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Black Heat Sinks- Better?

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January 16, 2007 2:28:47 PM

Ok so this came up in another thread, so I wanted to post about it.

In a nutshell, it was said that painting a HS flat black would increase its thermal efficiency.

I'm an engineer, so my knowledge of physics is mostly limited to electricity. What I do know is that "color" is an indication of the ability of a material to absorb or reflect certain parts of the color spectrum when ambient light hits the material. I can't see how a pigment would make a material radiate heat any better.

Although I'm skeptical, I'm genuinely curious. A heat sink is all about convection and surface area, and I would think that "paining" a heat sink would at least reduce the thermal effectiveness of the sink, and at most have no benefit (Even if the paint was a better conductor than the material, you're still limited by the material's conductivity).



At any rate, can someone fill in the blanks for me?

(Original Thread - Posts by NovemberWind and SirHeck) *edit*

More about : black heat sinks

January 16, 2007 2:32:41 PM

If it was laser-pitted black, then maybe, but not paint. Maybe black thermal grease.
January 16, 2007 2:42:05 PM

Ah, yeah, I remember reading something about that.

All metals can turn black - Article on ZDnet

Perhaps this is a rumor born from the above article.
Related resources
January 16, 2007 2:51:41 PM

(<-- Correcting myself)

Ah. Pitting is different than the article I read...
a c 487 à CPUs
January 16, 2007 3:19:29 PM

I'm no engineer or physicist, but wouldn't the black paint act as a layer of insulation, thus slightly reducing the heatsink's ability to radiate out the heat absorbed from the CPU?
January 16, 2007 5:44:55 PM

Whilst black paint would radiate the heat better as a function of it's colour, yes I'm pretty sure that that effect would be negligible compared to the insulating properties of a layer of paint.

The laser pitting looks like a far more promising prospect: If I'm interpreting that article right, it could not only turn the metal black, giving it better radiative properties by way of colour, but also dramatically increase the surface area of the treated metal. If it were possible to get an appreciable amount of air circulation around these microstructures, it could increase the cooling capacity of the material by a large amount. The problem would be in getting the air to flow around such tiny structures: at this kind of scale the viscosity of the air itself would probably prevent the extra surface area being useful and could even trap a layer of air (think of fur) at the surface of the metal, insulating it immensely. These are problems very similar to those pointed out in another thread about immersion cooling.
January 16, 2007 5:50:38 PM

Quote:
Whilst black paint would radiate the heat better as a function of it's colour, yes I'm pretty sure that that effect would be negligible compared to the insulating properties of a layer of paint.

The laser pitting looks like a far more promising prospect: If I'm interpreting that article right, it could not only turn the metal black, giving it better radiative properties by way of colour, but also dramatically increase the surface area of the treated metal. If it were possible to get an appreciable amount of air circulation around these microstructures, it could increase the cooling capacity of the material by a large amount. The problem would be in getting the air to flow around such tiny structures: at this kind of scale the viscosity of the air itself would probably prevent the extra surface area being useful and could even trap a layer of air (think of fur) at the surface of the metal, insulating it immensely. These are problems very similar to those pointed out in another thread about immersion cooling.


That was my thinking as well. I think manufacturers of high-end cooling should take a look at it.
January 16, 2007 6:42:55 PM

The original poster is partially right: a blackened HS surface has higher emissivity, which is the degree to which an object produces far infrared radiation (=heat radiation) compared to a perfect blackbody.

However, the energy radiated from the HS is also determined by the temperature of the 'receiving' surface, which is the inner surface of the computer's case.

You can 'easily' compute the radiation by yourself, it's

Q = 5.67e-8 x A x e x (Th^4 - Tc^4), Q in Watt

with A (in m²), the 'exposed' surface of your HS, this does not include the surface between the actual cooling fins,
e, emissivity which is equal to 1 for a perfectly black HS (i.e. carbon black)
Th, heat sink temperature (in Kelvin)
Tc, case surface temperature (in Kelvin)

Taking some standard values for these variables, I only get a heat dissipation due to radiation of 1 to 5 Watts.

Which is fairly negligible, considering your HS has to pull > 100 Watt under full load.
January 16, 2007 7:17:42 PM

The only way a black heatsink would help is if you are totally depending on radiative heat transfer. Today, most heatsinks in computers are using convective heat transfer (fan blowing air over the heatsink). In the case of where you are using convective heat transfer, it wouldn't matter much if the heatsink was black, silvery, or puce.
a b à CPUs
January 16, 2007 9:16:57 PM

Quote:
The only way a black heatsink would help is if you are totally depending on radiative heat transfer. Today, most heatsinks in computers are using convective heat transfer (fan blowing air over the heatsink). In the case of where you are using convective heat transfer, it wouldn't matter much if the heatsink was black, silvery, or puce.


This is EXACTLY right! And yes, the paint's thermal transfer rate VERY likely would make it act more like an insulator, completely negating any advantage for radiant heat dissipation (which is, as others have said, only a minor contributor to overall heat removal).
January 16, 2007 9:24:33 PM

I think the reflective silver works best for now.
January 16, 2007 9:46:15 PM

Better off buying a small low rpm fan can a can of spray paint.
January 17, 2007 2:06:46 AM

Thanks for the input.

I'm genuinely curious, but this is what I've found:

http://en.wikipedia.org/wiki/Emissivity

No flaming wiki, if you please.


A black body isn't something that's necessarily black; it's something that absorbs all electromagnetic radiation, and appears black as a result. Painting something black only means the surface will absorb more of the visible spectrum, and doesn't make the heat sink itself more of a "black body," and thus it shouldn't increase the amount of emissive radiation by any substancial amount.


My personal opinion is that this is a misinterpretation of physics rules, more specifically what defines an object as a "Black Body." Feel free to correct me if I'm wrong.
January 17, 2007 3:53:22 AM

The only way the black paint could aid in cooling would be if the sink was hollow with vent tubing exhausting outside of the case.
The heat transfer would cause an updraft effect such as in black solar chimney piping for natural solar venting.
This is dependent on the titanium content in the pigment for solid heat generating objects.
It has been used in the automotive industry for the past 20 years to aid in the thermal stability of the internal combustion engine helping to keep the temperatures at an optimum operating temperature of 223 to 227 deg F.
As jumping jack stated black absorbs heat energy and white reflects.
Under the right circumstances this theory can be applied.
January 17, 2007 4:24:12 AM

Quote:
Black will reabsorb radiative energy, what you want is to reflect that away --- so shiny and white is the best means of doing so....


This is incorrect. Black does absorb heat better, but it also radiates it better. It's all based on emissivity. Shiny white objects have low emissivity, and dull, more black object have high emissivity. Emissivity is to radiative heat transfer and conductivity is to conductive heat transfer. This is why the SR-71 Blackbird is black: because it flying at Mach 3.5 the air no longer cools the jet, but actually heats it (beginnings of hypersonic effects). Radiative heat transfer becomes the most effective way for the spy plane to keep "cool". The same goes for the X-15 and the heat shielding panels of the space shuttle.

As for a black heatsink, I don't think that it would be effective. Effective radiative heat transfer generally requires very large temperature differences between the object and ambient. You certainly don't want your heatsink to be that hot. Convective heat transfer is the norm for CPU cooling because it works very well.

Other posters noted about the laser pitting to make a metal black. This is a great idea, but I'm sure it would be expensive, at least for a while. The black colour might not give you a huge boost in cooling effectiveness, but it won't hurt. As for increasing surface area of the material with the nano-scale pits, the pits are on the order of the wavelength of visible light, which is too small for the fine structures of turbulence to fit into. Once an object is below the Kolmogorov scale, it will have no effect on the flow.

This brings me to a different idea. In wind tunnels, sometimes it is necessary to "trip" the flow from laminar to turbulent in a test. One way to do this is to create a region of small bumps, about the size of a grain of sand, over which the flow must go. This may be quite effective for CPU coolers, as turbulent flow has a much greater effectiveness for convective heat transfer.
January 17, 2007 4:42:26 AM

Sigh... I can see where you're going on the wrong path here. Putting the objects in sunlight exposes them to an environment with a high radiation energy... I'll say density. Black does absorb sunlight to get hotter better than white (or better still a mirror-finished piece of aluminium.

In the case of the CPU, it is not sitting in sunlight. The temperature of the CPU is higher than the vast majority of objects within line-of-sight of the CPU cooler, so that the balance of radiative heat transfer will be away from the cooler. It's all about equilibrium (you don't need university for this, high school will do). While heat is absorbed better by a black object, it is also radiated away faster. The heat radiated by the body is proportional to:

eATobj^4

where e is the emissivity, A is the surface area, and Tb is the absolute temperature of the object.

The energy absorbed by radiative heat transfer is proportional to:

eATs^4

where Ts is an averaged value for the temperature of the bodies within view of the object.

Therefore, the energy balance is the difference between the two, which amounts to ea(Tobj^4-Ts^4). This is the equilibrium part. If you put something that's at room temperature in view of the Sun, the overwhelming radiation coming out of the Sun will increase the average Ts so that the balance is heat being absorbed by the object. If the object is shaded from the Sun and it is warmer than it's surroundings, then it will emit radiation heat. The latter case describes the situation in which you should find your CPU, so the higher the emissivity of the cooler, the better it will be for radiative heat transfer.

All that said, radiative heat transfer will be an order of magnitude weaker than convective heat transfer.
January 17, 2007 4:43:53 AM

Its all about surface area. That's what laser pitting does. You can also get a very dramatic increase in heat transfer in cutting up your fins. The computers on the Space shuttle (the one responsible for engine - fuel management) have procupine fins.

Sorry this is not a direct answer, but one I think you will find more effective.
January 17, 2007 4:54:53 AM

Neither of your experiments demonstrate how a black-coloured body doesn't emit more radiative heat energy than a white-coloured body of the same temperature. Here's an experiment to prove that it does.

Take two bars of the same material, preferably something that can absorb a lot of heat. Paint one white and one black. Heat both bars to the same temperature in an oven, and then stand them on a mat or towel in a dark room. Measure the temperature of both over time, and you will find the black one cools faster.

A counter-example is a good coffee thermos. What keeps the coffee cool is a piece that's generally metal or glass that sandwiches a vacuum chamber. The vacuum cuts out conductive heat transfer, which is great. But the tube is always polished in the good thermoses to a near-mirror finish to keep radiative heat transfer to a minimum as well.

I'm quite sure that the folks at NASA and Lockheed's Skunk Works didn't have their heads up their A**ses when they went through all the trouble to make the X-15 and SR-71 black instead of just buffing up the metal.
January 17, 2007 5:02:55 AM

Quote:
See my edit above for emissivity (the constant your Raleigh-Jeans equation) and you will see that color (black, white yellow or red) has nothing to do with the effectiveness of an object to emit.


From that Table, under Paints:

Black, e=0.96 (I imagine this differs between flat, glossy, etc.)
White, e=0.95 in one case

Lower down on the page...

Oxidized Aluminium, e=0.11. (The range for e is from 0 for perfect reflectivity to 1 for perfect absorption).

It seems to me that it might not really matter what colour paint you choose, anything will dramatically increase the emissivity of a cooler (mine's got aluminium fins). In light of this, don't use white paint for the experiment with the hot rods. Use something with a metallic shine.
January 17, 2007 5:10:17 AM

Quote:
You are still trying to treat the color black as a black body.... this is incorrect, a black body radiator is a hypothetical beast.... furthermore, your T^4 law is actually incorrect, as mentioned by the ultraviolet catastrophe above. It only works for temperatures above the threshold to emit in the black body range. Below 200 or 300 deg C, emission is not the predominant heat transfer mechanism.


I just thought of a good counter-example to the high-temperature requirement you state for radiative heat transfer: the Earth. Radiative heat transfer is the only method for the Earth to balance it's temperature, since there's no conduction and no convection worth mentioning between Earth and space. If the Earth weren't able to radiate heat (mean temperature something like 15C), then the heat gained from the sun would continually build until the Earth reached a much higher temperature. It would have to be 200 to 300C to emit heat.
January 17, 2007 5:18:31 AM

So what you're saying is that an object of a certain colour, or material, whatever your determining factor is, will be more effective at either absorbing or emitting radiative heat, rather than have the same vale of emissivity for both cases. Also you're saying that radiative heat transfer is not in agreement with the ~eA(T^4) until a fairly high temperature is reached.

This is a little far-fetched for me to believe. Can you explain my previous counter example(s)?
January 17, 2007 5:42:04 AM

I thought I should put up some numbers so people can see the effect of changing emissivity on something like a CPU cooler. This should demonstrate that while emissivity increases can improve thermal radiative heat transfer, the total cooling power of radiation is an order of magnitude below the cooling provided by convective heat transfer.

All of this work is done on the basis of what is taught in a university course on heat transfer. If people want to try to prove that what universities are teaching in classical physics courses is so inaccurate as to be unusable, you should take that up with the Engineering accreditation council or other authority in your area.

I will use two coolers as examples. Both coolers have an effective surface area of 0.08 square metres. One is aluminium, for which I've used an emissivity estimate of e=0.11. As Jack says, e is a function of temperature, but it takes a large change in temperature to create a small change in e, so I'll be using this linearising simplification. The other cooler is painted or in some way made black, or as it turns out pretty much any paint colour will do, and I'll appoints its emissivity to be 0.95. Effective surrounding temperature will be 300K.

For the first one (aluminium surface), consider the case where the cooler temperature is 50C (323K). The power of radiative heat transfer emitted from the object is then 5.43 Watts. The power of radiation absorbed by the object is 4.04 Watts, so the total radiation cooling power is 1.39 Watts. This is pretty insignificant compared to the TDP of a CPU. If the cooler's temperature is reduced to 40C, the total radiative cooling power is reduced to 0.75 Watts.

For the second cooler (black surface), at 50C the power out is 47.4 Watts and the power in is 35.3 Watts, for a total balance of 12.1 Watts of radiative cooling. At a cooler temperature of 40C, the total radiative cooling is 6.5 Watts. Again, even with very high emissivity, this cooling is inadequate for a CPU.
January 17, 2007 5:43:28 AM

Painting anything black will make it behave more like the theoretical black body. Black is black because it absorbs radiation and that is exactly the main attribute of a perfect black body.


Imagine one black and one shiny body. Put them in a room at say ... room temperature :) . After a while they will reach thermal equilibrium.
Now as we all agree the black body absorbs more radiation than the shiny body. Still it stays at the same temperature as the shiny body. The only explanation is that it EMITS more radiation then the shiny body.
Assuming emissivity is a property of the body and it is bound to it and constant, whatever the temperature, we can conclude that a black body will always emit more radiation than a shiny body at the same temperature.

JumpingJack, you say that radiation is not dominant at lower temperatures. Thats true but it is also valid for absorbtion. If not much is radiated, not much is there to absorb.
January 17, 2007 5:46:18 AM

I never stated that radiative heat transfer would be adequate for CPU cooling in a PC. I stated the opposite. However, our disagreement was on the laws of physics. Yes, classical physics aren't exactly right. F does not exactly equal MA. But you have to be in pretty extreme circumstances to notice the difference between what's really happening and what Newton and his peers wrote.

I've merely been trying to defend the physics as I learned from my teachers and textbooks.
January 17, 2007 5:47:21 AM

Absence of evidence is not evidence of absence.

Stop worrying about the bullsh*t color of your heatsink and get laid. For the love of Garrett Oliver, won't somebody think of the children.
January 17, 2007 5:51:21 AM

Even if emissivity does depend on temperature, it proves that color does matter and black emits better.
January 17, 2007 6:20:37 AM

I can see how heat absorbtion is a function of wavelength, an that this also explains why emissivity of a material is a function of temperature. Is it true that emissivity peaks when the heat energy levels are high enough for the material to emit light that shares a characteristic wavelength with the material?

If emissivity for the purpose of absorption is a function of wavelength, then I would still expect that the effective emissivity of an object is the same for heating and for cooling in the example I gave above. The bodies in the room will be emitting radiation that is of a similar wavelength as those emitted by the cooler, since the temperature difference is small compared to absolute values.
January 17, 2007 6:25:01 AM

Quote:
Hey, I got one for you.... a real stumper. Lets say theoretically there is an infinite heat sink, and no energy to return. An object would continually radiate that energy.... over time, eventually, would it radiate all of it's energy to nothing?


So... an infinite piece of material in the vacuum of some fictional space where the absolute temperature of the background radiation is somehow 0 Kelvin... Well, I would say that if this object had a defined shape, such as an infinitely long cylinder, or a sheet of some defined thickness, then you could still calculate the temperature transience by reducing the order of the problem to two or one dimension. The cylinder would be a 2-D problem, and the sheet would be a 1-D problem.

While the body may have infinite total heat energy, and it may emit an infinite total amount of energy, any body with a defined shape will have a finite energy density.
January 17, 2007 6:42:14 AM

Yes, that makes sense to me. That answers my little question.

Now, do you not agree that the simple classical rules I've used in my example above are adequate enough to describe the situation of radiation power balance of a CPU cooler? What, do you suppose, is the margin of error imposed by the equation as opposed to the ability to measure such a phenomenon under these conditions?
January 17, 2007 6:47:14 AM

Is this "zero-point energy" of which you speak the spontaneous creation (and immediate destruction) of particle-antiparticle pairs in the vacuum of space? I was fascinated by this concept and I did manage to read a couple of Hawking's books.

And I have to think we have now officially hijacked this thread.
January 17, 2007 6:48:58 AM

Cheers
January 17, 2007 8:02:40 AM

Back to the original thread. BTW you are both very intelligent people. Take example, the red blood cell. Instead of coloring something black or hot ping or white. Just dimple the fins. Adding drag to the fins would increase the time the air would spend absorbing the heat away from the heatsink. At the same time make the surface rough, so that its even less aerodynamic. Add to that, make the fins thinner too.

The best heatsink for air cooling would be made of copper with each fin rough and dimpled with adequate space in between fins to increase airflow. This would add to the surface area of the fins, add resistence to the flow of air, and better cool the CPU using the same or less copper used in standard heatsinks.

Fill the heatpipes with mercury to increase conduction of heat, thus allowing the cpu to be cooled faster and allow the temps to raise at a slower rate.

Finally, in regards to the heatpipes filled with mercury. Leave room for the metaloid to expand when heated, and fill the void with a gas that cools as it compresses. Or, fill the heatpipes with a fluid that if compressed with a specific gas causes a endothermic reaction.

I'm not a university graduate. Thats why I would find it helpful if such bright minds could critique my idea(s). To conclude I will follow Jack's lead and sleep.

Eric Hanson
January 17, 2007 12:47:16 PM

Quote:
Hey, I got one for you.... a real stumper. Lets say theoretically there is an infinite heat sink, and no energy to return. An object would continually radiate that energy.... over time, eventually, would it radiate all of it's energy to nothing?


This sounds like the realm of Boltzman's Transport Equation without making any assumptions... :wink:
January 17, 2007 1:09:15 PM

Here's a question for the others here that have experience with heat transfer/transport about materials. Copper is listed as the best metal conductor of heat, but what material is actually the best conductor of heat (some may be shocked by the answer)?
January 17, 2007 1:18:14 PM

out of curiousity, wouldnt a black bottom on a heatsink and a white top be the best solution?

The black would absorb the heat from the chip and the white would reflect any external heat away from the heat sink.

kinda like a black heat pipe with white fins?
January 17, 2007 1:56:49 PM

isn't it diamond? I remember seeing a chart with diamond as the best thermal conductor.
January 17, 2007 2:12:56 PM

I dunno, are carbon crystals good heat conductors? Because that's all a diamond is...

I'm not sure what would happen to a perfect black body that would have no more energy to radiate... Wouldn't it turn its own mass into energy until there were no more mass to convert?

I think that's actually the crux of the third law Jack is citing: no more energy to radiate means the object doesn't -exist- anymore (the second law essentially describes entropy, if I'm not mistaken)
January 17, 2007 2:19:29 PM

Graphite is also a type of carbon crystal, but it doesn't conduct heat as well as diamond.
January 17, 2007 2:48:58 PM

Try sending that in to Mythbusters :D 
Im sure they'd try it out :lol: 
January 17, 2007 4:16:39 PM

Sorry I know this has been said, I just didnt have time to read though the thread.



Black paint would create an insulating polymer around the metal. The effectiveness of the paint's compound would determine the amount of convection. I would guess that water based paint (the color does not matter since heat is not being absorbed from an external souce affected by the pbsorption properties of color) would provide the best odds of increasing the thermal convection properties of the heatsink.

I have read much about the pitting of metals (by-product is turning the surface black).

This would increase overall surface area, which would inherently make it a better radiator then an equal, not surfaced alloy.

Either the original thread is thinking about nanostructure laser pitting, Blackbody radiation, or black paint.

Laser pitting is the correct method of turning a heatsync black, while improving thermal conductivity.
January 17, 2007 4:45:50 PM

Quote:
isn't it diamond? I remember seeing a chart with diamond as the best thermal conductor.


Yes, it is diamond. What makes it so unusual is it's the only odd ball substance out there that is a excellent heat conductor while being a very poor electical conductor. Diamond, as Jack noted, is a very poor conductor of electricity, but the best heat conductor known to man. Typically, if it conducts heat well, it also conduct electricity well (this is why Copper is chosen so often for both conducting heat and for conducting electricity as it is the best material for doing both).

And Jack, from what I've heard from people in the know at GE, you can make an artificial diamond that is indistiguishable from one that came out of the ground, DeBeyers is none too happy about this... :wink:
January 17, 2007 4:51:39 PM

Wow. Thanks to HotFoot and JumpingJack on this one. I've learned a lot. :!: This is exactly the kind of discussion I was looking for.

Ultimately, relating to the original post, I think this says it best:

Quote:
The original point is that emissivity is not a function of color but of the material.


I believe ^^^ is where the notion of painting a HS black was misconstrued, and unless I'm mistaken, that's been dispelled now.


I like the idea about (carefully designed) bumps on heat sinks. That makes a lot of sense.

Graphite and Diamond are both carbon-structures, no?



Back to the discussion, though, if Jack or HotFoot would like to enlighten me, I have some questions:

The emissivity determins how much the material is willing to gain/lose through electromagnetic radiation, right? If so, what determines how much the material is willing to gain/lose? Is the emissivity equal to both the radiation and absorbtion rates for a structure?

When an object changes color as a result of radiating a large amount of energy (i.e. glowing red-hot), does that change the emissivity of the object?

(I never took a physics class a day in my life, so please bear with me)


Please tell me what's wrong with this interpretation:

An object radiates energy in the form of photons (Anything else?). The radiation and absorbtion depends greatly on the difference in energy levels between the photon and the atom, and the emissivity of the materials. So when a heat sink gets hot, it radiates energy, and that energy is (mostly) absorbed by the cooler air, making the air hot. A "Black Body" has more emissivity, so in theory, a black-body would be a better conductor because it's more willing to forfeit its energy.

(That raises a question. What intices an atom to radiate energy? Do atoms constantly radiate and absorb energy?)

So if you have 2 objects in an absolute-zero environment (given Jack's riddle), the object that is more of a "Black-Body" would "Get colder faster," right?

I'm also assuming the physical properties of a material determine the preferred energy transfer, i.e. heat versus electricity?


There's so much I don't know :( 

Thanks.
January 17, 2007 5:21:04 PM

The simple answer to your question about a material's emissivity being equal for both radiation and absorption is yes. However, the more complicated answer is actually no: so "yes and no".

Emissivity is a function of wavelength of the radiation, or colour if you will. Therefore, a certain type of material will better absorb radiation of a certain wavelength versus radiation of a different wavelength. See the graph below:



That covers absorption. Emission, however, is more complicated. The preferred wavelength of emitted radiation is a function of the body's temperature. Therefore, if we follow the emissivity as a function of wavelength graph, the effective emissivity of the object will be higher at certain temperatures than at others. I believe effective emissivity generally grows with temperature, peaking when the thermal energy is sufficient to allow the material to radiate at a "natural" frequency.

This has implications which mean that the effective emissivity of a given material may be different for absorption than for emission. This is because it is not likely that the incident radiation being absorbed just happens to be the same wavelength as the radiation being emitted by the object.

Jack is also correct to point out that black paint doesn't mean you have a perfect black body. However, a rated emissivity of 0.96 is pretty high compared to aluminium's 0.11. The above argument, though, would indicate that a simple number isn't sufficient to describe an object's emissivity. You need to know the preferential wavelengths at which light is being absorbed and emitted and determine the effective emissivity of the material at these wavelengths.

EDIT: Oh yeah, and as for what causes the material to radiate, you'll get a better explanation from someone versed in quantum physics, but basically any oscillating charged particle (or maybe it doesn't even need to be charged) must emit radiation. Photons are emitted from atoms that have elevated energy states, which has something to do with electron orbits and energy levels. Photons that are absorbed by atoms, if the right wavelength, will cause the energy state of the atom to increase. I believe this extends to metallic structures where the valance electrons are shared by very large groups of atoms.
January 17, 2007 5:24:01 PM

Quote:



Graphite and Diamond are both carbon-structures, no?
Yes, but density has changed.


...

The emissivity determins how much the material is willing to gain/lose through electromagnetic radiation, right? If so, what determines how much the material is willing to gain/lose? Is the emissivity equal to both the radiation and absorbtion rates for a structure?

The type of atom (material) determines how long it holds on to energy before re emitting it in the form of electromagnetic radiation. The density changes how much energy is absorbed (as well as efficiency of the material to absorbs energy as specific wavelenghts).

When an object changes color as a result of radiating a large amount of energy (i.e. glowing red-hot), does that change the emissivity of the object?

The color change is the product of energy dissipation in the form of visible light spectrum. The material will still follow a set curve of emissivity at specific temperatures which accounts for it's present temperature.

(I never took a physics class a day in my life, so please bear with me)





ran out of time.... im sure the rest will finish answering y
our questions before I get back.


Took a second to look at Hotfoot.. looks like he should re answer what I said in a more technical form. Nice to meet you hotfoot.
January 17, 2007 6:06:30 PM

The subject gets back to one of Einsteins original papers he did I think it was back in 1905 or 06. He showed that light impacting a surface gave rise to a charge potential. What we know today as photovoltaic cells (Solar Cells) The effect is known as the photoelectric effect.

That is were all of this starts from.
January 17, 2007 6:22:37 PM

Quote:
The subject gets back to one of Einsteins original papers he did I think it was back in 1905 or 06. He showed that light impacting a surface gave rise to a charge potential. What we know today as photovoltaic cells (Solar Cells) The effect is known as the photoelectric effect.

That is were all of this starts from.


While Photoelectric Effect had the most far reaching affects of the three papers he wrote in that year (one on Brownian Motion, one on Relativity, and one on PE), PE has little to do with Black Body Radiation and more to do with the characterics of how Photons interact with matter. (Either raising an electron to a higher shell thus exciting the atom, ejecting an electron from the atom thus exciting it, scattering off the atom thus exciting it, the photons created by annihalation of an electron and an anti-electron, or spontaneously becoming an electron and anti-electron if the photon's energy was high enough, 1.022 MeV or greater.)
January 17, 2007 6:45:22 PM

For many years the standard heat sink for electronic was flat black anodized aluminum.

My questions is, was it black because that made it a better heatsink, or was it black becuase they wanted to anodize it to prevent corrosion and black what the only color they could anodize back then?

While you don't seen them black much anymore, I suspect this has more to do with cost rather than heat dissipation.

Maybe this is where the myth, if it is one, comes from.
January 17, 2007 7:45:24 PM

Actually Single Wall Carbon Nanotubes may be better than diamond
January 17, 2007 9:23:17 PM

Quote:
For many years the standard heat sink for electronic was flat black anodized aluminum.

My questions is, was it black because that made it a better heatsink, or was it black becuase they wanted to anodize it to prevent corrosion and black what the only color they could anodize back then?

While you don't seen them black much anymore, I suspect this has more to do with cost rather than heat dissipation.

Maybe this is where the myth, if it is one, comes from.


If it was due to cost, then you could buy enthusiast HSFs that were black. I dont remember why they were black in the first place though.
January 17, 2007 9:26:31 PM

Thats what I was thinking too. They are superconducters so if you coated copper with them, it should dramatically increase its cooling efficiency. The faster heat travels away from the CPU the fast you can cool it. I'd imagine coating an whole heat sink and fins would likely be rediculously expensive. In the next few years it may get cheaper though.
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