# Why does water get significantly colder while falling through the air?

I think we've all experienced this effect. You turn on the shower and the stream of water at the shower head is a fair bit hotter than the water towards the bottom of the stream. Alternatively, if you splash water out of the bath (I'm a child I know) the water will feel cold if you touch it before it lands back in the bath.

I want to know for sure why this is.

The obvious answer seems to be simple thermodynamic heat exchange, but I would point out a few things for why I think this is not the whole story (although someone can argue otherwise):

• The time it takes for a drop of water to fall from the head of a shower to the ground is really quite a short amount of time.
• Given the short amount of time falling, I don't think heat exchange will happen fast enough to explain the fairly large temperature change.
• Given the fact that water has a large heat capacity, I think there must be something else in play here to provide enough energy.
• The temperature of a room with a hot shower running is not very different from the temperature of the water itself.
• It seems (and this is impossible to know without somehow measuring the temp) that the temp of falling water at the bottom of the stream sometimes is colder than the surrounding air (imagine we extend the stream to the bottom of a waterfall then that certainly seems to be true).

My thought for the major contributor to temperature change, and an explanation for how the water could even overshoot thermodynamic equilibrium, is that the water droplet/stream has to do a pretty large amount of work on the atmosphere before reaching the ground. I do not know how to model this mathematically, however, and would like to get a general idea of the amount of work being done in this situation.

Another interesting thing to think about, and I'm quite conflicted by this, is if one could expect the temperature of a water droplet to increase as it gains speed. For instance, suppose the water were falling in a vacuum and it maintained its droplet formation, would the temperature of the droplet increase? It seems to me it would, although the gravitational potential energy side of this thought is troubling. Is that effect large enough, however, to be relevant to this question?

So, to generalize the question:

What is a good mathematical and thermodynamic model for explaining the temperature change of falling water? Particularly, what is the largest contributor, heat exchange, or energy lost due to displacing the atmosphere? Also, does the gain in kinetic energy of the water droplet increase (in a non-net type of way) the temperature of the droplet as a whole? Also, all these explanations I assume would lead to what the final temperature of a water droplet might be after falling some distance, so getting an idea of how much temperature change there is would be cool too.

• I'd rather disagree with your last point on the list. But hoping for some good answers. +1 for the thought. :) Sep 7, 2015 at 0:00
• Surely evaporative cooling is a major factor. Sep 7, 2015 at 0:50
• I would expect THE major factor to be the amount of water to fall on your hand. Close to the shower head, more water falls on your hand than near the floor. That means that your hand needs to distribute a larger amount of energy, and therefore will locally heat up more, which is what you feel. Mar 17, 2018 at 14:51

Firstly, it would be better to use actual, accurately measured numbers than the human 'experience': the human body is a poor thermometer and the mind plays tricks on us.

But that hot water droplets lose heat and thus cool down in cooler air is an established fact and a consequence of the laws of thermodynamics.

Regarding your three first bullet points, despite some limitations you point out, those do not mean a hot droplet of water doesn't cool in air: it does and partly in accordance Newton's cooling law.

As regards work done by the droplet (overcoming the viscous drag), if anything that would lead to heat generation, not cooling (but the effect is truly minuscule).

Kinetic or potential energy of the droplet have no effect on the droplet's temperature. Temperature is simply a measure of the average speed of the molecules of the water and that is not affected by these energies. Spinning water in an ultra-centrifuge does not make its temperature rise, for instance.

You have however overlooked one major cause of heat loss: evaporation. Your shower 'steams up' because hot water evaporates and that costs energy, known as the Enthalpy of vaporisation.

Millions of tons of water are cooled this way everyday in power plants world wide: the cooling towers drop hottish water from the top of the towers and evaporative heat cools down the water (the evaporated water escapes as steam clouds).

If your shower has been in operation for a long time and the bathroom's temperature is equal to the shower head's water temperature and the air is saturated with water vapour, then no cooling of the shower water would take place.

• I totally agree that evaporation must be a major contributor, but I think it's undeniable that the water must exert work on the atmosphere in order to pass through it. It just seems like that energy would be dissipated by lowering of temperature. And yes I agree that using actual data would be much better. I would particularly like some experiment to test if the work the water does on the atmosphere changes its temperature. Any suggestions on that? Sep 7, 2015 at 2:24
• @jheindel: hi. If anything, drag forces (i.e. friction) generate heat, just rub your hands together. Friction has negligible effects here though: the work done is mostly an increase in kinetic energy of the air in the wake of the droplet (turbulent flow). To measure temperature of the water use a styrofoam cup (fill it up and empty it several times, so the cup is at the same temperature as the water) and a decent domestic thermocouple based thermometer. Repeat all measurements a few times (noting all readings), then compare top and bottom water temperatures. Hope this helps.
– Gert
Sep 7, 2015 at 2:32
• @jheindel: I'll work out the work done by a droplet travelling through air tomorrow.
– Gert
Sep 7, 2015 at 2:43
• Another factor is that they are little droplets. Temperature drop by Newtonian cooling is proportional to surface area and temperature, but inversely proportional to mass. The ratio of surface area to mass for a spherical droplet is $3/(\rho r)$, where $\rho$ is density and $r$ is droplet radius. Decreasing droplet size increases Newtonian cooling (and also evaporative cooling as well). Some shower heads let you control whether the water is a fine mist or larger drops. The effect is much reduced when you use larger drops. Sep 7, 2015 at 16:38

Although you wouldn't be directly testing the thermal exchange, here's one possible way to at least test out the evaporation theory:

Find a way to make small droplets of water, either individually or in small enough groups that they wouldn't interfere with each other, and a surface on which to rest those drops. Make sure that you can control the temperature of the water reservoir the water comes from and that you can control the water temperature right up until the moment the water is formed into drops.

Create a temperature-controlled surface that gives decent contrast visibility with the water droplet(s) and on which a pre-measured grid was printed such that you could use it to measure scale. Using video, possibly even high speed video, you could then test out rates of evaporation for various sizes of water droplets and various combinations of initial droplet temperature and atmospheric temperature. You would likely want to control for atmospheric humidity across all experiments.

It would be ideal if you could actually just SUSPEND the droplets in mid-air, thus eliminating relative motion to the air as a variable, though that may be impractical. It could be possible to nucleate water droplets on a very thin string (dental floss or something like) so that the contact area with a flat surface is minimized as a variable that could hinder evaporation. I have heard of extremely strong electromagnets that have even been used to magnetize and levitate small and otherwise non-magnetic items, like strawberries and spiders, without any harm to those objects, so possibly that could be a means as well, if such is at your disposal.

Ultimately, this might be an excellent experiment for a shuttle/ISS mission. Wish I'd thought of it twenty years ago.