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Second law of Thermodynamics - does this violate it?

  1. Apr 7, 2009 #1
    As I understand it, the second law means that energy always tends to get delocalized. A blunt way to put it is that you can't make energy flow from a cold source to a hot one without doing work ( ie localizing a point by delocalizing another ) But here is an idea that struck me

    Suppose you have a hot body and a cold body. both sepereated by a long tube. Trough this tube , radiation goes back and forth between the two. Suppose now that we place a refractive medium between them ( glass or anything I don't know ) Since the hot body and cold body radiate different frequencies of light, the get refracted through this medium at different angles, and now we adjust it so that the radiation from the cold body reaches the hot one but the radiation from the hot body is refracted in such a way that it never meets the cold body.

    So haven't we successfully made heat travel from a cold body to a hot one without doing any external work?
     
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  3. Apr 7, 2009 #2

    Andy Resnick

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    Forget the tube and refraction- simply place an optical filter at the interface between the two objects- say two cavities held at different temperatures. Now that the geometry is simplified, can you see why it won't work?

    If not, here's a hint- look at the blackbody curves for two different temperatures.
     
  4. Apr 7, 2009 #3
    Doesn't radiation from the cold body go to the hot body anyway? Just to a very much lower extent. So the net transfer of energy goes from hot to cold, although some goes into the cold-> hot direction.
     
  5. Apr 7, 2009 #4
    Calculate the work done by the glass rod in refracting the radiations. I think there is a small energy loss for the radiation due to the refraction. So, some work is done there.

    But the second law of thermodynamics limits that energy cannot be transferred from a cold body to a hot body without doing some work.

    Hence, the transfer of radiations by refraction methods don't violate the Second Law of Thermodynamics, I suppose.


    If this is wrong, please kindly explain...:approve:
     
  6. Apr 7, 2009 #5
    That's not exactly true. They have different peak frequencies, but the hot one emits more energy than the cold one does even at the cold one's peak frequency. So no matter what the angle of the refractor, more energy will flow from the hot one to the cold one than in the other direction.
     
  7. Apr 8, 2009 #6
    Well i considered an optic fibre, but that would mean radiation from the hot body can reach the cold one, that defeats my purpose. My knowledge of black body radiation is rather limited, can you explain? - I'm a first year degree student
     
  8. Apr 8, 2009 #7
    Hmm ok i get what you mean. But is a hot body that does not emit the frequency of the cold one inconceivable ?
     
  9. Apr 8, 2009 #8
    Well I think that is wrong. The law states that heat cannot flow from a cold region to a hot one unaided by work, ie you have to do work on the cold region or substance. even if there is some energy loss at the refractor, there is still some energy transfer. But as the others have pointed out, the hot body is emitting in the frequency of the cold one too.
     
  10. Apr 8, 2009 #9
    I suppose an object could have an emissivity of zero at a particular frequency, but then it would also reflect all incoming radiation at that frequency (see Kirchhoff's law of thermal radiation) so there would still be no net transfer of energy from one body to a warmer one.
     
  11. Apr 9, 2009 #10
    Okay I see now. Thx for clearing that up :smile:
     
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