# What temperature is an evaporating cup of water at “room temperature”?

If I put 100 g of water in a cylindrical cup and I know it takes 2 weeks (lets say 1.2 × 106 seconds) at some RH/temp to completely evaporate in my room with a constant temperature, can I determine it's temperature?

I would know how much heat the water absorbed using the latent heat of vaporization ΔHvap = 2257 kJ kg-1 (at 100 °C...not sure how/if to correct), and the specific heat: CP = 4.1855 J kg-1 K-1 (at 15 °C).

So if it took t to evaporate m of water, it should be:

$$\frac{m\,\Delta H_{vap}}{t\,C_P} = \frac{0.1\,\mathrm{kg}\times 2.257 \times 10^6\mathrm{\ \frac{J}{kg}\ }}{1.2 \times 10^6\,\mathrm{s}\times 4.186\mathrm{\frac{J}{kg \cdot K}}} = \dots$$

But there's a left-over mass term...I could maybe square m...

The water pulled 225.7 kJ out of the room (lets say the room is sufficiently large to not change temperature due to that), so the average heat transfer was 0.188 W. My assumption is that because the surface area of the water is uniform (cylinder), that the rate of evaporation would be constant, which would mean constant heat...but given the same heat and a shrinking mass, the temperature would exponentially decrease: absurd. Where/what's the gap in my reasoning?

Does the 0.188 W of cooling need to be exactly balanced by 0.188 W of heating by the room, so they are actually at the same temperature? Would I need to consider something else (e.g. the thermal conductance between the room and the water)?

The rate of evaporation depends critically on the rate at which water vapor is removed from the water/air interface.

If your cup was closed, water would evaporate until you reach the saturated vapor pressure of water at the prevailing temperature, and then things would reach equilibrium - water vapor would go into the liquid as fast as it evaporated. The fact that water is evaporating at a constant rate tells us that a certain amount of moisture is escaping. Now there are two mechanisms for this: diffusion, and convection. If the air in the vicinity of the cup is perfectly still (imagine a very tall, narrow cup with just a little bit of water in the bottom) then you can compute the diffusion of water through the gradient (almost 100% at the interface, and equal to the relative humidity of the air at the opening). Now you are losing 0.1 mg of water per second (rounding) ; the mass diffusivity of water in air is 0.28 cm$^2$/s. For a cup 10 cm diameter and 5 cm deep, with a 50% RH at the opening, we compute the diffusion as follows:

Saturated vapor pressure at 20C (initial assumption... Let's see if we are in the ball park) is 2.34 kPa so the pressure difference across the gradient is half that - 1.67 kPa. That means the density difference is about 1/60th of the density of water vapor at STP which would be roughly 1/60 x 1.2 x 18/29 = 12 ug/ cm$^3$.

The mass flow due to diffusion would be $12 \cdot \pi \cdot 5^2 / 5 \approx 0.2 mg / s$. That is surprisingly close to the value we were getting - so we can assume that we don't need air flow to sustain the evaporation. Note that the assumptions about dimensions were pretty much pulled out of thin air...

If the above is valid, then we rule out the need for air flow in the vicinity of the cup (although it could be there, the assumption of "no air flow" will give the largest temperature difference. And that means we need to compute the rate at which heat flows into the water from air.

There are three distinct surfaces: the surface that the cup stands on, the wall of the cup, and the surface of the liquid. It is hard to find a good value for the thermal conductivity of a "cup", so we will use $20 W/mK$ as a representative number for a not-very-conductive ceramic. For the same cup as above, with water to a depth of 5 cm, the surface area is approximately 150 cm$^2$, and for a thinness of 4 mm we get a thermal resistance of 75 W/K. Now we need to add the thermal resistance of the air around the cup: the thermal conductivity of air is quite small at 0.024 W/mK so it will dominate the calculation. At the surface of the cup we would need a thermal gradient of 0.2 / (0.015\cdot 0.024) = 550 K/m which would be very significant - and in fact it would create a sufficiently large density gradient that convection would be set up.

If we assume a low air velocity, we find (from engineeringtoolbox.com) the h factor around 15 W/m$^2$K; for the same parameters as before, this gives a thermal conductivity of about 0.2 W/C, which implies a temperature difference of about 1C is needed to provide the necessary heat flow (0.2 W, as you had calculated). This is much less than the conductivity of the ceramic, which can therefore safely be ignored. We will also ignore thermal conduction to the liquid surface because the air above the liquid was assumed to be stagnant (talk narrow cup assumption). It's much easier to set up convection on the periphery of the cup rather than across the top.

Note that this calculation is quite sensitive to some very approximate assumptions, so the result can be wildly off. But is demonstrates the principle - the liquid would be about one degree C cooler than the surroundings if the air is quite steady at 20C and 50% relative humidity.

• My thought was that if you have a direct measure of the rate of evaporation (presumably easier to measure), it would seem like a very concise proxy for room temperature, air flow, humidity, etc, to find the dT versus room. – Nick T Jan 12 '16 at 4:04
• @NickT - unfortunately these things are intricately linked. You can get a (very) different answer for the temperature depending on the exact air flow speed. – Floris Jan 12 '16 at 5:21
• Nice analysis. I was going to do pretty much the same thing, but you beat me to it. I'm voting you an "answer is useful" on this. – Chet Miller Jan 12 '16 at 13:50

The heat is supplied mainly by the room air. The air temperature at the water interface will be slightly lower than that of the room air, and there will be a temperature gradient in the air in close proximity to the water surface so that heat can be conducted to the surface. Throughout most of the room, there are small air currents (e.g., from the ventilation system) that are sufficient to provide enough mixing to maintain a basically uniform temperature in the bulk of the room air.

The partial pressure of the water vapor at the interface will be equal to the equilibrium vapor pressure at the interface temperature. This partial pressure will be higher than in the bulk of the room air, so there will also be a mass transfer region near the interface where water vapor diffuses away from the interface. This will be driven by the gradient in water vapor partial pressure in close proximity to the interface. Throughout most of the room, the air currents are sufficient to provide enough mixing to maintain a basically uniform partial pressure of water vapor in the bulk of the room air.

So evaporative mass transfer takes place by diffusion away from the interface into the room air, and the heat transfer takes place by heat conduction toward the interface away from the bulk room air.

• All valid points but it doesn't calculate the temperature of anything... – Floris Jan 12 '16 at 2:43
• It's only slightly lower than room temperature throughout the bowl of liquid water. To estimate it accurately, one needs to do the model calculation. But I don't think anyone really expects the temperature of the water in the bowl to differ very significantly from the room air temperature. – Chet Miller Jan 12 '16 at 2:47