1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Heat Transfer and Scientific Reasoning

  1. Aug 7, 2011 #1
    Hello, friends:

    I just got done studying a chapter in a physics text that explains how heat transfers from higher-temperature objects to lower-temperature objects. Unfortunately, that chapter said little about why heat transfers from higher-temperature objects to lower-temperature objects, so naturally I set out to answer that question.

    On the macro level, a hotter object in contact with a cooler object will cool while the cooler object will warm up. After some time, the two objects will be the same temperature, and heat exchange will cease. I decided to consider what might be happening at the molecular level. Realizing that the energy levels of molecules vary directly with temperature, I concluded that the molecules in the hotter object must have higher energy levels than the molecules in the cooler object. How might energy be transferred between molecules?

    To answer this question, I decided to focus on the simplest system I can think of: two molecules of different energy levels. If a molecule of higher energy, like the molecule of a hot object, comes into contact with a lower-energy molecule like that in a cooler object, then energy is evidently exchanged between the two molecules. Since the hot molecule's energy is the greater of the two, its energy “overcomes” the energy of the cooler molecule, but only to an extant. The energy of the cool molecule “resists” the hot molecule's energy which lessens the energy of the hot molecule. The hot molecule's energy, on the other hand, increases the energy of the cooler molecule. The net effect is that the two molecules acquire an amount of energy that is about equal.

    If we zoom out back to the macro level, the trillions of molecules in the two objects are undergoing the kind of process I just described for two molecules. Thermal energy moves from trillions of hot molecule to trillions of cooler molecules. We experience this movement of energy as heat transfer or an increase in temperature in the hotter object and a decrease in energy and in temperature in the cooler object. Once the molecules' energy levels are about the same, heat transfer stops.

    A mechanical analogy may help to clarify the process of heat transfer on the molecular level. Say a car moving at 50 miles per hour is rear-ended by a truck moving at 60 miles per hour. If neither vehicle is damaged to the extent that they would slow down from friction, then the car would speed up while the truck might slow down a bit. Some of the truck's kinetic energy is transferred to the car increasing the car's energy while the truck's energy decreases by that same amount of energy. More energy for the car increases its speed; less energy for the truck decreases its speed. Molecules exchange energy in a similar fashion.

    Am I right or wrong about what accounts for heat transfer? At this point you may agree or disagree with what I'm saying. Whether I'm right or wrong is not terribly important to the point I'm trying to make. I'm trying to demonstrate basic reasoning that should be useful in answering scientific questions.

    I'd appreciate some feedback on this issue.

  2. jcsd
  3. Aug 8, 2011 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi Jagella,

    Thanks for posting this, I enjoyed reading it. These comments are intended to spark thought about the reasoning process:

    - What does it mean for a higher-energy molecule to "overcome" the lower-energy molecule? It is hard to visualize what is happening physically here.

    - Why do the two molecules need to end up with a temperature that is "about equal"? Conservation of energy would still be satisfied if the higher-energy molecule gained energy and the lower-energy molecule lost it. Essentially, the question still exists, although on a molecular level now, of why hotter objects heat colder objects.

    - In this model, all the molecules of a substance have approximately the same energy level. Is this necessary? Could they have a distribution of energy levels, with the mean or median being higher for the hotter object?

    - How do we explain radiative heat transfer through a vacuum, in which hotter objects also heat colder objects? No molecular contact exists here.
  4. Aug 8, 2011 #3


    User Avatar
    Science Advisor

    What is a hot molecule?
  5. Aug 8, 2011 #4
    I think that your ideas go in the right direction; however there should be no damage to gas molecules; the collitions must be elastic.

    For more, you could search the internet for "Kinetic Molecular Theory of gases", or start with the introduction here:

    For the same in a more advanced form, you can also have a look at Statistical Mechanics:

  6. Aug 8, 2011 #5
    And thank you for the response.

    It's interesting that you should ask about what I meant by “overcome” and how to visualize it. I had some problem there, too, and I realize that to say that one energy level overcomes another, lesser energy level is a bit vague. I used my drawing software to draw two molecules represented by circles, and I drew arrows coming from each molecule. The length of the arrows represents the amount of energy each molecule has. Longer arrows represent greater energy levels. When two molecules exchange some energy in a collision, the more energetic molecule will make the the lower-energy molecule move faster while the higher-energy molecule will move more slowly. I then redrew the molecules with arrows of equal length representing these equal energy levels. That's what I meant by “overcome.”

    By the way, using drawing software is a great way to visualize and understand concepts in physics, mathematics, and other disciplines.

    In a collision between two molecules, the higher-energy molecule will normally lose energy while the lower-energy molecule will tend to gain the same amount energy the other molecule loses. The higher-energy molecule gives the lower-energy molecule a push increasing the other molecule's energy and decreasing its own energy. This process brings the two energy levels closer in magnitude. This “zero-sum game” is a result of the conservation of energy.

    I understand it's possible for the higher-energy molecule to gain energy from a collision while the lower-energy molecule loses energy, but such an exchange is very unlikely. I think that if the two molecules are moving in the same general direction, then the lower-energy molecule might bump the higher-energy molecule from behind. In that case, the higher-energy molecule would gain energy while the lower-energy molecule would lose energy. If this situation happened for most of the molecules in two objects, then the hotter object would get hotter while the colder object got colder! This result is extremely unlikely, though.

    No. Heat transfer takes time. Prior to the time at which thermal equilibrium takes place, an object may have higher-energy-level molecules on one side and lower-energy-level molecules on the other side.

    Sure. If the energy levels of the molecules are randomly distributed, then most of the molecules in an object will have an energy level close to the mean value. If another object is colder, a similar distribution of molecular-energy levels should be present, but the mean value will be lower than the hotter object's mean value, of course.

    In that case heat is transferred via electromagnetic radiation which, unlike heat conduction or convection, needs to substance to move through. I'm not sure, but I believe this kind of heat transfer occurs through the fabric of space as described in Einstein's General Theory of Relativity.

    I am, admittedly, a little shaky on that last point. :confused:

  7. Aug 8, 2011 #6
    A "hot" molecule is a molecule that has a higher energy level than other molecules it comes into contact with. I was a bit vague in using that term, and I'm glad you asked for a clarification.

  8. Aug 8, 2011 #7
    Thanks for the advice and the link, but I'll need to read it later. At this point I'm more concerned with basic reasoning than with research. I could have looked up the answer to why heat transfers from hot to cold easily enough, but I'm trying to come up with the answers myself. That way I hope to be able to discover things that nobody else knows.

  9. Aug 8, 2011 #8


    User Avatar
    Science Advisor

    Higher kinetic energy?

    Let's say your procedure works. Now take the "final" state and reverse the directions of all velocities. The "reverse" collisions should still satisfy Newton's laws, and produce the initial state. No?
  10. Aug 8, 2011 #9


    User Avatar
    Staff Emeritus
    Science Advisor

    That isn't the way science works. You are effectively stuck in the 1800's and trying to come up with the kinetic theory of gases yourself. We already know what happens, and if you think you are going to find something that thousands upon thousands of people have missed somehow, then you are sorely mistaken and wasting time. And you have a much much better chance of finding something new if you understand the theory anyways. How can you find something new if you don't even know what is and isn't true in the first place?
  11. Aug 8, 2011 #10
    I don't believe merely reversing the directions of the molecules' motions can restore the initial state. In my rationale, the magnitude of each molecule's energy comes into play more than the direction of motion. To restore the initial state, the energy levels of the molecules would need to be reset to their initial values.

    That's an interesting question. Why do you ask?

  12. Aug 8, 2011 #11
    I think you misunderstand what I said. I didn't mean that I would discover something new about heat transfer; I meant that I might discover something new about something that is not yet well understood.

    What inspired my posting my initial comments is my recently learning about how Einstein came up with a theory that explains the photoelectric effect. Rather than trying to explain the photoelectric effect as being the result of wave motion, Einstein posited that light is made up of particles or photons. Photons strike the electrons in the metal, and the energy of the photons, understood as their frequency, impart the same frequency to the electrons they collide with. Einstein won the Nobel prize for this work.

    How did Einstein do it? It seems to me that his independent thinking enabled him to see things other physicists didn't see. He thought "out of the box," and outside of the box is where new knowledge resides waiting to be discovered.

    And as far as finding something new about heat transfer, I don't believe it hurts to take a look. History has demonstrated more than once that "thousands upon thousands" have missed discoveries. Ever hear of Copernicus? Da Vinci? Newton? Darwin?

  13. Aug 8, 2011 #12


    User Avatar
    Science Advisor

    Heat transfer is irreversible in time - you cannot get heat transferred from a cold object to a hot object by placing them next to each other. However, the laws of physics are reversible in time (with one exception, not relevant to this discussion). So the direction of heat transfer should not be derivable from the laws of physics. To take your truck and car example, there is no law of physics which says that the car cannot go backwards and front end a truck, causing the truck to start moving. It just so happens that the way the universe is set up (at the "beginning of time"), we see heat transfer in only one direction.
  14. Aug 9, 2011 #13
    That would be a good idea if the current theory was, as you probably thought, undeveloped or unsatisfactory. However, it's not very useful to try to reinvent the wheel. :tongue2:

    Meanwhile you received the feedback that you demanded: that type of scientific reasoning did lead to a highly satisfying theory.

  15. Aug 9, 2011 #14
    I think that that puzzle has been solved (derived from combining the laws of physics with logical reasoning), as also indicated in the second reference that I gave.

  16. Aug 9, 2011 #15


    User Avatar
    Science Advisor

    It depends on whether initial conditions are included in the "laws of physics".
  17. Aug 9, 2011 #16
    I am sorry. Can you explain your statements a little further. I am confused because i think you are mixing two unrelated things together (time and heat transfer).

    What the law of thermodynamics dictates is that heat transfer only occurs from a objects of higher kinetic energies to objects with lower kinetic energies. Nothing to do with time as far as i understand.

    Then there is time in physics, which you say is irreversible? In which you are mixing up heat with time travel?
  18. Aug 9, 2011 #17


    User Avatar
    Science Advisor

    Oh, I just mean something very naive. The second law of thermodynamics means heat flows from hot to cold. So if we see cold to hot, we assume time is running backwards, so as not to violate the second law. However, apart from the second law of thermodynamics, no other law of physics (with one irrelevant exception) prohibits the time-reversed version of a process from going forward in time.
  19. Aug 9, 2011 #18
    I see where you guys are going and why time and vectors were introduced into the topic.

    So quick answer to your question is Yes.

    I see from your two previous posts you are aiming for a revert the collision events that is why you included vectors and time. As you may already know, to revert a process, we must take into account the proper parameters and revert them accordingly in the right order.
    -Reverting only the directions without taking account the position at the given time will just create a new unrelated event.
    -Only reverting the time without reversing the vectors will just be a backward-playback of the original event.

    For the sake of the argument we will only use only one dimension:
    -Particle A sits on x=0m at t=0s with v(i)=4m/s
    -Particle B sits on x=4m at t=0 with v(i)=2m/s
    -For particle A at t=1s, x=4m, v=4m/s
    -For particle B at t=1s, x=6m, v=2m/s
    -For particle A at t=2s, x=8m, v=2.1m/s
    -For particle B at t=2s, x=8m, v=3.9m/s
    -For particle A at t>2s, v=2.1m/s
    -For particle B at t>2s, v=3.9m/s
    Code (Text):

    ..A>>B>.........   t=0s
    ..A...B.........   t=0s
    .....A.B........   t=1s
    .........AB.....   t=2s
    ..........A..B..   t=3s
    ..........A>.B>>   t=3s
    As you can see, the collision happens at t=2s.

    Now if we revert the process:
    -For particle B at t>2s, v=-3.9m/s
    -For particle A at t>2s, v=-2.1m/s
    -For particle B at t=2s, x=8m, v=-3.9m/s
    -For particle A at t=2s, x=8m, v=-2.1m/s
    -For particle B at t=1s, x=6m, v=-2m/s
    -For particle A at t=1s, x=4m, v=-4m/s
    -For particle B at t=0s, x=4m, v=-2m/s
    -For particle A at t=0s, x=0m, v=-4m/s
    Code (Text):

    .........<A<<B..   t=3s
    ..........A..B..   t=3s
    .........AB.....   t=2s
    .....A.B........   t=1s
    ..A...B.........   t=0s
    <<A..<B.........   t=0s
    Explaining heat transfer with particle collisions can be held true (without breaking the second law of thermodynamics) only if the following conditions are present:
    (1) If two particles of different kinetic energies collide, the resultant energy inside each particle must be closer to their average.
    (2) If two particles of the same kinetic energies collide, the collision must be completely elastic.

    Which becomes the new mysteries:
    (1) Why the resultant collision must be closer to their averages?
    (2) Why when particles have the same kinetic energies their collision becomes completely elastic? For example, can't two particles of same magnitude but opposite direction collide and result on a completely inelastic collision where both particles are put into halt?

    Edit: A possible explanation to (1) is because particles are aiming for completely elastic condition (2). But then we are stuck on why they aim for completely elastic collision.
    Last edited: Aug 9, 2011
  20. Aug 9, 2011 #19


    User Avatar
    Science Advisor

    It should be ok to use only elastic collisions. Fundamentally all interactions conserve energy. Let's assume the gas is pretty dilute, so most of the the time the inter-particle distances are large. We can assume that the potential energy falls off with inter-particle distance, so that at large inter-particle distances all the energy is kinetic. So conservation of energy is the same as conservation of kinetic energy.

    However, I still think that would not explain why the reversed collisions cannot take place.
  21. Aug 9, 2011 #20
    I guess i only complicated things by creating the second statement.

    I guess i forgot to explain some of the results of the example i presented before.
    -As you can see, by the time we reverse the process; the energy contents have already been swapped as well.
    -Originally particle A had more energy than particle B. After the collision, particle B had more energy than particle A.
    -So when we reverted the process, B had more content than particle A to begin with. Meaning that energy is still transferred from higher to lower energy content regardless.
    (Of course, my example is only valid if particle A and B had the same mass)

    Again, that goes back to statement (1) where we only allow energy of high energy objects to be transferred to lower energy objects. If we allow low energy content to become a lower energy content while transferring the energy to a higher energy content this explanation collapses. Because if statement (1) does not hold true, then even if we revert the process; then energy content will still be transferred from lower energy to higher energy contents.

    Edit: By the way, even in collisions, it is hard to imagine an object with little momentum to transfer energy to a object with higher momentum. The example would be an object metting another object at a 90 degree angle.
    Code (Text):


    Last edited: Aug 9, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook