# Microwave oven+black bodies

1. Aug 19, 2010

### fluidistic

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.

2. Aug 19, 2010

### Pete99

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]

Last edited by a moderator: May 4, 2017
3. Aug 19, 2010

### alxm

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.

4. Aug 19, 2010

### fluidistic

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.

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.

Last edited by a moderator: May 4, 2017
5. Aug 20, 2010

### Pete99

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.

6. Aug 20, 2010

### fluidistic

Yes that's what I meant in
. 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.

7. Aug 20, 2010

### brainstorm

My understanding is that infrared heat doesn't penetrate the surface of food. Microwaves, however, somehow do penetrate and vibrate the water molecules, thereby heating the food anywhere it is moist. So with a conventional (infrared) oven, much of the radiation is getting reflected off the food while conduction is taking place to get the heat from the outside to the inside of whatever you're cooking. I guess microwave cooking goes so fast because you don't have to deal with (as much?) reflection and conduction. I may not be completely correctly interpreting the book I read, though. I also, for example, don't get why microwaves interact with water molecules in particular when their wavelengths are from something like 3mm to 30cm (I forget the exact range). It does make sense that they don't penetrate the atmosphere, though, since they must get absorbed by water vapor before making it to the ground.

8. Aug 21, 2010

### alxm

Yes, food is opaque to infrared, and less opaque to microwaves. So the microwaves penetrate straight through the food and heat it evenly. The main reason this is faster is that it allows you to use higher power. With ordinary heating you're limited by the fact the surface will start to burn.

Microwaves interact with molecules that have an electrical dipole moment. (such as water) The varying electric field of the radiation causes them to re-orient themselves with it - causing molecular motion, i.e. heat.