just checking the temp of the walls behind the stove and it got me tinking. wouldn't they absorb more heat if they were black? and if so i don't think i would be comfortable with that as they are quite hot now.
wall colour (color) clearances
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The ones I have seen to purchase from Copperfield are black, and that could be why. You get a board and then a spacer kit.
I think we need to call the Mythbusters.
Does color matter in regards to radiated heat? I thought that only made a difference if light (specifically sunlight) was involved.
Does color matter in regards to radiated heat? I thought that only made a difference if light (specifically sunlight) was involved.
Definitely an issue that comes up from time to time, it would be neat to see some type of actual demonstration that we could all wrap our hands around. Don't hold me to this, but depending on how things go this weekend, I may be able to round up a black and white panel and some thermocouples and see how it goes.
As for the question, I suspect what you are actually concerned about is 'emissivity' or how much energy the surface emits. While many black surfaces absorb a substantial amount of visible light, and convert that to heat, they also shine brightly in the infra red portion of the spectrum - they can often approach emissivity values around .95 with a 1 being a "perfect" emitter. The wall behind your stove is probably absorbing very little visible light, so the visible color we see is of little importance. But being black with an emissivity of .95 means it is shining brightly in the infra red - which will actually keep the wall the coolest - that is the theory anyway.
As for the question, I suspect what you are actually concerned about is 'emissivity' or how much energy the surface emits. While many black surfaces absorb a substantial amount of visible light, and convert that to heat, they also shine brightly in the infra red portion of the spectrum - they can often approach emissivity values around .95 with a 1 being a "perfect" emitter. The wall behind your stove is probably absorbing very little visible light, so the visible color we see is of little importance. But being black with an emissivity of .95 means it is shining brightly in the infra red - which will actually keep the wall the coolest - that is the theory anyway.
In searching the internet, I found this quote:
"We call the sun's energy radiation. Unfortunately, we also call the heat coming from a fireplace "radiant", but they are two different things. Heat radiating from a fireplace cannot be reflected. A white sock drying in front of a fireplace will get just as hot as a black sock. But, if you put the two socks in the sun, the black sock will get hotter."
But I'm not convinced this is actually true. I will have to do the black/white sock expariment. However light from the sun is not the same wavelength as light from a woodstove, I believe most of the woodstove light is in the infrared range. This complicates things, because what appears light in color in our human visible spectrum, may in fact be dark in color in the infrared spectrum.
(broken link removed) — B
In principle, the brownie would heat up faster by radiation in a hot environment and cool off faster by radiation in a cold environment. A black object is better at both absorbing thermal radiation and emitting thermal radiation, so the brownie would soak up more thermal radiation in the hot environment and give off more thermal radiation in the cold environment.
In practice, however, most of the radiation involved in baking these desserts and letting them cool on a kitchen counter is in the infrared and it's hard to tell just what color a brownie or cake is in the infrared. It's likely that both are pretty dark when viewed in infrared light. Basically, even things that look white to your eye are often gray or black in the infrared. Thus I suspect that both the brownie and cake absorb most of the thermal radiation they receive while being baked and emit thermal radiation efficienty while they're cooling on the counter.
"We call the sun's energy radiation. Unfortunately, we also call the heat coming from a fireplace "radiant", but they are two different things. Heat radiating from a fireplace cannot be reflected. A white sock drying in front of a fireplace will get just as hot as a black sock. But, if you put the two socks in the sun, the black sock will get hotter."
But I'm not convinced this is actually true. I will have to do the black/white sock expariment. However light from the sun is not the same wavelength as light from a woodstove, I believe most of the woodstove light is in the infrared range. This complicates things, because what appears light in color in our human visible spectrum, may in fact be dark in color in the infrared spectrum.
(broken link removed) — B
In principle, the brownie would heat up faster by radiation in a hot environment and cool off faster by radiation in a cold environment. A black object is better at both absorbing thermal radiation and emitting thermal radiation, so the brownie would soak up more thermal radiation in the hot environment and give off more thermal radiation in the cold environment.
In practice, however, most of the radiation involved in baking these desserts and letting them cool on a kitchen counter is in the infrared and it's hard to tell just what color a brownie or cake is in the infrared. It's likely that both are pretty dark when viewed in infrared light. Basically, even things that look white to your eye are often gray or black in the infrared. Thus I suspect that both the brownie and cake absorb most of the thermal radiation they receive while being baked and emit thermal radiation efficienty while they're cooling on the counter.
I don't know the exact physics to recite chapter and verse, however I do know that when Elk and I were in the VC burn lab, the setup they use for clearance testing has black walls, hearth, and ceiling. Supposedly this was so that the wall would get as hot as possible so that the test would be determining the "worst case" minimum clearance.
FWIW, in case you were wondering how the clearances are determined, it is quite experimental - the walls are all movable and loaded with thermal sensors - they basically crank the stove and see how close they can shove the moveable walls to it before they exceed the allowed temp limits.
Gooserider
FWIW, in case you were wondering how the clearances are determined, it is quite experimental - the walls are all movable and loaded with thermal sensors - they basically crank the stove and see how close they can shove the moveable walls to it before they exceed the allowed temp limits.
Gooserider
I have been in the Heat & Glo test labs and they also have all their test walls painted black, loaded up with thermocouples.
jtp10181 said:
Maybe they use black walls because any other color might be more relective for the sensor equipment - for example many of us have those wireless temp sensors - these could reflect off a shiny wall and read a temperature from somewhere else in the room which would be a problem. The other thing is that black walls might absorb more radiation from other sources in the room including sunlight so to figure out the max temperature they would have to account for heat from multiple sources in addition to the woodstove.
But the woodstove also puts out plenty of infrared radiation which would be absorbed better by a black wall.
sounds correct. but have you ever used one of those survial blankets. its like a foil sleeping bag designed to reflect heat back towards the body. im going to stick a piece of tin foil on one wall and some dark paper on the other (i have a corner install) then see if i can notice a differece. i will tape all four sides to try to eliminate air flow. very primitive test, but if i can't notice a difference than its probably a mute point.
The best radiators are black and the best absorbers are black.
Here is a link about black surfaces.
https://www.hearth.com/talk/threads/3142/
Here is a link about black surfaces.
https://www.hearth.com/talk/threads/3142/
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