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Thermal radiation and copper

  1. Mar 15, 2007 #1
    I have a question. Assuming I have oil at 80 degrees in a copper calorimetry cup , will the heat loss by thermal radiation be very high and will it affect my values?

    I am thinking that it would not and that it would be negligible. But i do not know why. I heard somewhere that it would be high due to copper not having a dipole movement. Can someone please explain to me why?
     
  2. jcsd
  3. Mar 16, 2007 #2

    Mentz114

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    Gold Member

    Your own experience with hot drinks surely tells you that a liquid at 80C cools pretty quickly whatever its in.

    Copper is a very good conductor of heat and has a low heat capacity so it won't help much to keep the oil warm.

    The heat capacity of a substance depends on its internal degrees of freedom for vibration, that's where the atomic dipole movement comes in.
     
  4. Mar 16, 2007 #3

    Astronuc

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    Staff: Mentor

    See the discussion of radiation heat transport and the Stefan-Boltzmann Law
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

    Let Thot = 80°C or 353 K and Tcold = 25° or 298 K. One must use absolute temperature in the S-B law.
     
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