When adding Urea to water, the reaction is highly endothermic. I was playing with the idea of using that for a Desktop cooling system, but for that, I'd of course have to recover that Urea from the water somehow. Obviously, the easiest way would be to simply boil away the water, but given that this is supposed to feed into a cooling system involving sensible electronics, I unfortunately don't have that option. Since I'm myself better at engineering that at chemistry, I decided to ask here: Which other seperation methods are there that leave both the water and the urea "intact"? Ideally fast ones, but that is not really a requirement, it'd just make things easier for me. Can I use reverse osmosis? Electrochemical methods? I'd really appreciate some help. Thanks!

  • $\begingroup$ Reverse osmosis requires a rather high pressure. And it is a very slow process. Why do you refuse any boiling process ? $\endgroup$
    – Maurice
    Commented Jan 19, 2022 at 20:11
  • $\begingroup$ @Maurice because sensitive technology and high temperatures are typically a very bad match. I'm sure you could theorethically make it work with proper thermal insulation, but in the end my PC is pretty importent to me, and it's simply not something I'd like to risk if I can avoid it. $\endgroup$
    – Mat NX
    Commented Jan 19, 2022 at 20:36
  • 1
    $\begingroup$ Buying a cooling system on air fans only, or with an intermediate water cooling system; they have been around for quite some time. These could be an option ready-to-go. $\endgroup$
    – Buttonwood
    Commented Jan 19, 2022 at 21:44
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    $\begingroup$ I think this cooling method is a bad idea, I second Buttonwood. $\endgroup$
    – Poutnik
    Commented Jan 20, 2022 at 0:20
  • $\begingroup$ An interesting fact: Some datacenter-grade servers have even direct oil cooling, designed to be immersed in an oil bath ( from the local interview with a datacenter person ). $\endgroup$
    – Poutnik
    Commented Jan 20, 2022 at 8:28

2 Answers 2


@MatNX, as you state, "sensitive technology and high temperatures are typically a very bad match." However, there is no magic way to "make" cold, only to move heat from one location to another. Presumably, since you state you have an engineering background, you're familiar with the laws of thermodynamics. Any heat you remove from an IC will be displaced somewhere else, so you must provide a heat-sink, e.g., large volumes of air or lesser volumes of water.

While you could find some way to re-solidify the urea, that releases heat in the process -- more than was removed (2nd law: you can't break even).

There are many ways to move heat, though:

  • $\begingroup$ Thanks for the indepth answer! I probably should have clarified that I don't want to magically remove the heat, but in normal watercooling systems, theheated fluid is cooled using regular fans before it is cycled back. I want to replace that part specifically, having a container with urea that the water flows through. The basic idea is simple: Imitate an instant cold pack, having the endothermic reaction absorb heat, and then remove it in aome other form, be it pressure or electricity. Think fuel cell, H2 produces electricity instead of exploding. $\endgroup$
    – Mat NX
    Commented Jan 20, 2022 at 6:44
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    $\begingroup$ First, water has a higher heat capacity then a system of water + urea, so you'll lose cooling capacity after the water is saturated. Second, this would be a one-shot use of the urea -- again, once the water is saturated with urea, it can hold no more and needs to be dumped. Third, you want to regenerate the plain water and urea, so it's not a one-shot deal? That requires energy, with concomitant waste heat. Thermodynamics is a harsh science; TANSTAAFL, as Prof. Heinlein states. $\endgroup$ Commented Jan 20, 2022 at 16:42

The problem with a process as you have suggested is, plainly, energetics. Specifically, any physical or chemical process that involves a change of state follows the equation $$\Delta G = \Delta H - T\Delta S\tag{1}$$

where $\Delta G$ is the Gibbs free energy, $\Delta H$ is enthalpy, $T$ is temperature in kelvin and $\Delta S$ is entropy. A reaction or process will only occur if $\Delta G$ ends up being negative (exergonic). In a simplified model, the enthalpy part describes the heat that is transferred either into or out of the system, i.e. the temperature change you observe. If $\Delta H$ is positive – i.e. the reaction/process is endothermic or it absorbs heat from its surroundings – that must be offset by $T\Delta S$ in an appropriate manner to give an overall exergonic process.

The dissolution of urea in water is exergonic but endothermic, meaning $\Delta H>0$ and thus $\Delta S >0$. If you now want to reverse that process to recover your urea, this will necessarily mean that your reverse-process has $\Delta S<0$, meaning you need to find a sufficiently exothermic process to offset $-T\Delta S$ being positive.

Unfortunately for you, $G, H$ and $S$ are state functions, meaning their values depend only on the current state of the system, not the way the system took to get there. So any process of getting back from dissolved urea to solid urea and water will need to overcome the same barrier making the whole thing rather complex if not impossible at any small scale.

Furthermore, you are up against the Second Law of Thermodynamics, meaning that for any spontaneous process $\Delta S > 0$ – I hope you already see how this is negatively impacting your idea.

Of course, nothing in this answer states that it is impossible to separate a solution of urea in water. Obviously, distillation (evaporation and subsequent capture) will recover the pure starting materials. Reverse osmosis may help, I’m no expert on that. (Electrochemistry likely won’t help as that would involve chemical reactions which would modify urea to something else and you would have to reverse that &endash; difficult. Furthermore, all simple electrochemical processes I can think of in five minutes will lead to something that you won’t be able to turn back into urea in any simple way.) However, the first part of this answer shows that whichever process you use it is going to be really energy-intensive.

One potential solution to all this would be a one-way system where you constantly purchase water and urea and dispose of the dissolved urea. This isn’t the cheapest way of doing things but I would imagine that it would be more cost-effective than urea recovery.

On the other hand, cheap, non-energy-intensive cooling methods exist, have been engineered and used in the wild for decades such as water cooling or air cooling with fans.


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