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B How possibly can heat flow between equally hot bodies?

  1. Oct 2, 2015 #1
    An ideal Carnot engine is composed of two reservoirs and a working fluid. The hot Reservoir and the cold one have temperatures ##T_1## and ##T_2## respectively, with ##T_1>T_2##. The working fluid is in a phase transition and has temperature ##T_1## at the start of the Carnot cycle. It undergoes another phase transition at ##T_2## at the end of the cycle to return to its original state.

    This is a P-V diagram of the Carnot cycle which proceeds in fours steps:


    I'm particularly interested in the two stages (from 1 to 2) and (from 3 to 4). They can be described as follows:

    1) Stage (from 1 to 2) is an reversible isothermal expansion of the working fluid to transform from the liquid state to the gaseous one. The working fluid is at ##T_1## and it happens to have boiling point at ##T_1##; Hence heat ##Q_1## is supplied to the fluid from the hot reservoir until it transforms to a gas keeping its temperature constant along the whole process.(that the fluid's temperature is constant during the whole process is owing to it being in a phase transition).

    2) Stage (from 3 to 4) is an reversible isothermal compression, and its similar to what we have just described, with the difference being in this case, heat ##Q_2## is drawn out of the fluid and transfers to the cold reservoir and the fluid transforms from gas to liquid retaining a constant temperature of ##T_2## throughout the whole process

    I'm puzzled by the mechanism by which the working fluid undergoes phase transition. So at stage (from 1 to 2), both the fluid and the hot reservoir have the exact temperature, so that they're in a thermal equilibrium; Hence there should be no heat or energy exchange between the two bodies. The same can be said of stage (from 3 to 4).

    So how is it possible for heat to flow from two bodies having the exact same temperature?
  2. jcsd
  3. Oct 2, 2015 #2


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    Welcome to PF!
    Since you are correct that heat transfer cannot occur if there is no temperature difference, clearly there must be a temperature difference. What you are seeing in the diagram and description of the cycle is the temperature of the working fluid only. It tells you very little about the temperatures of the hot and cold reservoirs (only that they must be hotter and colder, respectively, than the evaporation and condensation stages, respectively).
  4. Oct 2, 2015 #3
    I beg to differ. Currently I'm reading Carnot's original paper: reflections on the motive power of fire.

    Here's a pdf link to it.

    In this paper, Carnot introduces the Carnot engine and Carnot cycle for the first time. And he assumed that the hot reservoir has the same temperature ##T_1## as the fluid. And the cold reservoir has the same temperature ##T_2## as that of the fluid after the fluid expands adiabatically. (Check from page 52 to 54 in the linked pdf).
  5. Oct 2, 2015 #4


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    Staff: Mentor

    Yeah, you're right. The Carnot cycle presents an ideal/limiting case for efficiency. The larger you make your heat exchangers, the closer you can get to having the temperatures the same, but you'll never quite get there. That would be part of the reason the ideal case is not achievable.
  6. Oct 4, 2015 #5
    In step 1-2, the reservoir temperature is slightly higher than the working fluid temperature and in step 3-4, the reservoir temperature is slightly lower than the working fluid temperature.

  7. Oct 12, 2015 #6
    Your PV diagrams for the isothermal processes are wrong. What you drew are the PV diagrams for an ideal gas. For a gas undergoing a phase transition from liquid to gas, the ideal gas law does not hold, and the shape of the isotherm is completely different. In fact, for the process you are interested in, the temperature AND the pressure are constant. There is a change only in volume as the liquid expands and becomes a gas. The heat absorbed by the system is the heat of vaporization. Similar statement is true for the process at T2.
  8. May 29, 2016 #7
    For the time being consider Carnot's cycle is possible.
    In Carnot's cycle: The cycle is done with piston cylinder; for adiabatic part insulation is made for the piston and the cylinder.

    Cylinder is made to contact with thermal reservoir (infinite heat capacity). And gas is obvious below the reservoir's temperature.
    As soon as temperature of gas inside cylinder becomes equal to the reservoir's temperature the piston moves and gas temperature become low. What is the consequence?? some thermal energy is transferred from reservoir to make the gas temperature again reservoir's temperature. So process seems to be Isothermal not because of heat transfer is occurring due to infinitesimal temperature difference.

    So why this Isothermal part is not possible; process need to made too slow; generally heat transfer with expansion or compression is polytropic not Isothermal.

    How could you maintain the thing that expansion will only happen when temperature of gas and reservoir becomes equal.

    But we all are sure at one point: Heat transfer needs temperature gradient.
  9. May 29, 2016 #8
    See my post # 5.
  10. May 29, 2016 #9
    I have seen your post #5
  11. May 29, 2016 #10
    Dear Omar Nagib,

    In January 3th 2013, Dr. Ulrich Schneider and a group of researchers at the Ludwig Maximilian University in Munich, Germany published a wonderfull paper, describing the creation of negative absolute temperatures quantum gas made of potassium atoms. Such systems would behave in strange ways, says Achim Rosch, a theoretical physicist at the University of Cologne in Germany, who proposed the technique used by Schneider and his team. For instance, at purely positive temperatures, the colder medium inevitably heats up in contrast, therefore absorbing a portion of the energy of the hot medium and thereby limits the efficiency. If the hot medium has a negative temperature, it is possible to absorb energy from both media simultaneously. The work performed by the engine is therefore greater than the energy taken from the hotter medium alone – the efficiency is over 100 percent.

    Rosch and his colleagues have calculated that whereas clouds of atoms would normally be pulled downwards by gravity, if part of the cloud is at a negative absolute temperature, some atoms will move upwards, apparently defying gravity.

    If such system one day could be construed, perhaps we could leave Earth easyly and cheaply, opening the way to the practical colonization of space.

    You can read the paper in Nature magazine (10.1038/nature.2013.12146) or in https://www.sciencedaily.com/releases/2013/01/130104143516.htm
  12. May 29, 2016 #11


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    If I remember correctly the reservoirs are considered "ideal" - zero thermal resistance between the reservoirs and the fluid - so that you don't need much temperature difference (approaching zero) to cause infinite power to flow from the reservoir to the fluid. So effectively you can assume the reservoir and fluid are at the same temperature.
  13. May 29, 2016 #12
    How does that relate to what you said in post #7?
  14. May 30, 2016 #13
    You had put your words precisely that heat transfer will be always from higher to lower temperature body.
  15. May 30, 2016 #14
  16. May 30, 2016 #15
    Correct me sir, if I am wrong
  17. May 30, 2016 #16
    I wan't able to understand exactly what you were saying.
  18. May 30, 2016 #17
    As OP asked "heat transfer between bodies have same temperature".

    I just explained with reservoir it is kind of equilibrium heat transfer which appears isothermal because dT reduction of temperature of fluid due to expansion is immediately compensated by heat transfer from the reservoir. But instantaneously there was temperature gradient to make the heat transfer happen.
  19. May 30, 2016 #18
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