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How does a black body emit as much energy as it absorbs?

  1. May 24, 2015 #1
    1. The problem statement, all variables and given/known data
    A black body absorbs all incident electromagnetic radiation, including visible light which has wavelengths from 380nm to 750nm. IR radiation has wavelengths that are so long they are measured in microns. That suggests that visible light has a higher frequency than IR, and, higher energy. So when a black body absorbs visible light and emits IR, how does it achieve equilibrium?

    2. Relevant equations
    Ein = Eout (Energy in = Energy out)
    E = hv (E = Energy; h = 6.63E-34 J-s; v = frequency in hertz)
    v = C/l (C = 3.00E8 m/s; l = wavelength in meters)

    3. The attempt at a solution
    I'm picturing in my mind the same amount of photons going in as going out, but this would imply more energy going in than going out. So what's up with the black body?
     
  2. jcsd
  3. May 24, 2015 #2

    marcus

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    That's the problem. Far more low-energy photons are emitted, so the total in and out energy balance.
     
  4. May 24, 2015 #3

    vela

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    Regardless of the number of photons involved, suppose the black body is absorbing more energy than it's radiating. It's not in equilibrium with its surroundings. What happens to the temperature of the black body?
     
  5. May 24, 2015 #4
    Thanks.
     
  6. May 24, 2015 #5
    I heard it goes up. Thanks for the input.
     
  7. May 25, 2015 #6

    vela

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    The way the question is worded implies that the blackbody isn't in equilibrium with its surroundings initially (otherwise it wouldn't have to do anything to achieve equilibrium), so it's not safe to assume the number of IR photons emitted is enough to maintain the energy balance. If you haven't already, you should consider what happens when the temperature of the blackbody goes up.
     
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