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Recently - motivated by the recent heat where we live - an acquaintance argued that a water cooled computer will not heat up the room as much as an air cooled computer will. His reasoning was that the processors never reach temperatures as high as they would with air cooling, so the room will not absorb as much heat.

This seems flawed to me; the processors have an energy consumption that is regulated by how much processing they do, and that doesn't depend on whether they are air cooled or water cooled. I figured the amount of energy that gets converted to heat should be the same in both cases, the water cooling just doesn't reach temperatures as high as the air cooling because water has a higher capacity for absorbing energy than air does. In the end, the same amount of heat energy would end up in my room in both cases, so I would expect the room to heat up the same, independent of whether we cool our computers with water or air cooling systems.

Which one is right?

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    $\begingroup$ Interesting question. Speculation: it's possible that a hotter processor has a higher internal resistance, and therefore requires more power to perform the same calculation as a cooler processor, so a processor forced to a cooler temperature would have lower power consumption and in fact would keep the room cooler for a given amount of computing. If the power required for a given computation is independent of processor temperature, then your reasoning is correct. I don't know enough about microprocessors to know which is right, though. $\endgroup$
    – Kyle Oman
    Commented Jun 12, 2014 at 0:31
  • $\begingroup$ I considered the cooler-processor-requires-less-energy-for-calculations possibility, which is one of the reasons I'm asking here; if that isn't actually the case (and I really have no clue - I also considered posting this question on Electronics instead) and a processor does its job with the same energy requirements regardless of temperature, I figured conservation of energy means that my room will heat up the same. Good to know I'm not that far off with my reasoning, at least. $\endgroup$
    – G. Bach
    Commented Jun 12, 2014 at 1:21
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    $\begingroup$ A water cooling system has a much higher thermal capacity. So for short uses the heat would mostly stay in the cooling system. You can have a more pleasant stay for short on times because of this. The energy would be same, under idealised conditions. $\endgroup$
    – WalyKu
    Commented Jun 12, 2014 at 10:49
  • $\begingroup$ The purveyors of water-cooled computer systems have made credible claims that, overall, water-cooled systems are much more efficient than air-cooled ones. Partly the effect is due to the absence of fans, and partly it's due to the more uniform temperature achieved, reducing the need to "over-cool". $\endgroup$
    – Hot Licks
    Commented Dec 2, 2014 at 18:03

4 Answers 4

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There are two parts to this question:

  1. if the fans consume more power than the pump

this depends mostly on the fan and pump, I don't have much of an answer here, but in any case the difference will be rather small relative to the total power consumed by the computer.

  1. if electrical components are more efficient at lower temperatures

This is true, and for proof, lets look at a power supply review that tests the efficiency of it both at room temperature, and also inside a hot box - notice how the efficiency is lower inside the hot box ( vs 88.1% at max wattage vs 87.9% at max wattage after a 11º C increase) - again, this is a negligible amount of power (.2%)

What about other components, such as the CPU? that test has been done here, and it seems like an increase of 10º C uses roughly 5W more - which seems significant relative to the others, however one must consider that a good fan & heatsink will be able to match a good liquid cooler to within a couple of degrees see here

So the answer is both will heat up the room by roughly the same amount, and the lead could go to either.

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  • $\begingroup$ Very interesting - I would have expected a water cooler to keep the temperature significantly lower than what a fan can do. $\endgroup$
    – G. Bach
    Commented Dec 2, 2014 at 11:35
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Your friend appears right to me.

What you are saying is that the cpu will release equal amount of heat to both systems (air-cooling and water-cooling). That is right. But with an equal amount of heat absorbed by water and air, water's temperature will rise far lesser than air.

So while the amount of heat absorbed by the cooling systems will be equal in both cases, the amount of heat RADIATED INTO THE ENVIRONMENT will be different, because after absorbing 100KJ, water (considering you are using reasonable amounts of it, not just 10 ml or so) will just be at some 50°C or so while air, after absorbing an equal amount of heat, will be at some 90 or 95°C (rough estimate, I haven't done the precise math).

Plus, this also depends on whether you are using the same water over and over again in the cooling system or you are replacing the heated water with cold/normal water every 30 mins or so. If you replace the water (spilling out the heated one in your bathroom wash basin) then your room temperature will rise very little. Unfortunately you cannot replace hot air with cooler air so easily (plus you will need to replace air so very often). It is interesting to note how water-cooled systems offer a much better cooling policy than air-cooled systems although the amount of heat dissipated by both systems is exactly the same.

(p.s. sorry for taking the question with a more "practical" approach rather than being more "analytical")

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    $\begingroup$ I think we should assume the computer is running long enough for the water to reach equilibrium, after which it radiates heat at exactly the rate it absorbs it. $\endgroup$
    – user27118
    Commented Dec 2, 2014 at 17:52
  • $\begingroup$ Right. Even in that case, you can go and spill out the hot water in your bathroom wash basin and fill the cooling system with fresh water at room temperature. This will keep the room from getting overheated. You cannot do the same with air. The total energy radiated by the cooling system is same in both cases, agreed. But how much the COMPUTER ROOM gets hot as a result of cooling system radiated energy differs. $\endgroup$
    – Imran Roy
    Commented Dec 4, 2014 at 6:43
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The heat that comes out of the computer box is the same as the power that goes into the box. Cooling just changes the temperature distribution inside the box, but energy is conserved. If you measure the power in the main input, that is all you need to know.

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  • $\begingroup$ Cooling also changing the process. Ohmic losses will be reduced in the core when it is cooler. $\endgroup$
    – Aron
    Commented Dec 2, 2014 at 9:39
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So talking in sense of your question the room will heat up exactly the same way because energy will always be the same and the heat will be the same. The Cpu may not heat up as fast but it still will produce the same amount of heat. This is due to something called specific heat. That is the amount of energy that it takes to change the temperature of a substance. The reason why a Cpu will get hotter when in air is because air has a much lower specific heat than water meaning that it is easier to heat up than water. They in total will move the same amount of heat they will just do it at different speeds. The only thing I can say on the whole water/air cooling thing is that if your running an insane cooler with a wimpy computer they computer cooler should exchange heat faster than what the cpu can heat it. Meaning that is will actually cool the room. So if the cooler is faster than the computer yes it will cool otherwise it is just a waste of time and money. For more info on speacific heat http://www.bbc.co.uk/schools/gcsebitesize/science/aqa/heatingandcooling/buildingsrev3.shtml and http://www.iun.edu/~cpanhd/C101webnotes/matter-and-energy/specificheat.html are great links to learn some more. Hope your lucks turns out to be good and the heat in your area goes away.

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  • $\begingroup$ Ok understand this post is pointless because you basicly said this in your question sorry. $\endgroup$
    – Pi_Co
    Commented Jun 12, 2014 at 1:09

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