 6:26
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    $\begingroup$ Why, if the relative humidity in the air is 100%, but the air is at 25 °C, and your water is heated to 45 °C, then it will evaporate. Though of course that would not be as fast as it could be in a less humid air, and you'll get moisture all over the place, too. $\endgroup$ – Ivan Neretin Sep 8 '15 at 10:32

The evaporation rate $E_r$ can be approximated by: $$E_r = C \times A \times (x_s - x) = 0.013$$ kg/h

$C$... evaporation coefficient $ = 25 + 19 \times v$

$A$... water surface $= 0.03 $ m$^2$

$x_s$... saturation humidity $= 0.027$ kg/kg (at 30 °C)

$x$... absolute humidity $= 0.011 $ kg/kg (at 30 °C and 40 % rel. humidity)

$v$... air speed above water surface $= 0.08$ m/s

If you mesure the air temperature above the water surface - which I approximated with 30 °C - you can get a way better approximation for your case. Also having the rel. humidity you can then find the values for $x_s$ and $x$ in a Mollier diagram (h,x-diagram).

PS: You will waste a lot of energy when you want to obtain the water temperature at 45 °C. Heating the air instead should be more effective.

Reference: Link (in German)

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  • $\begingroup$ I'm curious if you'd care to respond to a similar question I posted in SE Earth Science, here: earthscience.stackexchange.com/questions/5071/…. I never did get a response and don't want to cross-post. $\endgroup$ – BillDOe Sep 10 '15 at 5:21
  • $\begingroup$ @earthscience.stackexchange.com/questions/5071/…: Since your 1. Formula also takes the surface temperature into account, when calculating the evaporation coefficient, i´d say its more accurate. But if you were trying to calculate evaporation rates at room conditions, like in this post, then both of them should be inprecise. The only way to really be sure is to find out more about the specific test conditions, under which both formulas had been constructed. $\endgroup$ – AstronAUT Sep 10 '15 at 12:39
  • $\begingroup$ Formula 2 does use the pond surface temperature indirectly from the saturation pressure calculated in B7. $\endgroup$ – BillDOe Sep 10 '15 at 18:13
  • $\begingroup$ Thank you for the answer, @AstronAUT. So according to your formula (and given those approximations), it evaporates at a rate of 13grams per hour and will take a little over 3 days. Also, just to clarify... are you saying that the water will evaporate faster in an oven than on a stove (over direct flame)? $\endgroup$ – Mark Simpson Sep 11 '15 at 10:40
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    $\begingroup$ At the same temperatures, turned on convection and if you allow the oven to constantly receive fresh air, yes of course. $\endgroup$ – AstronAUT Sep 11 '15 at 10:59

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