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Heat Radiation in Thermodynamics

  1. Nov 30, 2013 #1
    Hello, first of all, I must say that I am quite knowledge lacking in the thermodynamics topic, I am just seeking some intuitions.

    From what I have read so far, heat can only flow from hot bodies to cold bodies, and not the way around, according to the 2nd law of thermodynamics.

    According to this image,
    which illustrates the transfer of heat between two objects, it seems that when a hot body is placed nearby a cold one, only the hot body irradiates heat to the cold body. Is this actually the case? I ask such because i've also read somewhere that every body irradiates some heat, hence it would make sense for the cold body to send some photons to the hot body.

    I would imagine that the cold body also irradiates heat, but since it irradiates less heat than the hot one, the image just ilustrastes that the "effective" heat movement goes from the hot body to the cold body. Am i correct in this point?

    My second major question, relates to the subject as a whole. Why is it that an equilibrium between two bodies with diferent temperatures has to be achieved? What in the nature justifies such? I mean, speaking in terms of atoms, electrons and the speed of these, how does the process as a whole work?

    I've found an answer to a certain degree in this link

    The relevant quote is the following "The answer lies in energy and momentum conservation in a collision - one can show, using these two principles, that in a collision between two objects which conserves energy (called an elastic collision the faster object slows down and the slower object speeds up. "

    This means that when the atoms of the hot object (fast ones) colide with the atoms of the cold object (slow ones), the slow ones gain momentum, while the fast ones lose momentum, hence the cold object will get warmer while the hot object will get colder. Still, i can't understand how the process will work to reach its end. What if the atoms of the cold object, the recently accelerated ones, bounce back and colide with the atoms of the hot object once more? Wouldn't they give back the gained momentum and hence keep the cold object cold and the hot object hot? What forces the gained heat to remain in the cold object and not return to the hot one?

    Also, when dealing with heat radiation, where the atoms of the objects are not coliding, how does the process work to prevent heat from returning to the hot object?

    I'm sorry if im having any kind of confusion or if i am missing something.

    Thanks in advance
  2. jcsd
  3. Nov 30, 2013 #2

    Simon Bridge

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    This is not correct - both bodies radiate.
    They radiate in proportion to their temperature, so hotter bodies lose heat faster than merely hot bodies.
    They also absorb radiation, gaining heat that way.

    A hot body close to a hotter body is radiated more by the hotter body than it gives back.
    If the hotter body is sufficiently hot, the it may heat the hot body faster than the hot body loses heat.

    There is a standard thermodynamic object called a "blackbody". This object radiates but is in thermal equilibrium with it's surroundings - so it is gaining heat at the same rate as it loses it by radiation. The radiation it gives off has a characteristic spectrum.
  4. Nov 30, 2013 #3
    You are absolutely correct, Simon. However, I fully understand the questioner's confusion.

    In strict thermodynamic terms, heat only flows from a hotter body to a cooler one. Thermal energy, in contrast, can go either way. Heat, of course, is a form of thermal energy; and both terms involve transfer of energy by photons, by mass-to-mass conduction, and by mass transfer (adding warm water to a tub of cold water--or vice versa!).

    I seem to remember an article a few years ago that examined how often the correct definition of heat was used in thermodynamic papers. The vast majority of papers used the term incorrectly at least once. Many papers used the term "heat" in more than three different and contradictory ways.

    If thermodynamic physicists can't get their terms straight, I can't feel too badly over my occasional lapse.
  5. Dec 1, 2013 #4

    Simon Bridge

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    Strictly, this is the net flow of heat. For the example of two radiating bodies, there is a net heat flow from the hotter one to the cooler one (with proviso) ... where "hotter" is taken to mean " higher temperature".

    You can have lots of fun with misunderstandings about heat.
    The main trouble is that it is also a common everyday word with a broader use like work, energy, force ...
  6. Dec 1, 2013 #5
    I think a lot

    I thank a lot for the answers so far. So, according to these, both bodies actually irradiate, as i initially thought.

    I'd just like to ask for some further help on this topic. Is the rate at which the bodies irradiate regular? I mean like, body A irradiates 10 photons each 1 nanosecond, or is it actually irregular, like a body irradiates 10 seconds now... 1 nanosecond later it irradiates 9, then it does not irradiate for 2 nanoseconds and then it returns to irradiating some.

    I ask this because i can imagine a scenario (which may be physically impossible, forgive my ignorance) where body A (hot body) irradiates a lot of photons to body B (cold body) and B grows hotter than A because meanwhile he did not release any of the absorbed photons.

    Hence we would reach a state where B actually got hotter than A. Do the bodies actually reach the same exact temperature? Or do small oscillations occur in the "equilibrium" state?

    On the other hand, if these emissions actually occur at regular time intervals, i can pretty much understand how equilibrium would be reached, like if all bodies emit photons at regular time intervals, say every 1 nanosecond, then such oscillations would not exist.

    Am i too far from the actual answer to this question?
  7. Dec 1, 2013 #6

    Simon Bridge

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    Depends how finely you want to look at it - bodies typically radiate continuously at a rate that depends on their temperature.

    See: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

    In the fine detail of very short times and low temperatures, there will be a quantum limit where one photon is radiated... then nothing... then another etc. On this scale we don't talk about rates of radiation but about probablities for emission.

    That would be the case for quantum-level fluctuations ... i.e. A and B differ in energy by one photon (of a particular wavelength) then there is a chance that A radiates that photon so that is subsequently gets absorbed by B ... thus swapping their states.
    The more photons are involved, the less likely that is.
    For the kinds of differences that we can measure in a lab for less than, say, a million bucks, the probablities are like the chance that all the air molecules in the room you are sitting in spontaneously fly to one side of the room. It could happen.

    Again - depend on how small you mean.
    For normal objects you can hold in your hand, entropy dominates and pretty flat equilibriums are reached in quite short times.... like a matter of minutes to hours depending on the material.

    You cannot get away from quantum fluctuations though.

    Is there a specific problem you are interested in?
  8. Dec 2, 2013 #7
    Thermodynamics is misunderstood by many. A simple and important point to remember in thermodynamics is that, time should not enter the arguments as time plays no role in thermodynamics. Quantum mechanics or statistical mechnics need not and should not be brought in, to understand the process of equalization of temperatures.

    Elastic Collisions are brought in to explain the mechanism of equalization of temperature in gases. In those collisions the difference in kinetic energies of two masses necessarily decreases as a result of elastic collision (Maxwell derived that result). It was a wrong result.

    It is that result which brought irreversibility into dynamics by giving a definite direction (thereby increase of entropy etc) to an otherwise reversible mechanical process of an elastic collision.

    When two bodies at different temperatures are brought into thermal contact they reach a common temperature because the system of the two bodies loses energy possessed by it by virtue of difference in temperatures. That difference cannot decrease unless the system loses the energy.

    Energy is lost from the system of the two bodies as the two temperature approach one another just as in the case of a battery where the potentials of the positive and negative terminals approach each other. as the battery loses energy (discharges).
  9. Dec 2, 2013 #8
    I can't quite understand how time does not take any role in thermodynamics. I mean, according to the link provided by Simon (thanks a lot) http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

    My main question lies not in thermal transfer through direct contact, but by thermal radiation emission. Is it possible to define a formula that gives the difference of temperature between two bodies irradiating one another at a given time T?
  10. Dec 2, 2013 #9

    Simon Bridge

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    So all those post-grad papers relating statistical mechanics and quantum mechanics to thermodynamics are a waste of time then?

    It is important to stick to just one model at a time though.
  11. Dec 2, 2013 #10

    Simon Bridge

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    It is sometimes possible to treat the radiation as a gas.

    The main trick with this sort of discussion is to pick a model ... your questions have spanned different scales and need different models to work out what is going on. I've been careful to stress this.

    If you have two balls the same diameter and material, one is hotter than the other (somehow) and they are all alone in an infinite Universe ... then they will both cool by radiation to absolute zero.
    There will be a net transfer of energy from the hot ball to the cooler one while this happens.

    If the Universe is not infinite nothingness - say, it is a big box with perfectly reflecting sides: then both will cool until they are in thermal equilibrium with the surrounding radiation (and each other).
    In this case it is easy to see how the cool one can gain heat and the hot one loses it ... though it is possible the equilibrium could be at a temperature cooler than they both started out.

    When we get to the scale of two molecules rather than two balls - each can be vibrating, which is one of the ways gasses store heat. But it does not really make sense to talk about the individual molecules having a temperature does it?
  12. Dec 3, 2013 #11

    Andy Resnick

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    Unfortunately, the answer is often 'yes'. Although SM and QM have been clearly correlated with thermostatics and partially correlated with thermokinetics (e.g. the fluctuation-dissipation theorem and Onsager's relations), a microscopic theory of thermodynamics (nonequilibrium) does not even have a clear conceptual formulation.

    Two good introductions to the problem are Jou, Casas-Vazquez and Lebon, "Extended Irreversible Thermodynamics" and Balescu's "Statistical Dynamics: Matter out of Equilibrium".
  13. Dec 3, 2013 #12

    Simon Bridge

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    Which works seem to echo my replies to the questions above - with details of course.
  14. Dec 4, 2013 #13
    In a way thermodynamics derives its simplicity by by not having 'time' in its formulation! By introducing time one can have as much dubiousunderstanding/confusion as possible.
    The second law simply states that energy in the form of heat flowsspontaneously from a high temperature body to a low temperature body. Advantage can be derived from such a natural tendency to derive benifit, for example, by way of obtaining mechanical work using a heat engine.

    When energy in the form of heat is ratdiated (spontaneously) by two bodies at different temperatures and each body absorbs energy coming from the other irrespective of the temperature of that other body, second law gets violated. To save the situation rates of energy transfer are brought in so as to make the net rate of transfer satisfies the flow from higher to lower temperature.
  15. Dec 4, 2013 #14

    Simon Bridge

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    But you still need to be able to answer questions like: "How long will it take to heat my a cup of water to boiling?"

    There are lots of dynamical situations where losing information about time is useful - just look at the simple pendulum. That does not mean that questions about time should not be considered.

    Note: time is not something that we add to a thermodynamic model (say, a state diagram) it is something we remove from another description to get the thermodynamic one you are used to. Real experiments take place over time.

    All you've done is point out that thermodynamic models are not going to be useful to answer the question.
    OK - so what can you use to answer the question?
  16. Dec 4, 2013 #15
    Thermodynamics gives us predictive ability. Given two equilibrium states of a system A, B thermodynamics enables us to predict whether the system goes spontaneously from state A to B or from state B to A.

    But it does not give us time it takes for such a process to occur.
  17. Dec 5, 2013 #16
    I guess you could split things up into thermodynamics (which is really about tracking energy in a given system) and heat transfer (which is more about time-dependent energy transfer).

    IMHO, thermodynamics is hard because it runs somewhat counter to our experiences growing up. For example, before I knew anything about thermo, I thought temperature was the same as heat energy, when it's really only related to the energy of a substance. Entropy was a totally foreign concept that I had no mental picture for, and the second law of thermodynamics is predicated on it. Statistics for me before I took thermo was numerical magic, so that explanation of things was just all hand-waving and magic tricks. It was a totally different way of looking at things, so it was hard to wrap my mind around.
  18. Dec 6, 2013 #17

    Simon Bridge

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    All you've done is point out that thermodynamic models are not going to be useful to answer the question.
    OK - so what can you use to answer the question?

    Please refer to post #1.
  19. Dec 6, 2013 #18

    Simon Bridge

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    This is all very well - but how about answering the question in post #1?
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