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I am currently thinking through evaporation over lakes, specifically the Laurentian Great Lakes (a complex subject, I know). Particularly, I am trying to wrap my head around why evaporation peaks in the fall and winter. Based on what I have read, this fact is due to the vapor pressure gradient that exists between relatively warm water and dry air (dry in an absolute sense because the air is cold) and the high winds which continually replace that dry air over the water.

I have also read that this evaporation from the Lakes has a cooling effect on the Lakes themselves, causing a temperature decrease in the Lakes. I have seen it insinuated that over-lake evaporation is synonymous with a latent heat-flux. Based on what I have read, latent heat is an exchange of energy to a substance without a change of temperature of the substance. For water transitioning from a liquid to a gas, the required amount of energy to cause this phase change is called the enthalpy of vaporization or the latent heat of vaporization.

When the air is colder than the water, where is the energy coming from to supply the latent heat of vaporization that causes evaporation (assuming its a cloudy day)? If the vapor pressure gradient is the main driver of the rate of the evaporation, how can the evaporation still be said to be a latent heat flux, i.e., how is the transfer of particles based on a pressure differential called a transfer of heat, albeit a latent transfer of heat? Does a latent transfer of heat mean that the water particles which evaporate do not change temperature, though the liquid water which they leave behind decreases in temperature?

Sorry, that's more like three questions! Hopefully they are not stupid ones :) it seems this is a complex topic, and my initial intuition that evaporation is always higher when the air is warmer than the water (thus transferring heat energy to the water, increasing the energy of the molecules, leading to increased evaporation rate since the water molecules are now higher energy) is simplistic and even wrong, as it depends on so many more factors than that. Our world is not a simple one, physically at least!

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The reason that this seems complicated is that there are two things happening at the same time. Both heat transfer and mass transfer are occurring.

The heat to provide the energy of vaporization is coming mostly from the lake water. There is a temperature gradient established within the water (with depth) as a result of the evaporation. The temperature at depth is higher than at the surface, and this causes heat to flow from depth to the surface. The rate of upward heat flow approximately matches the rate of evaporation times the heat of vaporization.

But, in order for vaporization to occur, the equilibrium vapor pressure of the water at the surface temperature must exceed the partial pressure of water vapor in the air above. This difference in vapor pressures provides a driving force for water vapor to diffuse into the overlying air and be swept away by the air currents.

So, in the situation you describe, both heat transfer and mass transfer are present to determine the overall rate of evaporation and surface cooling.

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  • $\begingroup$ Your answer and Andrew Steane's provide some great clarity. So the temperature gradient within the water is established due to evaporation, while also providing the energy needed for evaporation? Or does the evaporation not necessarily need an energy input to happen in the first place in the presence of a vapor pressure gradient, as the process occurs due to the increase in volume of water molecules from liquid to vapor leads to an increase in entropy? $\endgroup$ Jun 29, 2023 at 13:42
  • $\begingroup$ Well ti starts out with a uniform water temperature, with warm enough temperature at the surface for evaporation to occur. As evaporation occurs, the surface cools a little, and this sets up the temperature gradient for upward heat flow to occur so that the cooling at the surface does not effectively shut off the evaporation. $\endgroup$ Jun 29, 2023 at 14:15
  • $\begingroup$ So if the water were to start at a uniform temp, with the air colder than the water, the energy needed for the surface water to evaporate would first have to come from somewhere outside the surface water-air system, such as the sun or the lake bed? $\endgroup$ Jun 29, 2023 at 15:05
  • $\begingroup$ No. It would still come from the water at its surface region. You need to study heat transfer to understand how this comes about in detail. $\endgroup$ Jun 29, 2023 at 19:07
  • $\begingroup$ I have studied heat transfer in college. Upon thinking about this, it has been helpful to look at the temperature–entropy diagram of the vapor-compression cycle, which shows that a liquid can become a vapor and change temperature just through expansion, without a latent heat transfer. But when there is a temperature difference between the fluid and the environment, a latent heat transfer can occur in which the fluid transitions completely to a vapor in an isothermal process where it absorbs energy from a higher temperature environment outside of the system. $\endgroup$ Jun 30, 2023 at 14:57
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When the air is colder than the water, where is the energy coming from to supply the latent heat of vaporization that causes evaporation (assuming its a cloudy day)?

Both the air and the water have some thermal energy. The fact that the air is cooler doesn't change that. As you mention, this can have a cooling effect on the lakes. Thermal energy from the lake can supply the energy to allow evaporation.

Does a latent transfer of heat mean that the water particles which evaporate do not change temperature, though the liquid water which they leave behind decreases in temperature?

I wouldn't want to talk about the temperature of a particular particle. But yes, basically this. The water->vapor transition is endothermic. The heat transfer does not go into raising the temperature of the vapor, but of releasing it from the water.

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  • $\begingroup$ I do understand that the air still has thermal energy even though it is colder than the lake, but I am under the impression that thermal energy can only be transferred from a high temperature body to a low temperature body. But it makes sense that the energy would come from underneath the surface of the lake, where the water is slightly warmer than the surface water. $\endgroup$ Jun 29, 2023 at 13:26
  • $\begingroup$ It doesn't have to come from something "warmer". Imagine a dish of alcohol. It will cool significantly in air due to the volatility. Once cooler than the environment, it still has sufficient energy to continue to evaporate (and cool further). $\endgroup$
    – BowlOfRed
    Jun 29, 2023 at 15:30
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I will address the question about latent heat.

To evaporate a liquid we need to supply energy in order to overcome the attraction that keeps molecules together and convert the liquid into a vapor. What we call heat of vaporization is specifically defined as the energy to evaporate a liquid under constant temperature and pressure. This however does not mean that evaporation must necessarily happen under constant temperature or pressure.

Think about what happens when we sweat: the evaporation of the sweat makes us feel cooler because the phase transition draws energy from our bodies. If we blow air over the sweat it feels even colder. This is the same as the cooling effect of the lake: blowing air establishes a gradient between a sweaty body and the surrounding air. As a result of this gradient more sweat evaporates and the body feels much cooler.

More generally, when a liquid is brought to a state that thermodynamics dictates that its phase must vapor (or a vapor/liquid mixture), its temperature drops as it expends the energy needed for the evaporation. Most refrigerators and A/C units operate based on this principle.

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  • $\begingroup$ When you say that evaporation doesn't necessarily need to occur under constant temperature or pressure, do you mean that evaporation is not always a latent heat transfer/endothermic transition? $\endgroup$ Jun 29, 2023 at 13:28
  • $\begingroup$ @chaserone10 It's too things happening simultaneously, here is how to think about it: evaporation requires latent heat to break the attraction forces between molecules in the liquid. IF this energy is supplied by an external source it is consumed entirely to produce vapor at the same temperature as the liquid before evaporation. IF an external source is not available AND the system is brought to a state that is must be vapor, then this energy will be drawn from the system itself. It is this latter effect that is responsible for the cooling effect. $\endgroup$
    – Themis
    Jun 29, 2023 at 14:42
  • $\begingroup$ Ah, so in the absence of an external source, latent heat comes from within the system, producing a temperature change for the part of the system from which the energy came? And this only happens when there must be vapor, meaning the system is not in equilibrium? $\endgroup$ Jun 29, 2023 at 15:16
  • $\begingroup$ @chaserone10 Exactly! Rub some alcohol on your hand and blow air over it. Some alcohol must turn into vapor because blowing removes vapors from the vicinity of the liquid (the diffusion gradient Chet Miller talks about). This evaporation draws energy from the alcohol itself, and also from your hand, which is in thermal contact with the alcohol. That's why your hand feels cooler. $\endgroup$
    – Themis
    Jun 29, 2023 at 18:31
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I agree the answer by Chet Miller and here I will add some further information.

With evaporation it helps a lot if you understand the physical concepts of vapour pressure and partial pressure.

Vapour pressure (at a given temperature) is what the pressure would be above a liquid surface if that surface were in equilibrium with a vapour of the same substance, with no other gases present.

Partial pressure of substance X is the contribution to the total pressure made by substance X when a number of gases are intermixed in the same region of space.

The vapour pressure of water is 1 atmosphere at 100 degrees celcius. The vapour pressure of water is $0.006$ atmosphere at zero degrees celcius (so: low but not zero), and it roughly doubles with every 10 degrees increase in temperature.

The partial pressure of water in completely dry air is zero. The partial pressure of water in humid air is around $0.02$ atmosphere.

Water will evaporate if the partial pressure of water vapour near the liquid surface is less than the vapour pressure for water at the given temperature. The way this ordinarily happens is that evaporation makes the air near the surface become more humid, raising the partial pressure, and wind blows that air away. If the air coming in to replace it is drier (has a lower partial pressure of water vapour) then further evaporation will take place.

So far I merely said what happens. But the question also concerns the physical mechanism. The entropy required is brought to the evaporating liquid by heat flow from the rest of the liquid (it can also flow from the air but usually the flow from the body of the liquid water is the main route). But why, you might ask, does liquid even bother to evaporate? (After all, the air might be colder than the water!) It is because there is a great increase in volume on phase change from liquid to vapour, and this implies there is an increase in entropy too, even if the temperature falls. So it is a net entropy-increasing process, so it takes place.

In the movement from liquid to vapour the molecules move out of attractive potential energy wells so on average they slow down and arrive in the vapour with a lower kinetic energy than they had on average in the liquid. The liquid is also left with a lower energy per molecule on average (since the more energetic ones have escaped). So the temperature of the system as a whole may fall. However if there is also heat flowing in then the process may overall balance out with no net change of temperature.

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  • $\begingroup$ I hadn't thought much about entropy! I guess I didn't understand it well enough in my thermodynamics courses in college; it's a bit more nebulous to me than thinking strictly in terms of energy transfer, which I realize isn't wise in a world where the 2nd law of thermodynamics exists alongside the 1st :) if, in the movement from liquid to vapour, the water molecules lose kinetic energy, does this mean it is not a latent heat transfer? $\endgroup$ Jun 29, 2023 at 13:31
  • $\begingroup$ Thinking on this a bit more, if the partial pressure of water in the air equals the vapour pressure of the liquid water (meaning the air is saturated with water), wouldn't evaporation still be occurring? The rate of evaporation would just equal the rate of condensation. $\endgroup$ Jun 29, 2023 at 13:54
  • $\begingroup$ on the second: yes it's a two-way dynamic equilibrium. On the first: the latent heat is $L = T \Delta S$ where $\Delta S$ is the change in entropy. So when $\Delta S \ne 0$ we have $L \ne 0$. This is energy which moves into the system as a whole (both water and vapour together), ultimately from other things such as the lake bed or the Sun etc. $\endgroup$ Jun 29, 2023 at 14:20
  • $\begingroup$ So the T is constant for the system (water and vapour) as a whole, though the water part of the system decreases in temperature due to the evaporation? And that means when we describe something as a latent heat transfer, the temperature of the system does not change, but a change in temperature might have taken place somewhere in the system? I might be getting confused here between a closed and open system and whether mass is being transferred in addition to energy. $\endgroup$ Jun 29, 2023 at 15:24
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    $\begingroup$ I think reading up on the refrigeration cycle, particularly looking at the Temperature–entropy diagram of the vapor-compression cycle, helps me to start to wrap my head around evaporation in the Lakes better as well. $\endgroup$ Jun 29, 2023 at 15:47
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The following is a simple approximate model of the heat and mass transfer in this system. It is 1D in the vertical direction z.

Below the water surface, the heat transfer is governed by the transient heat conduction equation (neglecting convection): $$\rho C\frac{\partial T}{\partial t}=k\frac{\partial ^2 T}{\partial z^2}$$

The thermal boundary condition at the water-air interface z=0 is: $$-k\frac{\partial T}{\partial z}=\phi (\Delta H_{vap})+h((T_S-T_{\infty})$$where $\phi$ is the molar evaporation rate of water per unit area, $\Delta H_{vap}$ is the molar heat of vaporization, h is the heat transfer coefficient between the water and the air, $T_S$ is the temperature at the interface, and $T_{\infty}$ is the air temperature in the bulk air stream. This equation says that the heat flux entering the interface minus the heat flux leaving the interface (into the air) is equal to the rate of vaporization times the heat of vaporization.

The last equation in the model gives the molar flux of water at the interface: $$\phi=k_m(P_S(T_S)-P_{\infty})/RT_S$$Where $k_m$ is the molar mass transfer coefficient between the water-air interface and the bulk air, $P_S(T_S)$ is the equilibrium vapor pressure of water at the surface temperature, and $P_{\infty}$ is the partial pressure of water vapor in the bulk air stream.

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  • $\begingroup$ If the equilibrium vapor pressure is greater than the partial pressure of the water vapor in the bulk air stream, we would have a positive molar mass transfer coefficient, which makes sense. But what then happens if the temperature of the water is uniform, and the air temperature is lower than the temperature of the interface? Then the thermal boundary condition equation would evaluate to zero on the left hand side equaling the sum of two positive numbers on the right hand side. Or is this scenario I have described impossible? $\endgroup$ Jul 3, 2023 at 13:55
  • $\begingroup$ I guess my question is a bad one. The heat transfer to the surface from lower in the water does not drive the evaporation; the energy taken away from the water in evaporation and in convective heat transfer to the air drive a temperature change at the water surface, causing a heat transfer from lower in the water to the water surface. The driver of the evaporation is not an input of energy per se, but it occurs because it is a net entropy increasing process, as Andrew mentioned. $\endgroup$ Jul 3, 2023 at 14:21
  • $\begingroup$ Well, what happens is that, initially, there is a finite driving force for mass transfer. All you need for heat transfer from the lake to begin at this point is a discontinuous change in the temperature gradient to be established the surface. Look up in the literature the problem of constant heat flux at a surface of a semi-infinite medium. In our cases, the heat flux will be toward the surface. Byy studying this peoblem, you will get the idea of what happens in the lake in proximity to the surface. Even here, most oft the heat comes form the lake, not the air. $\endgroup$ Jul 3, 2023 at 17:56
  • $\begingroup$ To give you an idea what the temperature profile in the lake water is in the vicinity of the surface at short times, it looks something like $$T=T_{Lake}-\frac{q\delta }{k}(1+x/\delta))$$where q is the upward heat flux, k is the thermal conductivity, $\delta(t)$ is the boundary layer thickness at time t, x is the upward vertical coordinate, and $T_{Lake}$ is the lake temperature at $x\lt -\delta$ . At time zero, $\delta =0$ and increases with time. Note that the temperature gradient at the surface is constant at -q/k for all times. $\endgroup$ Jul 3, 2023 at 20:47

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