Using light from the sun to heat something above temperature of the sun?

This is a question I stole from another source. The answer given was that it was impossible. This seems counter-intuitive to me, could somebody explain what the answer is, and why?

Lets use ideal circumstances of an unrealistically large magnifying glass (or many?), on a small object which absorbs most light that has no atmosphere around it, as well as any other factors which I have neglected.

Think of it from the ant’s point of view. When the ant looks up, every direction it looks, it sees the surface of the Sun. It’s like it’s completely surrounded by the surface of a star. And hence, it heats up to the temperature of the surface of a star.

Well, the reason you’re having a problem with this question is that you’re mixing up some of the terms. And like you’ve already said, leaving some out.

Consider. The sun give up some heat. You take a portion of it, and concentrate it on a small area. How can you have more, when you took a portion? You mean you have more temperature per surface area? Well, sure, but not more heat, if you follow.

There was this old children’s book of experiments that had a caper where you could “Generate 100,000 pound per square inch of force.” What you did was push a pin through a cloth. The force your hand generated, per the tiny surface area of a pin point, generated the result.

I always though that experiment was lame. It still doesn’t seem to me to teach anything. Is it any wonder I get just as confused in this older thread as you are in this thread?

You might want to look at http://en.wikipedia.org/wiki/Heliostat.

This analogy was very useful, and it answers my question. Thank you for stating it in this way.

You can certainly do what the OP asks, by creating electricity through solar cells and then using it to drive the right kind of heating, perhaps an arc.

Doing it by heating a surface through absorption of the light, that’s impossible. I think.

It certainly is possible. All you have to do is collect solar energy over a wide area (or over a long period of time) and concentrate it in a small amount of matter. There is no law of conservation of temperature.

There is something related to the OP’s question, though. It used to be called the Brightness Theorem, but the People In Charge of Confusing Names have now decreed that it is the law of Conservation of Radiance. It states that the Radiance (which used to be called brightness) divided by the square of the refractive index of its medium is always the same:

http://www.its.bldrdoc.gov/fs-1037/dir-009/_1241.htm

As the Wikipedia page (or any text on radiometry) tells you, the Radiance is the Radiant flux (power emitted in photons) per unit area of the emitter per unit solid angle it’s emitted into divided by the angle between the surface normal and the direction.

See also Etendue:

http://en.wikipedia.org/wiki/Etendue
What this means is that, do what you will in a passive ootics system (one that can’t add energy to the system), if the index of the emitter and receiver are the same, you can’t get more than the radiuance of the emitter (You can always get less, by having elements that don’t perfectly transmit all the light). So your magnifying glass can’t make the radiance at the any exceed that of the sun.

What you can do is to change the solid angle, or the area under consideration. Suggestions of storing the power given above in the thread violate the requirement of passivity.

I’m pretty sure this is the wrong way to think about it.

If you collected all the light that was output from, say, one square foot of the sun’s surface and concentrated it on a well-insulated square inch of the Earth’s surface, that square inch would heat up to much hotter than the surface of the sun. The energy radiated from the surface of the sun is a function of its temperature. For the square inch to reach an equilibrium temperature, it would have to radiate as much energy as is being put into it (144 times as much as the sun). If it’s outputting energy the same way the sun is (black body radiation), it would have to get much hotter.

You both refer to collecting and concentrating. I thing the OP is disallowing any process that introduces energy into the system. So you can collect and concentrate with lenses, mirrors and any other passive system you want but the second law of thermodynamics will not allow you to use the sun to heat an object to a temperature hotter than the sun that way. If the object in question was hotter, it would cool itself by heating up the sun.

No… I’m with the walrus on this one.

Black body radiation alone, without any magnifying glasses or lenses, would never do more than bring the objects to equilibrium because at a certain point, they’re both radiating the same amount of energy.

But black body radiation is a function of surface area as well as temperature. (From Wikipedia: “The Stefan–Boltzmann law states that the power emitted per unit area of the surface of a black body is directly proportional to the fourth power of its absolute temperature”). If you can use a lens to take a square foot of the sun and focus it on a square inch of the Earth, the square inch will have to be much hotter before it can reach an equilibrium between incoming and outgoing energy.

See my post # 8 “Radiance” is radiant flux per unit time per unit area per unit solid angle and divided by the cosine of the angle.

I am with the walrus also.

I do have a question.
With the best lens humans can make today, what is the percentage of energy of the square foot of sun that is actually being concentrated?

As a kid, I used a fair dinkium sized magnifying glass to burn/blow up stuff and I know I was not using the total surface of the sun.

Sun rays are parallel as far as is practical for us so, with loss caused by the atmosphere, is not what I am using just a percentage of the power in that small diameter? :dubious:

With a 10 sq in glass, I am getting only a percentage of the power that the glass can see/utilize and then the losses of the imperfect glass get added, so do I have enough power on a spot 1/10th of a square inch, to get the target over the point source temperature of the sun? :confused:

Not what the OP really asked but I was wondering how hot the spot I made as a kid in the instant of maximum temperature? :smiley:

I’m clearly out of my depth so I’ll just ask. How is the Stefan–Boltzmann law reconcilable with the second law of thermodynamics?

Why would you think it wouldn’t be? The Clausius statement that you link to is true for the net exchange of energy.

So can a body with a large surface area transfer heat to a hotter body with a small surface area?

Hmm.

If you set up mirrors/lenses/whatever to channel radiation from the sun to your object, then as soon as your object got up to the temperature of the sun, there’s no more net heat transfer, since the object would be radiating as much energy back through those same mirrors/lenses/whatever to the sun as it was receiving.

Maybe? I’ll have to think about this some more.

No, because the smaller patch would have to radiate more energy per unit of area. You can only radiate more energy per unit of area if you at a higher temperature.

There certainly is an equilibrium point between the two objects, but in the given example (using lenses to focus light), the smaller object must have a higher temperature in order to reach equilibrium.

Put your object inside a greenhouse. Wouldn’t the object absorb the incoming energy as well as retaining the infrared heat that is trapped within the greenhouse? The object would heat up faster than heat could be radiated out from the greenhouse glass, no?

If the glass can trap heat, then the glass itself can also radiate.

And the Sun’s rays are not parallel: They vary by as much as half a degree.