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Simple question on Heat transmission.

  1. Oct 8, 2004 #1
    If I have a beaker and it is being lined with silver. I use this in an experiment and it is essential that as little heat is lost to the surroundings as possible. How effective would this be?

    Can I use the logic that since light passes through the transparent beaker, infra-red radiation should behave likewise? and I am I right to say that a significant amount of heat is being lost as IR radiation due to a transparent beaker? Thanks alot.
     
  2. jcsd
  3. Oct 8, 2004 #2
    If it helps, I know that glass is transparent from 400 nm to 2500 nm. This encompasses VIS and near IR, but the logic you suggest doesn't work in general. I do believe that a significant amount of heat is lost.

    If you enclose your beaker in aluminum foil, you will reflect the heat back into it, and this seems to be what you're looking for.
     
  4. Oct 8, 2004 #3
    Thanks, is there a way to calculate how much IR radiation would be kept in the beaker if I put use a beaker with silver sides? Will it be significant, the heat being kept inside the beaker by radiation, if we have say, a temperature of 1000 degrees?

    Also, compared to conduction, is there significant heat lost through radiation at a 1000 degrees ?
     
    Last edited: Oct 8, 2004
  5. Oct 8, 2004 #4
    Well, 1000 degrees will melt most metals. And it would probably be much easier to measure or find measured reflexion spectra than to calculate them. I'd have to check what's involved in the calculation.
     
  6. Oct 8, 2004 #5

    Integral

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    Houston, we have a problem!

    The melting point of Aluminum is 660C Since you do not specify the units of your temperature we can only guess, but 1000F = 1832C so you will not be able you use Al foil for an insulator at those temperatures. That is pretty hot! you will need a lot of insulation to maintain that temperature, irregardless of which scale you are using. Radiation is only one mechanism of heat transfer, both conduction and convection must be considered, and a simply layer of foil will not help those much. Especially at such high temperatures.
     
  7. Oct 9, 2004 #6

    Tide

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    Integral: 1000 deg F = 537.8 deg C

    You meant 1000 deg C = 1832 C
     
  8. Oct 9, 2004 #7

    Integral

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    :biggrin:

    1000C = 1832 F

    Yep, we'll get it right sooner or later!

    To much work, to little sleep!
     
  9. Oct 9, 2004 #8
    Oh, actually, not really aluminium foil that is gonna be used, just a beaker with silvered sides, something like a thermoflask. Definitely other methods of heat transmission would be taken into account such as conduction, convection, but is heat lost through radiation significant at a thousand degrees....celsius... :smile: in case you guys are confused.
     
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