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Question about heat transfer

  1. Dec 17, 2008 #1
    What force is responsible for the phenomenon of heat transfer?
     
  2. jcsd
  3. Dec 17, 2008 #2

    Hootenanny

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    What makes you think that heat transfer requires a force?
     
  4. Dec 17, 2008 #3
    Because it just seems to happen automatically - so don't you need a force when something happens automatically?
     
  5. Dec 17, 2008 #4

    Hootenanny

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    What seems to move? And what do you mean by 'automatically'?

    Before we go any further, perhaps it would be useful to clarify what heat actually is. See here for more information.
     
  6. Dec 17, 2008 #5
    I mean that the average translational kinetic energy of a hot space seems to get less when next to a colder space. And this seems to happen spontaneously or without any external interference.

    For example, I am in my apartment and it's all warm in here. Then I open the door to the hallway where it's cold. The hot air just rushes out the door. I am assuming this is heat transfer - am I wrong? If I am right, what I am asking is, what force if any is responsible for this behavior? And if there is no force responsible for this behavior, then why not?
     
  7. Dec 17, 2008 #6
    It is a statistical effect. A system in a closed volume with some fixed total energy E, has equal chance to be in any of the possible physical states with energy E. Suppose you have two systems, one has energy E1 and the other has energy E2. Then system 1 can be in Omega(E1) possible states, system 2 can be in Omega1(E2) possible states. So, the comboined system can be in Omega1(E1)*Omega(E2) possible states.

    Suppose that we allow heat to be transferred from one system to the other system, but the system1 and system 2 combined are still isolated, so thew total energy E1 + E2 remains constant. Then since all states are equally likely, you will end up with that energy distribution over the two systems for which you have the most states, i.e. for which:

    Omega1(E1)*Omega2(E2)

    is as large as possible. Take the logarithm, put E2 = E - E1, with E the total energy and differentiate w.r.t. E1 and set it equal to zero:

    d Log(Omega1(E1))/dE1 +d Log(Omega2(E-E1)) /dE1 = 0

    If you put back E - E1 = E2 and that by the chain rule

    d/dE1 = -d/dE2

    you get:

    d Log(Omega1(E1))dE1 = d Log(Omega2(E2))/dE2

    So, in the situation we eventually end up in a quantity that depends on only the system is the same for both systems. Now, we have already phenomenolgically defined the temperature to be such a quantity, so this must be related to the temperature. The temperature is then rigorously defined as:

    1/(k T) = d Log(Omega)/dE

    The fact that when you have thermal contact you have energy transfer intul the temperatures are equal is thus a purely statistical effect.
     
  8. Dec 17, 2008 #7
    Consider an imaginary boundary between a hot volume and a cold volume. For example, the plane of your doorway. Since molecules are always randomly jostling around and transferring energy from one to another, in a unit of time a packet of energy slightly inside your room has a certain probability to be transferred across the doorway and out of the room. Similarly, a packet of energy slightly outside has exactly the same probability to be transferred inside.

    However, since there are more packets of energy inside the room than outside, there is a net flow of energy packets across the doorway and out of the room. If P is the probability of a packet being transferred across the border (in any direction), and N_in and N_out are the number of energy packets inside and outside, then the flux (q) of packets across the doorway out of the room is the number of packets that leave minus the number of packets that enter:

    q = P*N_in - P*N_out

    If you take a few more steps, this basic reasoning gives Fourier's law of heat conduction.
    http://en.wikipedia.org/wiki/Heat_conduction#Fourier.27s_law
     
  9. Dec 18, 2008 #8
    Slightly linked to this follows the questions, why does heat not dissipate to an equlibric state across all space time, and how does heat transfer through vacuum in the lack of kinetic energy??
     
  10. Dec 18, 2008 #9

    Hootenanny

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    There is no conductive (or convective - thanks timmay) heat transfer in vacuum for precisely the reason you state.
     
    Last edited: Dec 18, 2008
  11. Dec 18, 2008 #10
    ...for convective and conductive heat transfer. Thermal equilibrium is achieved through radiative heat transfer in that case.
     
  12. Dec 18, 2008 #11

    Hootenanny

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    Whoops, missed convective. Thanks for the correction.
     
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