Correlation between watts and temperature

NuggetFlipper

Honorable
Feb 5, 2014
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10,520
Is there a function that describes this correlation?
For example, let's say a cooler cools a cpu towards an equilibrium of 50 C.

If I double the heat produced by the CPU, everything else remaining the same (except for watts, but that's out of the question), how much will the equilibrium temperature rise?

EDIT 1: crap, this seems really similar to a physics question I did in highschool...

EDIT 2: I pretty sure it's not thermal equilibrium because Q = mcT or T = Q/mc is a linear relationship, but we can see that's clearly not the case with coolers, is it? For example, if you take a look at the graphs here:
http://www.techspot.com/review/707-best-cpu-coolers/page6.html

The idle and stress temperatures don't differ too much considering that idle is probably produces 1/100th of the heat produced during full load. Maybe I'm overthinking and it is a linear relationship...
But then again, since ambient temp is 21C I can just subtract 21 from all the temps and get a more accurate set of data.

EDIT 3:
Ok, it's definitely not linear, but maybe I'll plot these data for individual coolers (or at least a couple)
http://www.anandtech.com/show/7738/closed-loop-aio-liquid-coolers/9

Sorry for the mass edits, I'm just stuck on a problem and trying to update you guys on my progress/thinking.
 
Solution


Short answer is you are right: The standard unit for the rate of heat transferred is the watt (W), defined as joules per second.
http://en.wikipedia.org/wiki/Heat#Notation_and_units

You probably could predict the CPU temperature based on Wattage and ambient temperature. I am not sure if you could predict how well each CPU cooler would work.
Or...


Ok, but if I were to give you two different temperatures and its associated heat produced, would you be able to graph the entire function for me?

I'm not too focused on the correlation between wattage and temps, there's too many variables to consider.
 


Damn, I really hoped this wasn't the answer. I guess there would be too many variables. Is it possible if I state how many joules of heat the cooler dissipates per second?

Are you sure there's no magic formula regarding this? Lol... I've watched temp graphs and they're all logarithmic, given nothing changes except for temperature and time. I figured this just had a single more variable to it so it would simply differ by a shift or factor or something :\

Btw thanks for being patient with me, I'm asking a lot at once
 


Pretty sure it wouldn't apply, since cpu uses some of the power to compute while the heat emitted is generated from wasted energy, right?

Btw your link died... It worked for a while but something went wrong.
 


Main point on there (works for me):
The specific heat of air is 716 J/kG K. The density of air is 1.3 kg/m^3. 1 watt is 1 J/sec.

So. In 60 seconds, you'd get a temperature rise of:
100*60/716/1.3=6.5 C



This the final formula, although it may be incorrect

Heat required to raise the temperature of the example copper wire:

Mass of the example wire = .045 kg
Specific Heat Capacity of Copper = .39 KJ/kg K
Final Temp = 100 C
Initial Temp = 40 C

Q = mc(T2 - T1)

Q = .045 * .39 * 60 = 1.053 KJ/sec = 1053 Watts
 
I see where you're going but there's no cooling involved so the temperature increases infinitely as time passes.
The thing with coolers is that they become more efficient as delta T increases, correct? Or else temps wouldn't stabilize at all?
But from the link from my 3rd edit, it almost looks as if wattage and delta T are in a linear relationship, which doesn't support my previous point.
 


Short answer is you are right: The standard unit for the rate of heat transferred is the watt (W), defined as joules per second.
http://en.wikipedia.org/wiki/Heat#Notation_and_units

You probably could predict the CPU temperature based on Wattage and ambient temperature. I am not sure if you could predict how well each CPU cooler would work.
Or the quality of the fans.
https://www.youtube.com/watch?v=VUbpb23yTK8

I'm not a physicist, but this might be the experiment you are looking for (using water block cooling)
http://www.procooling.com/index.php?func=articles&disp=138&pg=1
 
Solution