# Really dumb question

1. Dec 4, 2006

### Manchot

I know that this question is going to sound absurd, and the answer's probably obvious to most of you, but why can't you point a solar cell at a black body and get energy from its thermal radiation? Obviously, that violates the laws of thermodynamics.

2. Dec 4, 2006

### StatMechGuy

Well, you can, it's just that you can't get more energy out of the radiating black body than you put into it, so it's kind of inefficient.

3. Dec 4, 2006

### mathman

You can point a solar cell at the sun and get energy, so why not a black hole (you will get very little!)?

4. Dec 4, 2006

### HallsofIvy

mathman, he did not say a "black hole", he said a "black body".

5. Dec 4, 2006

### Repetit

Why would that be a problem? The sun is a black body (approximately) and that obviously works just fine with solar cells.

6. Dec 5, 2006

### Claude Bile

Radiated energy from any body, black or otherwise will generate a potential across a solar cell, provided the frequency of the radiation is high enough (i.e. the energy of the photons is sufficient to generate electron-hole pairs).

Claude.

7. Dec 5, 2006

### Staff: Mentor

Perhaps your confusion is about what happens to a black body: a black body loses energy as it radiates. If it isn't generating energy internally (ie, the sun), it cools down.

8. Dec 5, 2006

### Manchot

I realize that the black body cools off as it radiates: my problem with it is that you are getting heat energy without a temperature differential.

9. Dec 6, 2006

### lightarrow

Not at all. The surrounding environment's temperature must be lower than the body's temperature, or the heat you get is the same you lose.

Last edited: Dec 6, 2006
10. Dec 6, 2006

### Aero

A temperature differential is only required for heat conduction. WIth a solar cell, you are getting heat energy by radiation.

11. Dec 6, 2006

### Jheriko

I think perhaps the term "black-body" is the source of the confusion. When I first encountered black bodies at school I found them pretty confusing myself because I couldn't see how a black body could possibly radiate any amount of energy. I couldn't get myself away from the idea that it was actually black... and therefore didn't emit any light.

As far as I know, it is described as "black" because it doesn't emit any radiation on its own from internal processes (e.g. radioactivity, having lots of electric current flowing through it etc..) and because it doesn't reflect any radiation from external sources (i.e. absorbs all of it). This is an analogy to how something appears black if it doesn't emit or reflect any visible light. The source of the black body radiation is the heat of the body, so it doesn't really look black when it gets hot. For instance, the Sun is a bright yellow because the peak of its black-body radiation curve is in that part of the visible spectrum.

The black body radiates only because it has already been given energy to radiate away, and the body cools whilst this is happening from the loss of energy to radiation. I'm not 100% sure (I can imagine some pesky quantum effect breaking it) but a black body at 0K should radiate nothing at all since there is no energy left to produce the radiation.

So, since a black body needs to have energy put into it somehow, in order to radiate, there is no "break" in the laws of thermodynamics if we point a solar cell at a black body (the sun) and are then able to convert this radiation into electrical energy. The ultimate source of this energy is the source of the energy for the black body, in the case of the Sun this probably would have come from the gravitational collapse as it formed and/or the nuclear fusion in the core.

Hope this helps.

12. Dec 6, 2006

### lightarrow

Because the radiation's temperature is greater than the solar cell's temperature and this, in turn, because the body that has emitted that radiation had a greater temperature than the solar cell.
You always need temperature differentials.

13. Dec 6, 2006

### Repetit

Is this true? I mean, if we imagine the solar cell being at a higher temperature than the surface of the sun (I know this is impossible for various reasons, but theoretically speaking) it would still recieve radiation from the sun and convert this radiation to electric energy wouldn't it?

14. Dec 6, 2006

### Aero

How can radiation have a temperature? Radiation is just electromagnetic waves, but temperature only refers to the level of thermal vibrations in matter.

15. Dec 6, 2006

### Staff: Mentor

No, it wouldn't - the efficiency of the solar cell is temperature dependent. Look at the equation for heat transfer via radiation. It has terms for both temperatures: http://www.engineersedge.com/heat_transfer/black_body_radiation.htm

16. Dec 6, 2006

### Staff: Mentor

The frequency of that radiation is temperature dependent. That's why an infrared thermometer works (and why the color of the light from a light bulb changes as you change its output)...

http://www.omega.com/prodinfo/infraredthermometer.html

17. Dec 7, 2006

### lightarrow

So what would be, for example, Cosmic Background Radiation's temperature ( 2.725Â°K)?

Actually, it's possible to associate a temperature to radiation if this has a blackbody's spectrum; I assume a monochromatic radiation's temperature could be the same of a blackbody which maximum emission is on that frequency, or something like this.

18. Dec 7, 2006

### Aero

But isn't this temperature irrelevant, because light of any "temperature" will heat something up, no matter waht the temperature differential?

19. Dec 7, 2006

### lightarrow

No. If you want to heat, e.g., a lamp with another's equal lamp radiation, what happens is that the lamp absorbs the radiation but also emits radiation and, so, loses heat; the net result is: no heating. Every body emits radiation (not only lamps!) as more as greater is the temp.

Thermodinamics laws are still valid with radiations! They must be valid.

20. Dec 7, 2006

### Staff: Mentor

If two objects have the same temperature, they will be firing photons of the same energy at each other and neither will gain or lose temperature.