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Heat transfer question

  1. Sep 14, 2007 #1
    this question was posed by someone on a computer forum and is probably easy for some of you, so here goes :

    imagine water in an open container which is perfectly insulated on the sides and bottom. the water is stationary and heat is being added from the top (assume that the surroundings above the water are at a much higher temperature than the water itself). eventually, will the complete mass of water reach a higher equilibrium temperature? if so, what mechanism of heat transfer causes this to happen?
  2. jcsd
  3. Sep 14, 2007 #2
    Thats wrong on many levels. First of all, the 'mass' of water wont reach one temperature, there will be a temperature gradient across the water. Second, the water will reach an equilibrium temperature. That temperature wont be higher than the source temperature, or heat would flow back into the surroundings. This is all done by conduction and convection.
  4. Sep 14, 2007 #3


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    You may want to reread the problem statement. The container is insulated on the bottom and sides, no heat loss implies that the water temperature will stabilize at the same temp as the air.

    The mechanism is conduction.
  5. Sep 14, 2007 #4
    i guess my initial post was a little ambiguous. when i said "higher equilibrium temperature", i meant higher than the original water temperature before heat addition, i didn't mean higher than the ambient.

    why will there be convection? from my limited knowledge, convection is a result of varying densities, however since heat is added from the top, the colder denser fluid stays at the bottom and lighter warm fluid stays at the top.

    will there still be a gradient when time equals infinity or will the entire mass of water be the same temperature as the surroundings?

    i'm asking because i don't know, so correct me wherever i'm wrong.
  6. Sep 14, 2007 #5
    Right, the top surface of the water will be at the same temperature as the air; however, the sides wont. The sides will be at room temperature, because its a perfect insulator. You will have a thermal gradient in the water, and hence convection inside the water. Conduction at the top water/air interface.
  7. Sep 14, 2007 #6
    I think the sides will be at the same temperature of the water.

    And you forgot radiation and absotions of this radiation in/of the water. ( and reflection and absortion of the radiation at the sides and bottom of the container )
    And if the container is open the water will evaporate.
  8. Sep 14, 2007 #7
    Radiation will be negligible compared to everything else alvaros. Maybe 1% in magnitude or less. If the water at the insulator interface were the same temperature as the heated interface, there would be a temperature jump from room temperature to air temperature at that boundary, and hence an infinte slope jump. This makes no physical sense.
  9. Sep 14, 2007 #8


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    Is this a practial question of a gedanken experiment?
    In a real experiment you could simply use a thermos bottle with an insulation vacuum as the container. In this case the inner sides of the container will be, to a very good approximation, at the same temperature as the water and the outer sides the temperature of the room.
    So in a real experiment the "dynamics" would be almost complettely governed by the flux of energy at the surface.
  10. Sep 14, 2007 #9


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    Trying to define what you mean by "the temperature of a perfect insulator" is more about philosophy than engineering. And it does't really matter, because if it's a perfect insulator there is no heat flow whatever its temperature is.

    Re the "temperature jump" between the air and the water, it does make sense in terms of a kinetic theory model. The average KE of the the gas molecules can be different from the average KE of the water molecules, therefore the temperatures at the interface can be different - though different temperatures are not in equilibrium of course. If the temperatures at the interface were the same (on a practical size scale), Newton's law of convective cooling wouldn't make any sense.
  11. Sep 17, 2007 #10
    Can you explain that some more Aleph? It seems to me that a jump in temperature at the boundary would mean an infinite dT/dx. The only way to make delta(E) zero at the boundary is if k=0. This seems shady to me as you would have 0*infinity, which is not really zero mathematically since its indeterminate.

    Even if it was a really *REALLY* good material with k=10^-23, made from unobtanium, the fact that dT/dx is infinite would make the delta(E) blow up.

    I can only see the solution making any sense if the boundaries are at room temperature and the top surface is at the source temperature. That temperature difference at each face would cause convection due to the different densities.
  12. Sep 18, 2007 #11


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    With convection, there is flow in the fluids. Think of hot solid object in cold air. The air molecules get heated by colliding with the surface and rebounding with higher energy. It doesn't follow that each individual molecule heats up the same temp of the solid in just one collision, and even if it did it would cool down when it collided with another colder air molecule. The *average* temperature of the air, very close to the surface, is less than the surface temperature of the solid.

    That's what Newton's Law of Cooling says: across the interface, heat flux Q = h(T_1-T_0) where h is the heat transfer coefficient. H which depends on the surface finish, the flow configuration, and various nondimensional fluid parameters like Reynolds and Prandtl numbers.

    This is different from conduction within a solid (or between two solids in contact) where the atoms only vibrate, they don't move around on a global scale. For solid conduction agree with you, the temperature gradient has to be finite and the temperature distribution has to be continuous. At least, that's true for the "classical" model of heat flow - at the atomic scale, considering the temperature (or KE) of individual particles is a different story.

    For a solid that is an absolutely perfect insulator (K = 0, not 10^-23) then the diffusion/conduction equation just says dT/dt = 0 at every point. So there can (conceptually) be any temperature distribution you like in the solid, and it never changes. There's no reason why dT/dx has to be continuous because dT/dx doesn't come into the equation (or it's multiplied by 0, if you prefer). That's what I meant by saying the concept of "temperature in a perfect insulator" is not very well defined (or useful).

    But of course there is no such thing as a perfect insulator.

    The question of what happens in convection between a perfect insulator and a fluid is another "what is 0/0" type question I think. There can't be any finite amount of heat flux going into or out of the solid, because the heat can't go anywhere inside the solid. I think all you can really say is dT/dn is zero at the boundary of the fluid (where n is the normal vector) and the temperature in the fluid can be whatever it wants to be. There can be heat transfer along the surface, inside the fluid, so it doesn't mean the fluid temp has to be constant over the whole boundary.
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