Cute Physics Problem

I heard this nice physics problem today so I thought I’d try it out on here. You will probably be able to find the answer on the net somewhere but please try to figure it out yourself before doing so!

There are two identical chambers, A and B containing identical metal balls which begin the experiment at the same temperature. Apart from the balls, each chamber is a perfect vacuum and has thermally conducting walls at a lower temperature than the ball it contains.

In A the ball is resting on the floor, which is made of material which is a perfect thermal insulator.

In B the ball is hanging from the ceiling by a piece of light inextensible string, not touching the floor. Both the string and the ceiling are also made of perfectly insulating material.

Which ball cools down faster?

Please put your answer through the poll here. When enough people have voted, I’ll tell you the answer…


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52 Responses to “Cute Physics Problem”

  1. Richard Engkraf Says:


  2. Neither cool down at all?

    Vacuum, so no conduction or convection.

    In a perfect thermally insulated chamber, so any infrared emitted with bounce around within the chamber.

    Case B is the easiest to imagine: emitting IR isotropically.

    Case A: considering a perfect sphere and perfectly smooth chamber walls, the surface in contact is infinitesimally small, therefore acts just the same as case B.

    • telescoper Says:

      The floor and ceiling are thermally insulated, but you can’t assume the walls are….

    • Every object that has a temperature greater than absolute zero gives off heat in the form of radiation. Radiation does not require a medium to travel through, thus, even in a perfect vacuum both balls will cool down.

  3. Anton Garrett Says:

    My immediate reaction is that, supposing the walls of chamber A are at the same temperature as the walls of chamber B and that the spheres are at a higher temperature (as I think you mean us to suppose), the two spheres cool radiatively at the same rate. I have to say that I’ve reached this conclusion by ‘cheating’, since if the answer was anything else then it would depend on the shape of the chambers – which you do not specify.

  4. I put A… cos B has PE as well as it’s temp energy??

  5. telescoper Says:

    I’ve re-worded it to make it clear that the walls can transmit heat out of the chamber and that they are at a lower temperature than the spheres.

  6. I submitted B because I didn’t think it through, but now after giving it some thought, I think I’m wrong, and here’s why:

    [ !!! SPOILER ALERT !!! ]*

    If we take this to an absurd level of precision, because that’s at what level it needs to be to actually notice the time differences, than we’re talking about minute (as in small) fraction of time, in essence, this is a question of surface heat conduction.

    If our perfectly thermally insulated ceiling and floor are not accepting heat – so to say – than the amount of atoms that are giving of heat to the vacuum is smaller in the ball on the floor (ball A), hence a minute amount of heat will be kept in the ball for a longer period of time, and since the ball that is suspended in the air, or in our case vacuum, will lose its’ heat a little bit faster.

    Hence ball A will cool-down faster and wins the race.

    * The spoiler alert may or may not be false!

    • telescoper Says:

      There’s a much more basic thing that you’re neglecting.

      Think about what breaks the symmetry.

  7. telescoper Says:

    I’m holding a few spoiler comments in limbo until more people have had a go…

  8. If you are assuming that both chambers exist in a gravitational field then energy was input into Ball B in order to suspend it. Assuming both balls radiate isotropically, and the walls absorb the energy and syphon it off then Ball B will have the greater energy and the greater rate of cooling for small times after the experiment starts. I guess the question is whether which ball has the faster cool down rate? which would be Ball B, or which one reaches a certain temperature first? which would be Ball A?

  9. Ball B that is suspended in the air should cool down more quickly as it should do be able to radiate isotropically. Ball A on the other hand is in contact with the perfectly insulating floor and so there will be some small fraction of its surface area through which it cannot cool – also a higher fraction of reflected heat radiation from the floor will be reabsorbed in Ball A than Ball B as it is so close (less of the heat radiation reflected from the wall escape through the walls as ball A is on the floor).

    So my guess is Ball B (suspended in the air) cools quicker.

    But I suspect a trap……?!

  10. thasentinel Says:

    Classic problem! Great post telescoper! As a *small* hint for others, what is the big difference between ball A and ball B, that is, what does it mean for one to be suspended and the other to be resting on a surface?

  11. telescoper Says:

    I’m amazed how much interest this post has generated, and it still seems to be collecting votes so I won’t give the answer straight away. I will however give two more hints:

    1. What is the significance of the fact that the balls are made of metal?
    2. What breaks the symmetry of the problem?

    The answer has nothing to do with the rate at which energy is radiated by the two balls.


  12. John Peacock Says:

    Sounds like I failed the test. I assumed it was rather like coming out on a frosty morning and finding your car rimed up, or not, depending on the solid angle of the sky it sees. Taking the T=0 limit of your setup, the non-insulating walls might as well be transparent, and the ball radiates to infinity. But energy radiated to the insulating wall will heat it up, which in turn returns some energy to the ball. Since the ball in the middle of the box sees a smaller solid angle of insulator, clearly this cools fastest. The fact that the balls are metal seems irrelevant.

    • telescoper Says:

      In retrospect I could have made it clearer. I could have said that both balls were attached to inextensible insulating rods of the same length, one attached to the celing hanging downwards and one to the floor standing upwards so that the symmetry is absolutely clear. I just passed it on as I was told it, however…

  13. Bloody first past the post system…

  14. Until the latest post from telescoper I was thinking it is about what is greater, the diameter of the string (Ball B) or the surface area of contact between the ball and the floor (Ball A). But since the rewording… Hmm what links gravity and the fact that the balls are metallic.

  15. Dipak Munshi Says:

    Are these balls at the same height? or do we neglect gravity?

  16. Does upside down system A = system B? and vice versa

  17. telescoper Says:

    OK. Time for another hint, I think:

    What does metal do when it cools?

    • Ball A is losing gravitational potential energy from shrinking, whilst ball B is gaining it. So maybe the thermal energy in ball B is being lost through both blackbody radiation and the increase in potential energy. On that basis, I believe ball B cools down faster.

      Of course, it could be the other way around…

    • the iternal energy of the hanging ball falls more rapidly since it also has to do work to elevate it’s centre of mass, however I doubt whether this would result in a faster reduction of its temperature (compared to Ball A)? I had assumed that the thermal energy is caused by the jittering of the atoms? how is interlnal energy related to the thermal energy?

  18. The one on the floor (or lower one) will cool slower due to gravitational time dilation? Unless we are talking now about the scenario where it is not on the floor but on a rod and the length of the rod(s) is(are) greater than half the height of the box, then it will be the other way around. Something tells me this isnt the solution and gr is not meant to be accounted for in this problem ;) Although this is about the only link i can think of between gravity and rate of cooling; through the changing of the position of the centre of mass of the ball in a gravitational field due to shrinking while cooling.

    I do not see how a change in gravitational potential energy will directly affect the thermal energy.

  19. Or the other way around, the photons closer to the ground will be blue shifted, having more energy, therefore carrying heat away faster.

  20. I’m working on the basis that stuff usually heats up (in terms of K.E.) when travels down a potential and cools down when it travels up one.

    • The heat of something is based the internal KE of the atoms. If the macroscopic object moves in a potential well, the KE of the atoms inside the reference frame of the ball remains unchanged… i think ?

    • I’m admittedly using the word “heat” in a hand-wavy, wibbly-wobbly kind of way.

  21. telescoper Says:

    Energy is conserved…..

  22. For what it’s worth, I completely didn’t understand the situation until the comment:

    “In retrospect I could have made it clearer. I could have said that both balls were attached to inextensible insulating rods of the same length, one attached to the celing hanging downwards and one to the floor standing upwards so that the symmetry is absolutely clear.”

    I forget which way I voted, but no doubt I got it wrong. Ball A loses gravitational potential energy as it cools and shrinks, whereas Ball B gains it. So when both balls have radiated the same amount of energy, Ball A has more internal (thermal) energy than Ball B. That is, B cools faster.

  23. telescoper Says:

    Yes, that’s basically the right answer.

    The only remaining point to recognize is that both balls lose energy by radiation at the same rate (as they have the same size and shape and are initially at the same temperature). If there were no other form of energy involved, this would be accounted for by a change in thermal energy and therefore a change in temperature.

    However, as has now been established the centre of mass of one ball (B) rises while the other falls. This means B does work against gravity while A has work done on it. Since the rate of radiative loss is the same for both, this means that the thermal energy lost by B must be greater than A, so it must cool quicker.

    Sorry if people were confused by the wording but, as I said, I just passed it on as I heard it. It seemed obvious to me what was meant, but I know I’m wierd. :)

  24. John Peacock Says:

    Peter. this last argument of yours would be correct if all the walls of the box were isothermal. But you clearly stated that one of the walls was a perfect insulator. So I think that this must heat up under radiation from the ball, as I suggested above, and this breaks the symmetry far more effectively than the gravitational effects you discuss.

    • telescoper Says:


      I stated that the floor and ceiling are perfect thermal insulators in the respective cases.

      If the situation is symmetric, i.e. ball A is the same distance from the floor as B is from the ceiling, then the resorption of reflected heat will be the same in both cases. I quote from my addendum…

      … both balls were attached to inextensible insulating rods of the same length, one attached to the celing hanging downwards and one to the floor standing upwards …


  25. Garret Cotter Says:

    Oh Peter, you twist and turn like a twisty-turny thing… ;)

    • telescoper Says:

      Yes indeed. This is why I hate writing exam questions!

      Taking account of all the loopholes and ambiguities is so difficult even in simple cases…

  26. John Peacock Says:


    > I stated that the floor and ceiling are perfect thermal insulators.

    Indeed, and that’s why they will heat up on absorbing radiation.
    But to be more precise, the problem as originally stated had the
    floor (only) as an insulator in A and the ceiling (only) as an insulator
    in B. So the solid angle of insulator seen by A is much larger. But
    actually A still sees more solid angle even if both floor and ceiling
    are insulators.

    > If the situation is symmetric, i.e. ball A is the same distance from the
    > floor as B is from the ceiling, then the resorption of reflected heat will
    > be the same in both cases. I quote from my addendum…

    Agreed: in that case, and in zero-G, both clearly cool at the same rate.
    But as originally stated, and in zero G, I think B must cool faster. Not
    convinced that gravity would be the dominant effect, in any case.

    • telescoper Says:

      Your conclusion depends crucially on the length of the piece of string (!)….
      ..and also on the efficiency with which the spheres absorb the reflected radiation.

      It’s clearly not zero-gravity, otherwise the ball would not be “hanging” from the ceiling.

      Incidentally I went to our physics stores yesterday and it turns out they don’t stock light inextensible string. You’d think they would, since it appears in some many questions.

    • Anton Garrett Says:

      You can’t find an ideal world either.

      Personally I’ve always enjoyed the ambiguity in the phrase “a light beam” found in physics questions.

  27. telescoper Says:

    My semantic inexactitudes notwithstanding, I hope you agree this is a great question for a physics tutorial group – even with the ambiguities it’s pretty much guaranteed to lead to a discussion of interesting physics (radiation, energy, thermal expansion, you name it).

    I’ve got a few more like this, but I’ll lay off for a while before posting another…

  28. Anton Garrett Says:

    The art in setting such questions is in correctly specifying what you can and can’t assume without giving away the physical effect that the answerer is meant to take into account.

    • telescoper Says:


      One could set this up as a traditional problem by specifying:

      The mass and radius of the balls;
      The specific heat capacity of the material they’re made of;
      The coefficient of thermal expansion of the material they’re made of;
      The value of g…


      But that’s too many hints, I think. Incidentally, for steel the second number is about 500 (J/Kg/K) at room temperature while the third is about 10^-5. The P.E. versus thermal energy loss ratio with these numbers is pretty small for reasonably sized balls, i.e. you need very big balls to make it a large effect.

    • Daniel Mortlock Says:

      So you’re saying it depends on who’s making the measurement?

  29. Brendan Says:

    I liked this question; I was originally thinking along John’s line of reasoning, but couldn’t figure it out so ruined it for myself by turning to Google for the answer.

    “i.e. you need very big balls to make it a large effect.”

    Clearly we just need an experimentalist with big enough balls to write the grant application.

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