I read some questions such this for example and I asked me another question. Instead of only calculate how much time does it takes to have a cup of tea at room temperature, I'm asking how much time does is take to evaporate a part of the water in a cup of tea.

Just to be clear

For this question I am considering that the cup is full of water at 100°C and the cup too. The cup is consider to have no effect on heat transfert and has a high of $12 \mathrm{\text{ cm}}$. During all the experience the temperature and the pressure in the room are constant and respectively are $T_0 =20^\circ \mathrm{C}$ and $P_0 = 101325 \mathrm{\text{ Pa}}$.

From the beginning of the experiment to the time where the temperature of water and the cup will be the same than the room it is not difficult to imagine that some water will evaportate and escape the cup and goes in air. But don't forget that their is an equilibrum between water in the liquid water and the water in the air. We will considering during all the time of the experiment the hygrometry of the air is zero. So then by the diffusion phenomena very very slowly the water in the cup will escape.

At the end we assume the pressure of water in the air is the saturation pressure of water in the air and there is only $5 \text{ cm}$ of water inside the cup.

My Question

We know how to calculate the time for water to cool from $T_0$ and room temperature. How should I know then during this time how much water goes out of the cup in gas phase ?

"this time" refers to the time for the water to goes from $\text{100}^\circ\text{C}$ to room temperature. The time during the water at room temperature will diffuse in the air is given in the next part with details for you.

Some details for you

I'm asking this question because I know how to calculate the time to evaporate water using law of diffusion but this time depends on the high of liquids there is in the cup ...

My calculus give :

$$t=\frac{\left(z_f^2-z_i^2\right) \rho RT_0}{2M_{H_2O} P_0 \mathcal{D}_{w/a} \ln\left(\frac{1-y_i}{1-y_f}\right)}$$

Where :

$z$ is the difference lengh between the top of the cup and the top of the water, $\rho$ is the density of water at $T_0$, $M_{H_2O}$ is the molar weight of water, $\mathcal{D_{w/a}}$ is the diffusion coefficient of water in the air and $y$ is the molar fraction of water in the air.

If we consider for example that until room temperature only $0.5 \text{ cm}$ of water evaporates then $t=5426025 \text{ s}$ I just need know the real value of how much centimeters really escaped.

Sorry for the lengh of the question and thank you in advance for your answer !

  • $\begingroup$ OK, maybe you are over complexifying the problem. If the cup does not allow heat transfer (completely insulated) then would the heat lost by the remaining amount of liquid water have to equal the heat gained by the amount of water that vaporizes? $\endgroup$ Dec 30, 2016 at 18:20
  • $\begingroup$ @JosephHirsch you might haven't understood something. During the time the water will cool from 100°C to room temperature, maybe only one millimiter will escape maybe two. My question is how much millimeters of water will escape during this time. When I say the cup have no effect on heat transfert I mean for the heat transfer you can consider there is no cup between the water and the air (but not for the mass transfer). :) $\endgroup$
    – ParaH2
    Dec 30, 2016 at 18:25
  • $\begingroup$ Got it. So it is really a rate problem. $\endgroup$ Dec 30, 2016 at 18:33


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