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Apparent Violation of 2nd Law

  1. May 14, 2010 #1
    1. Keeping the temperature constant, does the rate of emission from a black body depend solely on its surface area?

    2. If it does, let's consider the following: we have two black bodies in our system, A and B lying close to each other. Both are at same temperature but the surface area of A is twice that of B. Now, the rate of emission from A should also be twice that of B. This implies that over a certain time, A loses more energy than it gains. So its temperature should drop. But doesn't that violate the 2nd law of Thermodynamics?

    (it's not really a HW question, just a thought that popped up in my mind)
     
  2. jcsd
  3. May 14, 2010 #2

    Borek

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    Staff: Mentor

    If the system in question consist just of those two bodies, and if the energy they emit is lost in space - there is no problem. They don't have to cool at the same speed.

    If the system in question consist just of those two bodies, and if the energy they emit is reflected back - larger body absorbs the energy in speed proportional to its surface, that is, twice faster, so its temperature doesn't change. There is no problem.
     
  4. May 14, 2010 #3
    Thanks for replying.

    For the two possibilities that you mentioned, indeed there wouldn't be any problem.

    But what if we somehow ensure that all of A's radiation falls on B (without any of it reflecting back to A), and vice versa? Then the temp of B would increase and that of A would decrease.
     
  5. May 14, 2010 #4

    Borek

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    If we "somehow ensure" you are right. But 2nd law tells that it is impossible. It may sound like a circular reasoning, but it is not - it just just tells you that you have to actually design and implement system that "somehow ensures" before you can state 2nd law is violated.
     
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