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Thread: Violation of 2. Law of Thermodynamics?

  1. #1 Violation of 2. Law of Thermodynamics? 
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    I made up a thought experiment recently, in which I can not find the flaw (if any )

    The idea is that you have two objects, a cold one with temperature Tc and a hot one with temp. Th. (The objects are as well denoted as Tc and Th, respectively)
    You place them in a special designed box, shaped like a peanut, so that they do not touch each other. The inside walls of the box is covered with insulating material, (reflecting infrared radiation), so that in theory the system is isolated.

    In between the objects you place a semi-reflective mirror that lets through infrared radiation from Tc to Th, and reflects as much radiation as possible in the direction of Th.

    I am aware of photons interacting with the mirror as they are reflected, and that the mirror is emitting heat radiation as well, but if the temperatures Tc and Th are close, I suggest that there will be a spontaneous net heat flow from a cold reservoir to a hot reservoir.

    This conflicts with the 2. Law of Thermodynamics - Where am I wrong?
    See rather simplified illustration below.



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  3. #2 Re: Violation of 2. Law of Thermodynamics? 
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    Quote Originally Posted by Andro
    I made up a thought experiment recently, in which I can not find the flaw (if any )

    The idea is that you have two objects, a cold one with temperature Tc and a hot one with temp. Th. (The objects are as well denoted as Tc and Th, respectively)
    You place them in a special designed box, shaped like a peanut, so that they do not touch each other. The inside walls of the box is covered with insulating material, (reflecting infrared radiation), so that in theory the system is isolated.

    In between the objects you place a semi-reflective mirror that lets through infrared radiation from Tc to Th, and reflects as much radiation as possible in the direction of Th.

    I am aware of photons interacting with the mirror as they are reflected, and that the mirror is emitting heat radiation as well, but if the temperatures Tc and Th are close, I suggest that there will be a spontaneous net heat flow from a cold reservoir to a hot reservoir.

    This conflicts with the 2. Law of Thermodynamics - Where am I wrong?
    See rather simplified illustration below.

    A semi-reflective mirror is not an optical diode. What you have constructed is an argument why no such optical diode is possible.


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  4. #3  
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    What is a semi-reflective mirror, then, if it is not what I have drawn?
    And what is the difference?
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  5. #4  
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    Quote Originally Posted by Andro
    What is a semi-reflective mirror, then, if it is not what I have drawn?
    And what is the difference?
    What you have drawn is a cartoon. Cartoons violate physics regularly.

    If you don't know what semi-reflective mirror is, then this is an excellent opportunity for you to do some research and learn for yourself. Hint: It cannot permit a violation of the second law.
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  6. #5  
    Forum Radioactive Isotope MagiMaster's Avatar
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    One hint, consider why the room behind those mirrors, where the police are, is always so dark.
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  7. #6  
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    Why not compress the cold article to a point mass then show it to the hot one, that will also transfer heat from the 'cold' to the 'hot' body. You have added a third object the the law of heat transfer thus it is no longer the third law. you are allowing heat to travel in only one direction thus you would need to re-write the law to become

    "In any two body differential temperature system where heat from the higher body is prevented from reaching the lower temperature body, heat transfer can only occur from the lower to the higher temperature body"

    What you did was the equivalent of altering one side of an equation without altering the other.
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  8. #7  
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    So,
    If I try to make this setup, factors like the optical diode absorbing and emitting heat will make the system obey the 2. law?

    Is it then theoretically possible to make an optical diode exactly so effective that it will not violate thermodynamics?

    And, lastly, would it be a violation if the diode was effective enough to keep the system in equilibirum?
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  9. #8  
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    Quote Originally Posted by Andro
    So,
    If I try to make this setup, factors like the optical diode absorbing and emitting heat will make the system obey the 2. law?

    Is it then theoretically possible to make an optical diode exactly so effective that it will not violate thermodynamics?

    And, lastly, would it be a violation if the diode was effective enough to keep the system in equilibirum?
    You are putting something in the way thus the law no longer applies.

    As an aside can you give me an example of an 'optical diode'?
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  10. #9  
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    Quote Originally Posted by Andro
    So,
    If I try to make this setup, factors like the optical diode absorbing and emitting heat will make the system obey the 2. law?
    No. The fact that there is no such thing as an optical diode is what will get in your way. The setup cannot possibly be made to begin with.

    You might as well put a magical unicorn there that blocks heat from one side and allows it to pass from the other. You'd have just as much luck finding one.
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  11. #10  
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    I suppose a diode will only work where you have motion along a potential gradient. Other than an electrical diode, where the potential gradient or voltage has a step, permitting current flow down the step but not up the step , the other one I can think of off the top of my head, is a waterfall, where the potential gradient is gravity and water will flow, due to the potential difference, down but not up the waterfall.

    Since photons do not move due to a field or potential gradient, they are constrained to move at c at all times, an optical diode would seem to be an impossibility.

    Maybe this is easier to understand than using conservation laws or 2nd law of thermodynamics.
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  12. #11  
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    On the other hand, the event horizon of a black hole could be considered an optical diode since it is a one way door, your experiment would still fail, however, as the temperature of the inside of the event horizon cannot be defined.
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  13. #12  
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    Quote Originally Posted by MigL
    On the other hand, the event horizon of a black hole could be considered an optical diode since it is a one way door, your experiment would still fail, however, as the temperature of the inside of the event horizon cannot be defined.
    nope

    But it is complicated.

    http://en.wikipedia.org/wiki/Black_hole_thermodynamics
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  14. #13  
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    I do know that there is an associated entropy to black holes and Hawking radiation gives a temperature to work with, but this temperature is due to black-body radiation from the event horizon, ie on the outside of the hypothetical optical diode which is the event horizon. I maintain that temperature 'inside the event horizon cannot be defined and so the experiment we are discussing involving a cold body, the event horizon as an optical diode and a 'hot something' inside the event horizon will still not violate the 2nd law.
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  15. #14  
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    Quote Originally Posted by MigL
    I suppose a diode will only work where you have motion along a potential gradient. Other than an electrical diode, where the potential gradient or voltage has a step, permitting current flow down the step but not up the step , the other one I can think of off the top of my head, is a waterfall, where the potential gradient is gravity and water will flow, due to the potential difference, down but not up the waterfall.
    I don't think a waterfall is a good analogue to a diode, really. If gravity suddenly reversed, water would go back up to the top. If the electric potential of a wire with a diode in it reverses, however, the current would not flow back through the diode. That's the point in diodes. For water, the equivalent construct is a valve.

    The trouble is, you can't use a diode to force electricity to flow from a positive to a negative potential, you can't use a valve to make water flow up hill, and the light diode situation will run into a similar problem. You can't make heat flow from cold to hot, or light flow from dark to bright. You can prevent it flowing from bright to dark, and that's pretty much all you can do.


    Now, if the temperature on one side were constantly fluctuating from hot to cold and back, that might work, since that is how a water pump also works, but in order to make it constantly fluctuate, you would have to do work on it.
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  16. #15  
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    Okay, can one assume the heat to be electromagnetic radiation?
    Since there are different possible methods of energy transfer without doing work.
    I am.
    You can't deny it.
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  17. #16  
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    Quote Originally Posted by mastermind
    Okay, can one assume the heat to be electromagnetic radiation?
    Heat is, by definition, a transfer of energy. That energy being transferred in this scenario will be electromagnetic, yes... Infrared IINM.
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