[Bored chemist]

Batroost, I think you might well find that Kirchoff's radiation law is not an odd thing to say. Feel free to research it.

Lightarrow's point is that if all you see is, say, green light, then you don't know what temperature the object is at. If you have light at 2 wavelengths, say red and blue, then you can asign a temperature to the object (in general the brighter the blue is compared to the red, the hotter the object). What I'm not sure about is how colour temperatures are calculated. I don't think it can be as simple as saying  "the object emits more light at 500nm than at any other wavelength. A black body at x Kelvin emits more light at 500nm than at any other wavelength so the object has a colour temperature of x Kelvin."
It would be ridiculous to try to asign colour temperatures to any sort of gas discharge/ fluorescent lamp this way so I guess it's matched by eye (in which case a green laser would be impossible to match to any black body radiation) or by some sort of curve fitting- the spectrum is closest to that of a black body at x Kelvin- again this wouldn't work for a monochromatic light source.

[another_someone]

Batroost, I think you might well find that Kirchoff's radiation law is not an odd thing to say. Feel free to research it.

I suspect that there is some truth in both sides of this.

In principle, I would read this as an extention of the idea in electronics that an antenna that is good for transmitting is equally good at receiving (and visa versa).

It assumes that the processes involved in transmission and absorption of light are indeed reversable.

One area where that would seem dubious is where one is talking about fluorescence, where the absorption spectra may be different from the emission spectra.  An extention of this is the famed 'greenhouse effect', where radiation emitted back out into space is of a different spectra from that being absorbed, and consequently an atmosphere that is inert to the incoming radiation can block outgoing radiation, thus allowing a buildup of temperature to a higher level than would be otherwise the case.

[Bored chemist]

The greenhouse effect is about the transmission of the atmosphere and the emission from the earth's surface. These are not the same object so the law doesn't apply.
it seems that the proof of the law has been debated at some length.
http://www.mzwtg.mwn.de/arbeitspapiere/Schirrmacher_2001_1.pdf

"It assumes that the processes involved in transmission and absorption of light are indeed reversable."

Fair enough, they are. a photon goes in and promotes an atom to an excited state or that excited state decays and gives the photon back. Of course, if the excited state changes then it is not strictly the same body so, for example, if it loses energy to thermal vibration, it no longer exists to have it's absorbtion or emission measured.

[lightarrow]

Just because it doesn't have a blackbody spectrum. I can find a specific wavelength, let's say 550 nm, in the spectrum of a blackbody of any temperature. I can take a tungsten block at 20°C or at 3000°C, put a very good colour filter in front of it, which let pass only 550 nm light, and how could we establish this light comes from a 20°C or 3000°C block of metal? Certainly at 20°C the intensity of radiation at 550 is less, but this could also be ascribed to source's distance and surphace properties.
It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature.

This is certainly true, but I am not sure how it relates to the argument.

That's very strange, because it was you to make the original question about a laser beam's temperature, to which I have answered!

We say that the surface of the Sun is approx 6000K, because that is the temperature of the light emanating from the surface.  We know that the temperature beneath the surface can reach millions of K, but that is not the radiation we receive, so we know that the temperature on the surface is not as hot an the depths of the Sun are.
Yes, you can put a filter in-front of an object, and that will cause the system as a whole to cool on its exterior (beyond that filter), but it does not prevent the interior of the system from being hotter.
The question is, should not the radiation emitted from a 6000K body should be able to induce a temperature in a black body that absorbs that radiation that is equal to the colour temperature.  OK, the real world is not composed of ideal black bodies, but it should at least create a benchmark.

Actually I haven't understood what you mean; however, about the max temperature it's possible to achieve concentrating sun's light with lenses or mirrors, I would agree with you (indeed I had never said you were wrong about it); I think we cannot change the sun's spectrum shape (and so, the radiation's temperature) in that way. Not sure, however. Interesting question.

[Foxman]

On a sunny day, i can burn paper with my magnifying glasses. If i had a really good glass, could i burn through steel? or maybe the earths core!!!

What a bright idea! " Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. "Archimedes once said. But does the long lever exist? In the same way, i am afraid that we can not find such a magnifying glass big enough to burn the earth. However, many children often burn ants with magnifier in sunny days.  []

[syhprum]

The power available depends on the area of the lens and temperature depends on the inverse of the f number, the problem of obtaining a sufficiently small f number can be overcome by using a two stage focusing system.
The melting of steel has been demonstrated both by this method and by mirrors.
References to follow when I hunt them down.
Whether or not 6000°K can be exceeded is an open question I would like to see comments.

http://gizmodo.com/5069043/solar-furnace-melts-steel-our-minds

[yor_on]

A

Why would a monochromatic LASER not have a colour temperature (it may not have a black body spectrum, but the light still has an energy associated with its frequency, and that energy equates to a temperature)?

Just because it doesn't have a blackbody spectrum. I can find a specific wavelength, let's say 550 nm, in the spectrum of a blackbody of any temperature. I can take a tungsten block at 20°C or at 3000°C, put a very good colour filter in front of it, which let pass only 550 nm light, and how could we establish this light comes from a 20°C or 3000°C block of metal? Certainly at 20°C the intensity of radiation at 550 is less, but this could also be ascribed to source's distance and surphace properties.
It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature.

Lightarrow?
If I get it right you need a whole spectrum to make a decided temperature.

But if it comes to a lasers light, how would you go about to define a temperature for it, isn't it the wavelength/frequency you would use then?

They have an energy, they must have a temperature?
What am I missing?
==

Or should I ask about the heat instead?

"Temperature is a number that is related to the average kinetic energy of the molecules of a substance. If temperature is measured in Kelvin degrees, then this number is directly proportional to the average kinetic energy of the molecules.

Heat is a measurement of the total energy in a substance. That total energy is made up of not only of the kinetic energies of the molecules of the substance, but total energy is also made up of the potential energies of the molecules."

[lightarrow]

A

Why would a monochromatic LASER not have a colour temperature (it may not have a black body spectrum, but the light still has an energy associated with its frequency, and that energy equates to a temperature)?

Just because it doesn't have a blackbody spectrum. I can find a specific wavelength, let's say 550 nm, in the spectrum of a blackbody of any temperature. I can take a tungsten block at 20°C or at 3000°C, put a very good colour filter in front of it, which let pass only 550 nm light, and how could we establish this light comes from a 20°C or 3000°C block of metal? Certainly at 20°C the intensity of radiation at 550 is less, but this could also be ascribed to source's distance and surphace properties.
It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature.

Lightarrow?
If I get it right you need a whole spectrum to make a decided temperature.

Not only this, you have to be sure that your spectrum is part of a blackbody spectrum.

But if it comes to a lasers light, how would you go about to define a temperature for it, isn't it the wavelength/frequency you would use then?

Absolutely not. As I wrote (3 years ago!) in the post you quoted, you could have two equal light beams, the same intensity, the same wavelenght and coherence, everything the same, but one coming from your laser and the other coming (after adequate filtering equipments) from a big bulb lamp. How do you establish the temperature?

They have an energy, they must have a temperature?
What am I missing?

To define the temperature of an EM radiation you must have a photon gas with a blackbody spectrum. Energy and temperature are two completely different concepts.

Or should I ask about the heat instead?

"Temperature is a number that is related to the average kinetic energy of the molecules of a substance. If temperature is measured in Kelvin degrees, then this number is directly proportional to the average kinetic energy of the molecules.

Heat is a measurement of the total energy in a substance.

No, heat is a way to transfer energy between two bodies: through a difference in their temperature.

That total energy is made up of not only of the kinetic energies of the molecules of the substance, but total energy is also made up of the potential energies of the molecules."

...at least. [] There is also another form of energy: mass of particles.

[yor_on]

Thanks LA, one mystery less

[Geezer]

Everyone: Thank you for your contributions on this topic. I think it's clear that solar energy can be concentrated sufficiently to melt steel, and a lot of other things too. In the presence of oxygen, steel will burn at those temperatures.

I think this horse has been officially flogged and it's time to move on. Does anyone have violent objections to locking or abandoning this thread?

Of course, I think there are some very important questions and concepts associated with the original question, but it might be more productive if we were to initiate new topics that focus on those specific points.

Wadyafink?

[lightarrow]

...
The melting of steel has been demonstrated both by this method and by mirrors
....
http://gizmodo.com/5069043/solar-furnace-melts-steel-our-minds

In this video we can see a metal plate which is melting in a circular region and producing a hole. But why it doesn't become red, then orange, then yellow, then white? Is it steel or tin.... []
(Then the image change, then it comes again on the melted plate with the hole, and this time it's orange-hot...strange []).

[Bored chemist]

My guess is that the filters they must have used in front of the camera (to stop the incredible brightness washing out the image) must have distorted the colour. Afterwards they show a conventional image as the metal cools down and you see it's still red hot.

[lightarrow]

It could be, but it doesn't convince me a lot. Why we don't see any smoke? If there were put filters for brightness, why we don't see the various colors (red, orange, yellow) in the different regions of the plate, even if the entire plate is perfectly visible?

[yor_on]

Are you thinking coherent (lasers) light versus incoherent (sun) light Lightarrow?
Could it be that the effect is so fast, the heat so instantaneous that the atoms around it doesn't have the time to react?

[Bored chemist]

The melting takes a few seconds so it can't be that.
It's  a good question. The molten metal doesn't seem to be glowing as it melts; and it should be.
I still think it's glowing, but you can't see it because of the glare of the multiplied Sun. We normally think of molten steel as bright but, in this context, it's quite dull compared to the incident light.
The same dark glass filters that protect the camera are hiding the glow.

[yor_on]

My link is so phreaking (G3) slow so i usually stay away from movies, but I guess i will need to see this one. Just guessing, I think you're right though Bored Chemist, If it takes seconds there is no possibility of the atoms not being able to react and then it seems as it must have something to do with the light caught in the camera.
==
Very cool effect and over quite a large area too, I was expecting it to be more like a laser, a needle sort of But I agree, I think it's the camera that gets 'overloaded' by the light even with the filter on, or possibly that it gets melted so quickly and over such an large area that the redness sort of gets hidden under it and . . Awh But now that I've seen it I got to admit that I don't know, you can retract the imagery and compare the plates as they melt and after and the plates color seems almost the same even though that the 'after' should be without any Sun-beam working?

[lightarrow]

How could we explain the absence of smokes?

[Geezer]

It heats up so rapidly there is little time for any oxidation. The steel is melting rather than burning. When you cut steel with oxygen, you are burning it in an exothermic reaction.

BTW - you can take a trip to France to see one in action (if it's still in operation.)

http://en.wikipedia.org/wiki/Solar_furnace

Edit: Come to think of it, the answer to the original question is, technically, no.

Unless you grind up the steel into small particles, it won't burn. A steel plate can only be burned in an atmosphere that is very rich in oxygen.

[Bored chemist]

How could we explain the absence of smokes?

Because the steel is only just melting,not burning.

[lightarrow]

How could we explain the absence of smokes?

Because the steel is only just melting,not burning.

There's no oxygen in the atmosphere there? [] At ~ 1400°C Iron, carbon, iron carbides do burns and generate smoke, IMHO.