# Microwave oven+black bodies

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## Main Question or Discussion Point

First of all I'm not sure this thread should be in this section or in the Thermodynamics one.
I don't understand something. If I'm not wrong, the wavelength of photons used in microwave oven cooking is around $$122 mm \sim 10^{-1} m$$, much greater than visible light photons (450 to 750 nanometers, from my memory, which is about $$5x10^{-6}m$$. So the frequencies of micro wave is much lesser than the one of visible light. Since $$E_{\text{photons}}=h \nu$$ where $$\nu$$ is the frequency, it means that photons of visible light are much energetic than the microwave's ones, in fact by a factor of $$10^5$$. I've heard that food (especially the ones containing water) can be cooked by microwaves because the wavelength of the waves "matches" the water molecules or so (though I don't understand this since it's they are at a totally different scale) and they enter in resonance so that in fact water molecules absorbs well the microwave. So far so good.

Now comes my problem. If I paint any body of black so that it absorbs say around 70% of visible light, how come it doesn't heat much more than water in a microwave? Photons of visible light are so much more energetic and are mostly absorbed by the material (by its surface at least) that I can't understand why the painted "black" body doesn't heat up more than water in a microwave.
I know Stefan-Boltzmann's law of radiation so I can understand that a black body wouldn't heat that much because it also emits part of what it absorbs... but what happens in the microwave, I have no idea.

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You have to take into account both, the energy of the photons and the number of them in order to calculate the amount of power absorbed.

The power of a normal microwave is around 1.5 kW, which would correspond to the power from the sun over a surface of 1 m^2. I guess that if you can concentrate all this power into a glass of water it will get hot too.

Actually there, it is being used already: http://en.wikipedia.org/wiki/Solar_cooker" [Broken]

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alxm
Now comes my problem. If I paint any body of black so that it absorbs say around 70% of visible light, how come it doesn't heat much more than water in a microwave?

Because these are two quite different parts of the electromagnetic spectrum (as you seem aware). There's no reason to assume that something more absorbing in one area of the spectrum is more absorbing of radiation in other areas; radiation from different regions of the spectrum cause different physical effects. (which is in fact more or less how the different regions are characterized, with the natural exception of the visual range)

After all, a black radio doesn't get worse reception than a white one.

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Thanks for the replies!
You have to take into account both, the energy of the photons and the number of them in order to calculate the amount of power absorbed.

The power of a normal microwave is around 1.5 kW, which would correspond to the power from the sun over a surface of 1 m^2. I guess that if you can concentrate all this power into a glass of water it will get hot too.

Actually there, it is being used already: http://en.wikipedia.org/wiki/Solar_cooker" [Broken]
So it means that, if I'm not wrong, the intensity of EM waves inside a microwave is about 10^5 the one on Earth caused by the Sun.

Because these are two quite different parts of the electromagnetic spectrum (as you seem aware). There's no reason to assume that something more absorbing in one area of the spectrum is more absorbing of radiation in other areas
Yes exactly. Water seems to absorbs microwave and not much visible light. That's why it needs so many photons to heat up since microwaves photons caries only $$10^{-5}$$ times the energy of visible light photons. On the other hand, a painted black body will absorbs light in the visible spectrum at least (and maybe even microwaves, I don't really know) and so it needs $$10^{-5}$$ times less photons to heat up compared to water, approximately. That's my understanding now... I hope it's right.

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Thanks for the replies!
So it means that, if I'm not wrong, the intensity of EM waves inside a microwave is about 10^5 the one on Earth caused by the Sun.
I am not sure what you mean by that. The number of photons per second to get the same power will be 10^5 times larger in the microwave frequency than for visible light.

What I meant is that the energy of all the sun light photons incident on a surface of 1 m^2 will be equal to the energy of all the photons in the microwave. The number of photons will not be equal though.

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I am not sure what you mean by that. The number of photons per second to get the same power will be 10^5 times larger in the microwave frequency than for visible light.

What I meant is that the energy of all the sun light photons incident on a surface of 1 m^2 will be equal to the energy of all the photons in the microwave. The number of photons will not be equal though.
Yes that's what I meant in
fluidistic said:
Water seems to absorbs microwave and not much visible light. That's why it needs so many photons to heat up since microwaves photons caries only LaTeX Code: 10^{-5} times the energy of visible light photons. On the other hand, a painted black body will absorbs light in the visible spectrum at least (and maybe even microwaves, I don't really know) and so it needs LaTeX Code: 10^{-5} times less photons to heat up compared to water, approximately.
. I get it, thanks a lot! I think the intensity of light is directly proportional to the number of photons hitting a given area. So in a microwave, the intensity is roughly $$10^5$$ times greater than the light coming from the Sun, on Earth. (Assuming that the surface area of a microwave is of the order of a square meter, which seems to be OK).
It wouldn't occur to me to heat up food with photons $$10^{-5}$$ times less energetic than visible light's ones because of the intensity required. It seems interesting to learn how a microwave works. Specially how does it get a so big intensity. I'll read some information.