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  • A Study in Themodynamics - For the love of thermodynamics

    Shashank Moghe

    Shashank Moghe

    @shashank-94ap1q
    Updated: Sep 11, 2014
    Views: 2.4K
    I have this habit of delving deeper into seemingly unnecessary details. But to me, it has born more fruit than skimming over the smaller (seemingly unimportant) details. Here is a thought experiment that once occurred to me. I also later discussed with my professor who provided a beautiful insight I had been missing.

    It should serve well for any person fascinated (or may be in love) with Thermodynamics/Physics. Let us discuss ideas/opinions. Do not be bothered to outrightly dismiss any of my ideas either.
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  • Shashank Moghe

    Member • Sep 11, 2014

    *A small correction, I meant PMM in the PDF, not PPM.
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  • Geredd Hayes

    Member • Sep 11, 2014

    i would think that the force of the gas stops the metal from expanding. When the metal and gas are at the same temperature, they would both be under the same pressure.
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  • Shashank Moghe

    Member • Sep 11, 2014

    Geredd Hayes
    i would think that the force of the gas stops the metal from expanding. When the metal and gas are at the same temperature, they would both be under the same pressure.
    Well, when the temperature of the gas and the metal bar is the same, the process would stop. You got that part correct.

    As for your statement on the pressure being the same, I guess that is a given (it has to be equal and opposite) since the metal bar is applying pressure on the gas. No discoveries there.

    This brings us to the next question in the problem- How does the process come to a halt? What brings the system to a situation where the temperatures would be equal?
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  • Geredd Hayes

    Member • Sep 12, 2014

    the energy that it takes for the expanding metal to push the gas is more than the heat energy that is transferred to it from the gas
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  • Shashank Moghe

    Member • Sep 13, 2014

    Geredd Hayes
    the energy that it takes for the expanding metal to push the gas is more than the heat energy that is transferred to it from the gas
    Well, you are so correct there. The process should stop when the energy required for expansion should equal the heat transferred. This ofcourse has to be a process where lesser and lesser quantum of heat is successively transferred to the bar at each time step.
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  • Ramani Aswath

    Member • Sep 29, 2014

    I love thermodynamics as well. It is the bread and butter of ChemEs.
    A closed multi-component system consists of various masses with various specific heats. The total thermal mass is the sum of the products of each mass and its specific heat.
    When a quantity of heat Q is added to the system either in one shot or continuously, the temperature rise is given by:

    (Delta T) = Q/[Sum(m1Cp1+m2Cp2+.......mnCpn)]

    That is all.
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  • Shashank Moghe

    Member • Sep 29, 2014

    A.V.Ramani
    I love thermodynamics as well. It is the bread and butter of ChemEs.
    A closed multi-component system consists of various masses with various specific heats. The total thermal mass is the sum of the products of each mass and its specific heat.
    When a quantity of heat Q is added to the system either in one shot or continuously, the temperature rise is given by:

    (Delta T) = Q/(Sum(m1Cp1+m2Cp2+.......mnCpn)

    That is all.
    Sir, agreed. But under what conditions the system comes to an equilibrium, that is the question here (although it should be extremely simple for you). It might serve as a good exercise for the newly introduced mind, of course a seasoned engineer like you can crack it in a second 😀

    But I am fascinated how much Thermodynamics can help us learn 😀
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  • Ramani Aswath

    Member • Sep 29, 2014

    Even assuming that the entire heat was transferred to the gas initially, as soon as it tries to heat the metal its temperature starts to fall. The second law ensures that the receiving object will always be at a temperature lower than the giving object.

    Try replacing the metal by another gas and separating the two by a super elastic membrane and consider what would happen. The ideal gas law will result in just a pressure increase given by (delta P) = n x R x (delta T) the delta T itself given by the equation in my earlier post.
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  • Shashank Moghe

    Member • Sep 29, 2014

    A.V.Ramani
    Even assuming that the entire heat was transferred to the gas initially, as soon as it tries to heat the metal its temperature starts to fall. The second law ensures that the receiving object will always be at a temperature lower than the giving object.

    Try replacing the metal by another gas and separating the two by a super elastic membrane and consider what would happen. The ideal gas law will result in just a pressure increase given by (delta P) = n x R x (delta T) the delta T itself given by the equation in my earlier post.
    Sure the temperature would fall on the gas. But having the other material as a metal with a very high Coefficient of Thermal expansion is what makes it tricky. The metal wants to expand now, thus compressing the gas and raising its temperature further.
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  • Ramani Aswath

    Member • Sep 29, 2014

    Shashank Moghe
    Sure the temperature would fall on the gas. But having the other material as a metal with a very high Coefficient of Thermal expansion is what makes it tricky. The metal wants to expand now, thus compressing the gas and raising its temperature further.
    Thermodynamics is quantitative. gases have the highest thermal expansion coefficient. Please think through your problem using two gases (the worst case).
    Q = mCp(delT) and pv = nRT will give the solution.
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  • Shashank Moghe

    Member • Sep 29, 2014

    Sir, I am aware of that. That is why this is a thought experiment. Which, even though it is, has a unique solution.
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  • Ramani Aswath

    Member • Sep 29, 2014

    There is nothing wrong with a thought experiment. However, the phenomena will still be restricted by the laws of thermodynamics. Take it that your metal has a TEC equal to a gas. Think the experiment through. Assuming the gas gets all the heat initially, its pressure increases as also the temperature. When it transfers heat to the metal it cools by a matched amount and the pressure drops. The metal expands by the same volume and restores the pressure. The total volume remains the same.
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  • Shashank Moghe

    Member • Sep 29, 2014

    Well, there is nothing that should stop it from compressing the gas beyond the initial pressure. Of course, if you apply the second law, entropy considerations will clearly tell you that it would be impossible/possible, but for that you would need actual data, which we do not have. Hence I said, assuming the TEC is extremely high, let us think the experiment through. Certainly, pressure being restored to the initial value is not the restricting factor here.
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  • Ramani Aswath

    Member • Sep 29, 2014

    TEC cannot be very high if a metal has to remain a solid becausse of increasing intermolecular distance. Also one has to take into account thermal masses without which meaningful conclusions cannot be drawn. While heat transfer requires a temperature difference the resultant values dpend on the thermal masses of the constituents.
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  • Shashank Moghe

    Member • Sep 30, 2014

    Of course thermal masses are required. I was just saying that the pressure restoration, as you said, is not the restricting factor to this process. That is not what shall bring the system to a thermodynamic equilibrium.
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