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I enjoy the occasional hot drink, but place it below a small fan in order to cool it to a drinkable temperature. Unfortunately, as expected, I commonly forget about my drink, and it ends up very cold.

In fact, it ends up so cold that it feels much colder than I would expect given my relatively warm room ($\sim \mathrm{25^\circ C}$). However, this could be an illusion caused by room temperature still being less than body temperature, or the mug being fairly cold.

I’m wondering if the fan is able to cool the liquid below room temperature. I’m aware that evaporation works on the warmest molecules, and leaves the remaining liquid cooler, but I’m not sure that it can ever become cooler than the surrounding air temperature?

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    $\begingroup$ See en.wikipedia.org/wiki/Evaporative_cooler $\endgroup$ – PM 2Ring May 31 at 22:24
  • $\begingroup$ Re, "this could be an illusion caused by..." and also, the heat capacity of the liquid. If you touch styrofoam that's colder than your skin, it won't feel cool at all because it's surface will almost instantly warm up to your skin temperature. Styrofoam has a very low heat capacity (i.e., it takes little energy to warm it up,) and it is a good insulator, so the heat at it's surface won't be "sucked" away. Touching cool water is a whole other story because it takes a lot of energy to raise the temperature of the water, and it conducts heat better than the foam. $\endgroup$ – Solomon Slow May 31 at 22:24
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    $\begingroup$ I would be interested if the poster could get an (accurate) thermometer and actually perform the experiment for themselves, and return the result here. I am not so convinced there is actually likely to be any practical cooling (of say more than a degree) of the liquid as the answers suggest, but I have been told my assumptions on this are wrong on "several levels", so I would be more than happy to be proven wrong. $\endgroup$ – MC2k May 31 at 23:05
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    $\begingroup$ @MC2k I do have both an internal air thermometer and a meat thermometer (which should work for the water). I’m unsure on the accuracy of these, so the effect might be masked, but tomorrow I’ll give it a go. $\endgroup$ – Tim May 31 at 23:16
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    $\begingroup$ If this didn't work, neither would sweating when it's +37C outside. Humans would just die in that temperature otherwise. $\endgroup$ – J... Jun 1 at 16:40

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Yes, as other answers have stated, the temperature could drop below room temperature through evaporative cooling. In fact it could get as cold as the wet-bulb temperature of the air in the room. If you know the temperature and humidity of the air, you can figure out the wet-bulb temperature by using a psychrometric chart:

Psychrometric chart

Find your room temperature on the green "Dry Bulb Temperature" scale and the room relative humidity on the red "Relative Humidity" scale. Locate the point where they meet, and read off that position on the light blue "Wet Bulb or Saturation Temperature" scale. For example, if the room temperature is 25°C and the relative humidity is 30% (which you indicated), the lowest temperature your drink could reach by evaporative cooling would be roughly 14°C. If the air was completely dry however (relative humidity 0%), it could reach about 8°C.

This does not necessarily mean that your drink will reach that temperature. It may require a large air flow, a large evaporation surface, and considerable time to actually reach the wet bulb temperature.

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Yes, it can become cooler than room temperature. The fan blowing on the liquid surface acts like the evaporation stage in a refrigerator. There's no recycling of the vapour by condensation as in a refrigerator, but that's clearly not important here!

It's no doubt possible even to freeze the liquid, but this is more easily demonstrated by blowing air at room temperature over a volatile liquid such as ether to freeze a small amount of water. [Safety warning: Don't try this except, possibly, outdoors and with extreme caution; ether is highly flammable and an anaesthetic.]

No need to worry, though, about the laws of thermodynamics: this is not a self-acting cyclic process and the second law is not violated!

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    $\begingroup$ I would argue that in this case it is not the fan that causes the cooling, but the evaporation of a separate fluid, which is indeed refrigeration, but not cooling with a fan. You have suggested using a fan to induce evaporation in a volatile liquid such as ether to remove heat, however one could simply (if they had some to hand) use an even more volatile liquid without a fan and achieve much the same result, indicating that the fan has no role in actually removing heat. It certainly does not pertain to the question of cooling a liquid below room temperature by blowing air over it with a fan. $\endgroup$ – MC2k May 31 at 22:35
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    $\begingroup$ @ MC2k (a) I don't think I'm claiming that the fan is essential. The fan was in the original question, which I was trying to answer in its own terms. (b) "It certainly does not pertain to the question of cooling a liquid below room temperature by blowing air over it with a fan." What is your "it" referring to? $\endgroup$ – Philip Wood May 31 at 22:40
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    $\begingroup$ link If you take a look at this pretty generous model, and then consider the energy removed from latent heat vaporisation, and then apply some heat transfer methods, one can probably find a steady state for water with a few reasonable assumptions using the heat flux equation. It is late where I live, so I do not have the will to do so, but I would be surprised if the temp drop is more than half a degree. $\endgroup$ – MC2k May 31 at 22:58
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    $\begingroup$ @MC2k - No model neede. We only need to check a wet bulb - dry bulb temperature chart like engineeringtoolbox.com/dry-wet-bulb-dew-point-air-d_682.html to see that a few degrees are easy to get and a few tens of degrees still possible in very dry air. $\endgroup$ – Pere Jun 1 at 13:11
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    $\begingroup$ Furthermore, the càntir or botijo en.wikipedia.org/wiki/Botijo (and probably several similar devices worldwide) had been invented as evaporative cooler long before the invention of the first fans and thermometers. $\endgroup$ – Pere Jun 1 at 13:14
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I used my hand sanitizer(thanks to COVID) which is the alcohol of 70%v/v will surely an easy experiment to demonstrate it, and my result was yes the temperature of the liquid is below the room temperature, and I feel it depends upon volatility of the liquids which cause the process of evaporation , and cool down the liquid , in your example, the temperature change would not be significantly much as it in the alcohol one but yes it can be cool down.

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  • $\begingroup$ Interestingly I found it hard to get the hand sanitiser to evaporate with the thermometer in it. I think the remaining 30% non-alcohol (moisturiser?) is affecting it. 100% ethl alcohol would likely work better. $\endgroup$ – Tim Jun 2 at 16:10
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There are two types of liquid i.e. volatile & non-volatile liquids. Consider two cases for these liquids

Case 1: Let the liquid be volatile say water. The liquids are mostly volatile. The fan blows air over the liquid surface. This makes the liquid to evaporate & taken away in form of vapors by blowing air. In this process, the liquid molecules on free surface of liquid takes enthalpy of vaporization partly from neighing molecules and partly from blowing air to change in vapor. This results in heat extraction & cooling of volatile liquid and also possibly partial cooling of blowing air depending on temperature of liquid. The minimum temperature that a volatile liquid can achieve is the Thermodynamic Wet Bulb Temperature (TWBT) of ambient air. This is one of reasons why water (i.e. abundant volatile liquid with high enthalpy of vaporization) is used in water cooling systems.

The condition of a volatile liquid to cooled by air blowing over its surface is that the temperature of volatile liquid should be greater than dew point temperature (DBT) (i.e. temperature at which water vapor in air starts condensing) & less than wet bulb temperature (WBT) (i.e. temperature at which air is saturated with vapor) of blowing air.

Case2: Let liquid be non volatile (i.e. can't evaporate) say Glycerin. The air blown by fan extracts heat from liquid surface by (forced) convection process if the temperature of (ambient) blowing air is less than liquid temperature & it can cool non-volatile liquid (Glycerin) otherwise not.

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Typically a convective boundary condition is of the form $$ Q = h \cdot A \cdot (T(t) - T_\text{env}) = h \cdot A \cdot \Delta T(t) $$ (where h can be a function of fluid velocity if necessary) so it is not really possible that heat can be transferred past the environmental temperature(if T is bigger than Tenv we have cooling if the reverse we have heating). Reading the other answers here there is a endothermic reaction of evaporation here that is the cause of cooling below the environmental temperature. Raoults law states that at equilibrium $$ x_i = \frac{y_i p_\text{total}}{p_i^\star}$$ so the relative humidity is basically a measure of how far from equilibrium we currently are. By blowing air past a volatile liquid we keep it further from equilibrium (faster reaction)and speed up the endothermic reaction causing more heat loss

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  • $\begingroup$ I am curious, but does the humidity prevent actual evaporation, or does it just mean that an equilibrium is reached between escaping and returning molecules? In the second case, if the liquid escaping is at a higher temperature than that returning, could cooling not still occur, even with no net evaporation? $\endgroup$ – rghome Jun 5 at 7:49
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    $\begingroup$ @rghome please see this link for more information on reaction kinetics en.wikipedia.org/wiki/Chemical_kinetics. At equilibrium the forward and reverse reactions are happening at the same rate so there is no net condensation or evaporation $\endgroup$ – ChemEng Jun 5 at 18:39
  • $\begingroup$ @rghome Typically with chemical reactions we use something called the rate equation that relates the reaction rate (moles/sec) to the concentration of the reactants. In this case this will not suffice as the concentration of water is constant and it is dependent on things like surface area, perhaps an opposed reaction rate equation can be used where both the forward and reverse reactions are considered thus we do not neglect the water in air concentration $\endgroup$ – ChemEng Jun 5 at 18:40
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A quick explanation of how evaporation cools your coffee. Let's assume that enough time has already passed so that the coffee is at room temperature. How can it get even cooler?

Although the coffee may be at room temperature overall, not all of the molecules will be at the same temperature. Some will be hotter (have more energy) and some cooler (less energy) so the overall temperature is just an average.

The molecules that break free of the liquid and evaporate will tend to be those that have the most energy. If those molecules escape completely then the average temperature of the coffee will obviously be less than it was before (as you have taken away some of the hottest molecules).

The escaping molecules will heat up the air around the coffee making it more likely that some of the escaped heat will simply be returned back into the coffee by conduction, but using a fan (and also convection) will help remove that heat from the vicinity, so helping prevent this. Overall there is a continuous flow of heat away from the cup. The room heats up slightly as the hot molecules are added to it and the coffee cools down.

In the case of an air conditioner, the liquid that evaporates is kept in a closed system. The liquid is forced to evaporate (which cools it) and then the gas is condensed (which heats it up). The two processes are done in separate parts of the system, so that the cold part has air passed over it that cools the house or car and the hot part has air passed over it that takes the excess heat and blows it out into the environment.

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Other answers here are interesting, but here is a simple version, without formulae.

Consider the hypothetical situation that the drink has already cooled to reach equilibrium, ie it has already cooled to room temperature of 25C. The fan then facilitates evaporation of one more water molecule, which will be one of the more energetic (hotter) molecules. So the drink now has a lower mean temperature, just below 25C. So that answers the question, but more interesting ...

The drink will continue losing heat in this way until the rate of heat loss by evaporation is equal to the rate of heating of the cup from the surrounding air, then there is a new equilibrium. So the better insulated the cup, the colder the drink will become before this equilibrium is reached.

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I don't see how it can be cooled below room temperature.

But if the environment is dry and large enough, the fan can remove molecules from the liquid, increasing the water vapor content in the air. In that process, the air may cool, transfering energy to evaporate the water.

Eventually the room can be cooled. As a consequence the drink can reach temperatures below the previous room temperature. But the effect will probably be negligible. Normally it is necessary special nozzles to produce a fine water spray, increasing the kinetics of the process.

I have worked in a factory where fans with water sprays cooled the environment. It could be measured by a thermometer, and was not only a refreshing feeling due to the evaporation of water on the skin.

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  • $\begingroup$ Read the answers above and check the links to understand why this is wrong. $\endgroup$ – rghome Jun 2 at 7:41
  • $\begingroup$ @rghome As for myself, I checked the link (and read the answers), and I still do not understand what's wrong with this answer. Could you elaborate ? (the link "Evaporative cooler" states that the air can be cooled, but doesn't say anything about the water ; that seems to me to pretty much support Claudio Saspinski's answer). $\endgroup$ – lmsteffan Jun 2 at 18:23
  • $\begingroup$ I added an answer explaining evaporative cooling in layman's terms. $\endgroup$ – rghome Jun 2 at 21:06
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As it's been asked about "fan", the fan can't cool anything by itself (imagine using a fan inside vacuum XD), it brings the fluid outside the container to motion hence helping to transfer the heat from container to the moving fluid(air in usual cases).

If we use much more cooler fluid we must be able bring down the temperature significantly.

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To address something nobody yet has, though not answering the question (as I can't top the others):

In fact, it ends up so cold that it feels much colder than I would expect given my relatively warm room ($\sim 25^\circ$). However, this could be an illusion caused by room temperature still being less than body temperature, or the mug being fairly cold.

This is indeed an illusion, but not quite the one you describe. If you walk into your kitchen after nobody's been in it for a while, it's reasonable to expect items in the kitchen are of roughly equal temperature. If you touch your wooden chopping board, it won't feel very cold at all. If you touch the metal spoon resting next to the chopping board, it will feel much colder.

This is because we sense heat (or lack thereof) based on how fast our own temperature changes. Because the wood is a poor heat-sink, it doesn't sap heat away from us very quickly, so we don't perceive it as cold. Conversely, the metal spoon saps heat away from us much faster, so it feels colder.

Tea is a good heat-sink.

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