Can't Get Energy from Black Body with Solar Cell

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Discussion Overview

The discussion centers around the feasibility of using a solar cell to extract energy from a black body through its thermal radiation. Participants explore the underlying principles of thermodynamics, the nature of black body radiation, and the conditions necessary for energy transfer via radiation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants question why a solar cell cannot extract energy from a black body, suggesting that it violates thermodynamic laws.
  • Others argue that while energy can be obtained from a black body, the efficiency is limited as one cannot extract more energy than is input into the system.
  • A participant points out that the sun, which is an approximate black body, successfully provides energy to solar cells.
  • Some participants clarify that thermal radiation can generate a potential across a solar cell if the radiation frequency is sufficient to create electron-hole pairs.
  • Concerns are raised about the necessity of a temperature differential for effective energy transfer, with some asserting that it is not required for radiation, while others insist it is essential.
  • There is a discussion about the concept of temperature in relation to radiation, with some participants questioning how radiation can possess temperature and its implications for energy transfer.
  • Some participants highlight that if two bodies are at the same temperature, there will be no net energy transfer between them, emphasizing the importance of temperature differentials in thermodynamic processes.

Areas of Agreement / Disagreement

Participants express differing views on the role of temperature differentials in energy transfer via radiation, with no consensus reached on whether a differential is necessary. The discussion remains unresolved regarding the implications of black body radiation and its interaction with solar cells.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about black body radiation, the definitions of temperature in relation to radiation, and the conditions under which energy can be extracted from thermal radiation.

Manchot
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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.
 
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Manchot said:
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.

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.
 
You can point a solar cell at the sun and get energy, so why not a black hole (you will get very little!)?
 
mathman said:
You can point a solar cell at the sun and get energy, so why not a black hole (you will get very little!)?
mathman, he did not say a "black hole", he said a "black body".
 
Why would that be a problem? The sun is a black body (approximately) and that obviously works just fine with solar cells.
 
Manchot said:
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.

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.
 
Manchot said:
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.
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.
 
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.
 
Manchot said:
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.
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:
  • #10
A temperature differential is only required for heat conduction. WIth a solar cell, you are getting heat energy by radiation.
 
  • #11
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
Aero said:
A temperature differential is only required for heat conduction. WIth a solar cell, you are getting heat energy by radiation.
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
lightarrow said:
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.

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 receive radiation from the sun and convert this radiation to electric energy wouldn't it?
 
  • #14
How can radiation have a temperature? Radiation is just electromagnetic waves, but temperature only refers to the level of thermal vibrations in matter.
 
  • #15
Repetit said:
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 receive radiation from the sun and convert this radiation to electric energy wouldn't it?
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
Aero said:
How can radiation have a temperature? Radiation is just electromagnetic waves, but temperature only refers to the level of thermal vibrations in matter.
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
Aero said:
How can radiation have a temperature? Radiation is just electromagnetic waves, but temperature only refers to the level of thermal vibrations in matter.
So what would be, for example, Cosmic Background Radiation's temperature ( 2.725°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
But isn't this temperature irrelevant, because light of any "temperature" will heat something up, no matter what the temperature differential?
 
  • #19
Aero said:
But isn't this temperature irrelevant, because light of any "temperature" will heat something up, no matter what the temperature differential?
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
Aero said:
How can radiation have a temperature? Radiation is just electromagnetic waves, but temperature only refers to the level of thermal vibrations in matter.
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.
 
  • #21
I do believe that you have just stated one of Kirchoff's laws for radiation.
 
  • #22
sicjeff said:
I do believe that you have just stated one of Kirchoff's laws for radiation.
Of course!
You see? We don't need to study on books, to make physics!:smile:
(Of course I'm joking!).
 

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