On 12/19/2010 1:05 AM, Adrian wrote:
> Ok a black body absorbs infra red radiation very well, but how does it
> radiate it?
> For it to radiate infra red radiation then it becomes red with rise in
> temperature and is no longer black.
No it does not have to change color. *Color* is a DIRECT indication of
*temperature*, not the amount of heat (IR) radiated.
IR by it's nature is not visible. Although the tube plate may glow red
or even white, the "heat" radiated is not visible. You can even look up
the specific temperatures for given colors.
Black is a very efficient radiator of heat. Hence it maintains a lower
temperature. Yes I know that is counter intuitive but it's the laws a
physics.
I calibrated Pyrometers for years. I spent 26 years in maint before
going to college. A good portion of that time was spent doing instrument
calibration, working with and helping develop standards as well as
specializing in Chromatography for about 10 years.
His take on black body radiation is correct. IOW " a black body absorbs
heat perfectly and therefor must radiate it perfectly."
> The reason black gives off heat so well is due to the immediate
> convection cooling with air contact and its higher surface temperature
> than other colours.
>
It helps but that is not the reason. Surface area also aids in getting
rid of heat.
> The problem with your theory is that convection heat transfer has been
> misrepresented as radiated heat transfer.
>
No it hasn't. Although forced air is not quite the same as convection
the basic mechanics are the same. (Cooler air picking up heat from a
warmer surface) But we are talking about heat absorbed and re-radiated
from a black surface.
> To demonstrate my point please make a black heat radiator that stays
> black, that I can feel the (radiated) heat from several feet away. use
> whatever energy source you like, but it must stay black
> at the heat source. ( No fan or air movement involved> convection).
>
That's true.
73
Roger (K8RI)
> Adrian ... vk4tux
>
>
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