Energy lost through photon emission

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Hey guys, new here. Here is my first question for the forums: Let's say that I have a controlled environment for an experiment whereby I want to heat, say, one cubic centimeter of steel until it is white hot. Assuming that I have perfect containment set up for it such that no energy can be transferred out of the material after it has been heat-charged, thus not allowing it cool, how long would it emit photons before going dark? Thanks.
 

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  • #2
berkeman
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Welcome to the PF. :smile:
Hey guys, new here. Here is my first question for the forums: Let's say that I have a controlled environment for an experiment whereby I want to heat, say, one cubic centimeter of steel until it is white hot. Assuming that I have perfect containment set up for it such that no energy can be transferred out of the material after it has been heat-charged, thus not allowing it cool, how long would it emit photons before going dark? Thanks.
So photon emission does not carry any energy away?

And are you familiar with the topic of block body emission yet?
 
  • #3
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Would Stefan–Boltzmann law suffice? As for how long, It will always emit photons. As long as T > 0K.
 
  • #4
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The ideal setup would be to only allow photon emission to carry the energy away, with photons being the only source of energy transferred out of the material. I am not familiar with block body emission.
 
  • #5
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such that no energy can be transferred out of the material
Do you mean “no energy transferred out of the material except by thermal radiation”?
 
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  • #6
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So, for as long as T>0K, it will emit. Any ideas on what that might be for visible light, or would it be the same? Thanks again.
 
  • #7
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Hi Dale. Assume that thermal radiation is also contained, and I just want to measure the loss of energy through photon emission. Or can I not have one without the other? Thanks.
 
  • #8
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Hi Dale. Assume that thermal radiation is also contained, and I just want to measure the loss of energy through photon emission. Or can I not have one without the other? Thanks.
Thermal radiation is just a fancy way to say photon emission in this case.
 
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  • #9
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Hi Dale. Assume that thermal radiation is also contained, and I just want to measure the loss of energy through photon emission. Or can I not have one without the other? Thanks.
Radiation is photon emission (or emission of other particles, but just photons in this case). Thermal radiation is photon emission that is based on an object’s temperature. It is also called black body radiation for historical and technical reasons.
 
  • #10
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Cool beans, BearY. It's only a hypothetical query at this point, as my focus is on researching various materials and their insulating and conductive properties. I like to invent stuff. But I like to research to see what can be done and what can't, and to push the limits as far as I can. :)
 
  • #11
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Dale, if I wanted to separate the thermal mechanism from the photonic, whereby the heat was not transferred but the photons were, could that be done? Or would I have to go a different route other than heating a material? Thanks.
 
  • #12
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Ok so we heat 1 cubic cm of steel to white hot, no conduction or convection so the only energy transfer is due to emission of thermal photons. And you want to know how long it will continue to glow in the visible spectrum, correct?

What temperature is the environment?
 
  • #13
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My background is pretty much computer aided drafting. I took intermediate physics way back in college, but forgot the stuff I never use lol, so thanks for putting up with the questions. :)

Ideally, the molten steel would be enclosed, say, in a soda lime glass container as an inner shell, separated magnetically from an outer shell, and the space between the shells is a vacuum to prevent as much transfer as possible to where you would be able to touch it and it would feel cool. The surrounding environment outside of that would be regulated at 72dF, my favorite temperature. How long would it continue to glow? Thanks.
 
  • #14
gneill
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If something is actually "white hot", would that not correspond to the temperature of the Sun's surface? About 5500 C? Steel melts at something like 1370 C. Iron boils at 2870 C. Methinks there will be a containment problem for more than heat conduction.
 
  • #16
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Containment problem for sure :)
 
  • #17
russ_watters
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We can do a specific scenario like @Dale proposes, but FYI, for objects that hot the vast majority of their heat loss is due to thermal radiation anyway. So we don't need to pretend we can insulate it from conduction and convection (though we could say it's in space...).
 
  • #19
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Hi Russ. Ok, so it would continue emitting photons until its thermal energy is dissipated. How long would it continue to glow in the visible spectrum before going dark? Thanks.
 
  • #20
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Hopping off here. Have to get up early. Thanks very much for all the replies and help guys, and keep it coming. My original thought was like a desktop paperweight or fancy gift item that is transparent with a white hot drop of steel at its center. Later on. :)
 
  • #21
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The surrounding environment outside of that would be regulated at 72dF, my favorite temperature.
So we could use the Stefan Boltzmann law to calculate the rate of heat loss from 1200 C to 500 C in a 72 C thermal bath. We would need to look up the specific heat capacity of steel and its emissivity. But other than that it would be a fairly straightforward calculation. Steel has a high thermal conductivity so I think it would be ok to approximate it as being the same temperature throughout the sphere.
 
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  • #22
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Ok. I'll try to be more specific lol. So assume I now have a space-grade radiant barrier material, with the air evacuated from the two shells, with the inner shell magnetically suspended, such that neither convection nor conduction transfer can occur. Then assume that the radiant barrier is semitransparent, allowing the emission of photons only within the wavelengths of the visible spectrum, and reflecting the thermal radiation back into the container at better than 99%, with less than 1% absorption into the material itself. Would it be possible to set up an experiment for this, so that only energy losses via visible spectrum emissions can be measured? Could I then attempt to make my pretty self-lit white-hot iron core desktop paperweight? Thanks again. :)

Edit: Any modified form of Alon can be ruled out because it is very new and therefore not yet economically feasible to work with.
 
  • #23
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Yes, you certainly could still do this but now you would need to use Planck’s law instead of the Stefan Boltzmann law.
 
  • #24
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I'm afraid I never got that far in physics. I ended up having to work instead lol. My knowledge is limited to intermediate physics at best. I never got as far as Planck's law and won't pretend to understand it. I know my basic newtonian stuff: forces, vectors, diagrams, that sort of thing. However, fundamental concepts about how things COULD work come fairly easily to me as I'm very logical about things. And so as an inventive person, quite often my ideas will take me beyond what I know mathematically, but at the same time make sense to me. This is why I'm on here pretty much...to see what will work and what won't.
 
  • #25
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Planck’s law describes the amount of energy radiated at each wavelength for a given temperature. So you would want to integrate that across the visible spectrum. That would give you your ideal semi-transparent material.

Then that is the rate of energy loss as a function of temperature. So you would take that and the specific heat and area and mass and so forth and you would get a rate of temperature loss.
 

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