What is the hottest you can make a focused beam of sunlight?

[kmarinas86]

Some have said it was 6000 degrees K.

I'm not so sure.

I know for sure that you can increase solar intensity arbitrarily by using lenses and mirrors.

What I don't get is why this doesn't increase temperature beyond a certain point.

Is it somehow related to the photoelectric effect? I guess, the quantum effects can explain the lack of temperature rise beyond that. But what happens to the excess of power density that does not lead to a temperature rise?

Once the object reaches a temperature of 6000 K, if additional power density is sent from the 6000 K spectrum, then it must be incapable of increasing the kinetic energy per degree of freedom per particle of the heated object. If we would see a rise in this kinetic energy, then it would mean that focused sunlight can be hotter than the light of the sun's surface, which would violate the 2nd law of thermodynamics.

Yet the light will be on that object. What choices do we have then? Well, we can:
* reflect the light, or
* let the light simply pass through, or
* force the material to dissipate the energy by contributing to a change of phase resisting increases in temperature (i.e. increase specific heat)

How does the material adapt to the spectrum of light imposed on it so that it can:
* reflect the light, or
* let the light simply pass through, or
* dissipate the energy by contributing to a change of phase resisting increases in temperature (i.e. increase specific heat)
...at a rate compensating for the power density whereat it exceeds the brightness of the source of question?

 

[Dale]

Once the surface you are irradiating reaches 6 kK it is in thermal equilibrium with the sun. If it gets any hotter it would heat the sun instead of vice versa.

So any additional power raises the temperature of both. Since the sun has a lot of thermal mass, it won't be detectable and the temperature will stay fixed at 6 kK.

 

[kmarinas86]

Once the surface you are irradiating reaches 6 kK it is in thermal equilibrium with the sun. If it gets any hotter it would heat the sun instead of vice versa.

What happens if the object is not a blackbody? What happens if it cannot emit the heat it had absorbed immediately? What happens if the heat is trapped there for an extended duration? How does such trapped radiation energy flux fail to increase the temperature?

Or is the object forced to take on a blackbody spectrum anyway when its temperature exceeds that of the spectrum of radiation forced upon it? If so, how does the imposed spectrum imprint anything so as to require that object to take on the characteristics of a blackbody?

So any additional power raises the temperature of both. Since the sun has a lot of thermal mass, it won't be detectable and the temperature will stay fixed at 6 kK.

If both objects would get hotter, what's heating both?

 

[Dale]

What happens if the object is not a blackbody? What happens if it cannot emit the heat it had absorbed immediately? What happens if the heat is trapped there for an extended duration? How does such trapped radiation energy flux fail to increase the temperature?

Or is the object forced to take on a blackbody spectrum anyway when its temperature exceeds that of the spectrum of radiation forced upon it? If so, how does the imposed spectrum imprint anything so as to require that object to take on the characteristics of blackbody?

I am not completely certain of this, but I do know that blackbodies are the best thermal absorbers as well as the best thermal emitters. If it is a graybody then the object will not absorb the heat as much, so I think it will heat to less than 6 kK.

I don't believe that the kind of delayed thermal response you described is possible, but I don't know for sure.

If both objects would get hotter, what's heating both?

Whatever the source of the additional power is. Presumably the thermonuclear reactions in the core of the sun.

 

[kmarinas86]

I don't believe that the kind of delayed thermal response you described is possible, but I don't know for sure.

What about a greenhouse?

 

[Dale]

AFAIK, a greenhouse doesn't delay its thermal response. It is simply a gray body which is transparent at visible wavelengths and opaque at infrared wavelengths.

 

[kmarinas86]

AFAIK, a greenhouse doesn't delay its thermal response. It is simply a gray body which is transparent at visible wavelengths and opaque at infrared wavelengths.

Well, it sure seems that the energy comes in as visible, and so it is not delayed going in, but then some converts to longer infrared, and the ones converted to infrared are delayed on the way out.

 

[Dale]

They aren't delayed, they are just absorbed.

 

[Ken G]

Well, it sure seems that the energy comes in as visible, and so it is not delayed going in, but then some converts to longer infrared, and the ones converted to infrared are delayed on the way out.

I don't think there is any way to arbitrarily increase the intensity of sunlight using lenses. There is something called the "brightness theorem", which basically says that all you can use a lens to do is make the Sun appear bigger to something looking at it through the lens. Once the Sun appears to fill all directions, you will be at 6000 K, and can go no hotter, nor will the intensity of the light ever appear greater than that.

The greenhouse effect does not change this. The reason sunlight is converted to infrared is expressly because the Earth is much cooler. If you used lenses in space, you could at best heat the Earth to 6000 K, and then the Earth would look like the Sun too-- it would emit in the visible, not the IR, and there would be no greenhouse effect.

It's true that the Sun is not a perfect blackbody at 6000 K, so there are some detailed ways you could take advantage of deviations in its spectrum, but I doubt you could do anything in practice that would be particularly useful or that would substantially modify this 6000 K limit, though I don't rule out something you could do in theory. Presumably you could pick whatever frequency the Sun has the highest brightness temperature, and lock down your emission and absorption to that frequency, and get a surface to that brightness temperature. Some chromospheric lines have brightness temperatures that are more like 20,000 K, so in principle you could get that kind of temperature, but I doubt it would really work in practice because the frequency band is too narrow.

 

[mrspeedybob]

Once the surface you are irradiating reaches 6 kK it is in thermal equilibrium with the sun. If it gets any hotter it would heat the sun instead of vice versa.

So any additional power raises the temperature of both. Since the sun has a lot of thermal mass, it won't be detectable and the temperature will stay fixed at 6 kK.

That does not make sense to me.

If I take all the radiation of 2 square meters of the suns surface and focus it onto 1 square meter of target the target should get hotter then the sun. Energy radiated by the 1 square meter would get diverged by the focusing device onto 2 square meters of the suns surface so in this case thermal equalibrium does not mean equal temperature.

 

[Matterwave]

You certainly CAN heat a piece of material to >6000K using the Sun's rays, but you need to DO WORK in order to do it. You can't SPONTANEOUSLY do it. That is the second law of thermodynamics.

If your scheme doesn't involve work done, then it's violating the second law.

So, if in your question, you simply mean that you do it via a lens of some sort, then 6000K would be correct. However, if you are allowed to do work to essentially pump all the solar radiation into a small area, then you can "focus" the beam to significantly higher temperatures (if that's what you mean by focus).

For example, I could take an extremely large solar panel, and then pump all of the electrical energy to heating a small sample of water. I can heat that sample to >6000K, but I need to do work in order to pump that energy.

 

[Andy Resnick]

DaleSpam is correct- the limiting temperature is 6000K. The final temperature you can heat an object to will depend on the ratio between power absorbed by the object and power radiated/conducted/transferred from the object.

One persistent myth involved here is that sunlight cannot be focused to an arbitrarily small point- the sun is an extended object, and so the focused 'spot' is not an actual point, or airy disc for that matter.

 

[Dale]

thermal equalibrium does not mean equal temperature.

Thermal equilibrium does mean equal temperature. It is the definition of temperature.

http://en.wikipedia.org/wiki/Temperature
"No heat will be exchanged between bodies of the same temperature; such bodies are said to be in 'thermal equilibrium'."

If I take all the radiation of 2 square meters of the suns surface and focus it onto 1 square meter of target the target should get hotter then the sun.

Every optical ray that you can trace which goes from the surface of the sun to the target also has an opposite optical ray which you can trace which goes from the target to the surface of the sun. When the target is hotter than the sun then along each ray there will be a net transfer of energy from the target to the sun. Integrate over all possible rays and you get heat transfer only from hot to cold, never the other way around. That is regardless of the details of your optical device.

 

[Astronuc]

What happens if the object is not a blackbody? What happens if it cannot emit the heat it had absorbed immediately? What happens if the heat is trapped there for an extended duration? How does such trapped radiation energy flux fail to increase the temperature?

As heat content increases within matter, the temperature increases, except during phase change, e.g., when solid becomes liquid and liquid becomes vapor. If more energy is input, then the gas becomes ionized. All the time, the hot object radiates heat.

Realize that the photosphere of the sun is about 5500-6000 K. The chromosphere is hotter ~ 6000K to ~20000 K, and the corona is even hotter, on the order of 1e6 K.
http://www.uni.edu/morgans/astro/course/Notes/section2/new5.html
http://solarscience.msfc.nasa.gov/chromos.shtml
http://solarscience.msfc.nasa.gov/corona.shtml

And the core is still hotter - ~ Central temperature: 1.571 x 10⁷ K - according to models (the core is unobservable).

So how can the cooler photosphere and chromosphere exist between much hotter core (solar interior) and corona?

It has to do with mass in addition to specific energy, and the flow of energy.

The corona has a very low density compared to the chromosphere, which has a lower density than the photosphere.

Matter heated by concentrated solar systems has apparently attained temperatures ~3500 C. The problem then is that most solids melt and become liquid or vapor, which would have to be contained either by another solid (like quartz or graphite), or a magnetic field. Magetic fields can levitate liquid metals up to a point. Even hotter, the liquid becomes vapor/gas, and magetic fields do not confine neutral gas - only plasmas. But plasmas are transparent to visible light.

 

[reasonableman]

A few people have mentioned brightness theorem. This states that it is not possible to illuminate something with a higher intensity than the source - if the optical indexes at the source and image are equal.

If this is considered perhaps a higher temperature can be attained? I believe achieving this would require the space between the Sun and the Earth to be filled with a material with a high refractive index but that's just an engineering problem!

 

[AlephZero]

Matter heated by concentrated solar systems has apparently attained temperatures ~3500 C.

If anybody wants to test some of the propositions here by experiment, you can reproduce the same effect quite easily using a halogen light such as a car headlight bulb. These bulbs have a reflector behind the filament so they produce a narrow parallel light beam without any additional optics.

You only need a very simple setup to heat small objects to a temperature comparable with the bulb filament (e.g. 1000C) in a few minutes. Any reports of heating something significantly above the filament temperature would be "interesting".

 

[haruspex]

But plasmas are transparent to visible light.

Did you mean that? I would have thought an ionised gas would absorb at pretty much any frequency.

 

[mfb]

A few people have mentioned brightness theorem. This states that it is not possible to illuminate something with a higher intensity than the source - if the optical indexes at the source and image are equal.

If this is considered perhaps a higher temperature can be attained? I believe achieving this would require the space between the Sun and the Earth to be filled with a material with a high refractive index but that's just an engineering problem!

I think you can get a higher energy density of the light, but the lower velocity gives you the same intensity again?

So how can the cooler photosphere and chromosphere exist between much hotter core (solar interior) and corona?

By doing work. In case of the sun, with electromagnetic fields and jets. I know that the details are still unclear, but the principle is easy to understand.

But plasmas are transparent to visible light.

Plasmas have a lot of free charges, which can absorb photons.The surface of the sun is the temperature limit if heating is done by light only, as shown here in the thread.

 

[DrDu]

There is one point I never quite figured out completely: Assume you have two spherical blackbodies of different radius in the two foci of an (ideal) ellipsoidal reflector. Then all the light emitted by the larger one is absorbed by the smaller one and vice versa. Nevertheless the power of the emitted light is proportional to the surface, so there would be a constant flux of energy from the larger one to the smaller one. Hence the system will actually never reach equilibrium.

 

[nasu]

There is one point I never quite figured out completely: Assume you have two spherical blackbodies of different radius in the two foci of an (ideal) ellipsoidal reflector. Then all the light emitted by the larger one is absorbed by the smaller one and vice versa. Nevertheless the power of the emitted light is proportional to the surface, so there would be a constant flux of energy from the larger one to the smaller one. Hence the system will actually never reach equilibrium.

Unless you have point sources, none of the emitting points will be in the focus.
The larger body will be more "defocused" than the smaller one.
It may be that some of the radiation emitted by one body does not reach the second body?

 

[K^2]

I am not completely certain of this, but I do know that blackbodies are the best thermal absorbers as well as the best thermal emitters. If it is a graybody then the object will not absorb the heat as much, so I think it will heat to less than 6 kK.

Gray body also radiates back less. The limiting temperature is still 6k.

 

[DrDu]

Unless you have point sources, none of the emitting points will be in the focus.
The larger body will be more "defocused" than the smaller one.
It may be that some of the radiation emitted by one body does not reach the second body?

Yes, meanwhile I thought further about it and came to the same conclusion. Specifically I was considering the following situation: Two spherical blackbodies(r1 and r2) in the two foci of a very large reflecting ellipsoid which is nearly spherical (radius R). Then only light emitted into a cone of opening angle r2/R or r1/R will reach the sphere r2 or r1, respectively after 1 reflection. This at least makes plausible, that there will be no net energy exchange when the two bodies are at the same temperature.

 

[kmarinas86]

Unless you have point sources, none of the emitting points will be in the focus.
The larger body will be more "defocused" than the smaller one.
It may be that some of the radiation emitted by one body does not reach the second body?

What if instead you had two spheres with effectively the same radius, but one has more area than the other in the form of hairs and/or bumps?

 

[Dale]

Gray body also radiates back less. The limiting temperature is still 6k.

Thanks for the correction, that makes sense.

 

[Antiphon]

This is actually a very old problem in classical thermodynamics. It might be due to Maxwell, I can't be sure.

The argument is that using a lens you can get work out of a thermal equilibrium and make a PMM.

In an equilibrated universe (say filled with visible light) you have a lens. The focal point of the lens concentrates the energy and you'd locate a heat engine there, extracting work from equlibrium. It doesn't work for obvious reasons.

 

[DrDu]

Unless you have point sources, none of the emitting points will be in the focus.
The larger body will be more "defocused" than the smaller one.
It may be that some of the radiation emitted by one body does not reach the second body?

I repeated my crude calculations and corrected some error. So now I find that it is possible in principle to focus all of the light emitted by the smaller of the two spheres onto the larger one (which will absorb it completely if it is a black body), but only the fraction r1^2/r2^2 of the radiation emitted by the larger body can be focussed onto the smaller one. With Stefans law this ratio can be seen to be the ratio of the powers of the radiation emitted by the two spheres. Hence it is clear that they will stay in equilibrium at that temperature.
In fact I was considering only the artificial example of the two spheres lying inside each other and the reflector to be also spherical. I think the situation would change but little for a slightly elliptical deformation of the sphere which produces two foci.
To calculate the general elliptic case one should start from blackbodies whose shape is limited by an ellipsoid and a hyperbola. From the discussion of trajectories in the ellipsis, http://mathworld.wolfram.com/Billiards.html
it should be clear that there are trajectories starting from the larger of the two blackbodies which never reach the smaller one.

 

[mfb]

What if instead you had two spheres with effectively the same radius, but one has more area than the other in the form of hairs and/or bumps?

Radiation from the bumps can hit the same body again.

It is easier to consider these problems from a different perspective: Assume that the whole volume is filled with isotropic radiation corresponding to whatever temperature. In that case, every surface with this temperature is in equilibrium with the radiation, and nothing happens. Geometry does not matter.

 

[kmarinas86]

Radiation from the bumps can hit the same body again.

But not 100% of it. So this statement doesn't really answer my question.

It is easier to consider these problems from a different perspective: Assume that the whole volume is filled with isotropic radiation corresponding to whatever temperature.

That's not a different perspective on the same problem. That's a different perspective on a completely different problem, though, I'm sure that you have your own reason why you think that this isn't irrelevant.

 

[russ_watters]

But not 100% of it. So this statement doesn't really answer my question.

No. All that matters is the cross section the objects are presenting each other. It is easiest to see when you focus on the uniform shape instead of the non-uniform shape: what it "sees" is just a cross section of the emitting object and added surface irregularity does not open any new paths for the light to take.

 

[kmarinas86]

No. All that matters is the cross section the objects are presenting each other. It is easiest to see when you focus on the uniform shape instead of the non-uniform shape: what it "sees" is just a cross section of the emitting object and added surface irregularity does not open any new paths for the light to take.

This is just like the previous response. Why is it always along the lines of...

"It is easier to consider these problems from a different perspective"
"It is easiest to see when you focus on the uniform shape instead"

...followed by a statement of the obvious, without any evidence to show equivalence?

 

[kmarinas86]

Let's just say that the paths that the light travels in the ellipsoidal sphere is reversible.

Therefore, any path taken from the sphere at one foci is the same exact path that can be taken by the other.

The effective radii of both spheres are equivalent. So "focus" of the light field at the foci is not the issue.

The areas are not the same for both spheres. One is rough, and the other is smooth.

A ray from the rough to the smooth sphere travels the opposite path as a ray from the smooth one.

Are these rays are going to be equal in power? They clearly will not be equal in number, because one surface is larger than the other.

Let's look at a similar example:

Let's have a reflective cone with a tip cut off. The top and the bottom parts of the cone are circular plates of the same temperature. The larger area emits more power, and the smaller area emits less power. Once this happens, the smaller area gets hotter, and the larger area gets colder. However, then the smaller area gets hot enough at some point to overcome this, and the flow reverses. We then have the thermodynamic equivalent of a pendulum - and, I ask you all, why shouldn't there be a thermodynamic analog of a pendulum?

I mean, if we look within a thermodynamic fluid, we see the same thing happening. Slower particles do not always get faster, sometimes they slow down, and faster particles can go faster, instead of slowing down. It does this by reducing the velocity difference between things that move slower (such when a sailboat moves over the ocean faster than the wind, while both the wind and the ocean are slower than the sailboat, compared to the common frame of reference - the Earth).

Questions about DDWFTTW
https://www.physicsforums.com/showthread.php?t=562993&page=11

Shouldn't a thermodynamic analogue of DDWFTTW be possible?

 

[russ_watters]

This is just like the previous response. Why is it always along the lines of...

"It is easier to consider these problems from a different perspective"
"It is easiest to see when you focus on the uniform shape instead"

...followed by a statement of the obvious, without any evidence to show equivalence?

What evidence do you want here? It isn't like we're trying to trick you or anything. Are you really trying to learn or do you just want to argue? It is often easiest to learn by simplifying rather than over-complicating. Over-complicating causes the errors you are making!

 

[kmarinas86]

What's wrong with the existence of a thermodynamic pendulum where objects of similar temperatures oscillate over a common value?

Perhaps all these examples alleging cold heating hot, or of objects of equal temperatures becoming unequal are missing the big picture. Maybe all of these scenarios are seeing only one-half of the picture. The other half is the other object getting hot enough to send power faster the other way.

How "fast" would a thermodynamic pendulum swing? Maybe a "thermodynamic pendulum" swings so fast that it would explain not being able to observe a hot plate get heated by a cold plate. Could such a pendulum be made to swing more slowly?

 

[russ_watters]

What's wrong with the existence of a thermodynamic pendulum where objects of similar temperatures oscillate over a common value?

This is exactly the question you already asked and had answered: It requires a delay between absorption and manifestation of the heat.

How "fast" would a thermodynamic pendulum swing? Maybe a "thermodynamic pendulum" swings so fast that it would explain not being able to observe a hot plate get heated by a cold plate. Could such a pendulum be made to swing more slowly?

Please respect the forum rules regarding personal theories.

Regarding shape issues, you should read into the concept of "shape factor" or "view factor": http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node137.html

 

[Dale]

A ray from the rough to the smooth sphere travels the opposite path as a ray from the smooth one.

Which is the general geometric reason that the net energy flux along the ray will never be from cold to hot.

Are these rays are going to be equal in power? They clearly will not be equal in number, because one surface is larger than the other.

No, this is incorrect. They must be equal in number because they are the same rays, just opposite directions. It is simply not possible to draw a ray that goes from large to small that does not also go from small to large.

Let's have a reflective cone with a tip cut off. The top and the bottom parts of the cone are circular plates of the same temperature. The larger area emits more power, and the smaller area emits less power. Once this happens, the smaller area gets hotter, and the larger area gets colder.

That assumes that all of the power emitted from the larger goes to the smaller. Tracing a few rays should convince you that is not the case. Some of the rays leaving the larger trace back to the larger.

We then have the thermodynamic equivalent of a pendulum - and, I ask you all, why shouldn't there be a thermodynamic analog of a pendulum?

An oscillator occurs whenever you have a second derivative of a quantity which is proportional to the quantity with a negative constant of proportionality. In radiative heat transfer I don't know of any second derivatives of temperature.