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Work or Heat transfer to system?

  1. Aug 6, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    a) Liquid is stirred at constant volume inside a container with adiabatic walls. The liquid and container are regarded as the system. Is heat being transferred to the system? Is work being done on the system?

    b) Water inside a rigid cylindrical insulated tank is set into rotation and left to come to rest under the action of viscous forces. Regard the tank and water as the system. Is any work done by the system as the water comes to rest? Is there any heat flow to or from the system?

    2. Relevant equations
    First Law of thermodynamics

    3. The attempt at a solution
    I think I have the correct answers but I want to check my reasoning for each case:
    a)Work is being done on the system. The stirring produces a mechanical exchange of energy to the water. No heat flow since the walls are adiabatic. I think you could also say given that the stirrer is in contact with the water, after a time, no object is hotter than the other so this may not be regarded as heat transfer.

    b)'left to come to rest' implies no work is done by the water. The water comes to rest solely by the viscous forces which act as a retarding force to the rotational motion of the water. No heat transfer because the system is insulated.

    Are these ok responses?
     
  2. jcsd
  3. Aug 6, 2013 #2

    rude man

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    That is right, except that the spoon would transfer heat to the water if it's hotter than the water initially, but I don't see why that should be the case. The heat is produced by friction with the container sides and I guess internal friction as well.

    No. How do you account for the rise in internal energy of the water? The kinetic energy of the rotating water is changed to heat. Since there is no heat transferred the rotating water must be doing work on the water. Again, by the 1st law.

    Both a and b are examples of adiabatic dissipation of work into internal energy of the system.
     
  4. Aug 6, 2013 #3

    TSny

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    I think your answer to (b) is correct: W = Q = 0.

    In problems like this where there is bulk movement of the system, the first law should be stated in a general form ΔE = Q - W where E is the total energy of the system including kinetic energy of bulk motion. W is defined as work done by the system (water and container) on the environment. Since the volume of the container is constant, the system does zero work on the environment. Likewise, Q is defined as heat transferred to the system from the environment. Since the walls are adiabatic, Q = 0.
     
  5. Aug 6, 2013 #4

    CAF123

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    Hi rude man,
    I presume since the heat produced by the container sides is within the container, (i.e internal to the system), we do not regard this as heat transfer to the system?

    Mathematically, we have the first law as follows: $$\Delta U + \text{bulk}(KE + PE) = Q + W$$ The right hand side is zero, so this gives ##\Delta U = - \text{bulk}(KE + PE)## i.e the rotational kinetic energy of the water is converted to the internal energy of the water.
    Is it what you meant?
    Thanks.
     
  6. Aug 6, 2013 #5

    rude man

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    Last edited: Aug 6, 2013
  7. Aug 6, 2013 #6

    rude man

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    Correct.
    Yes it is. Absolutely correct.

    @Chet - join in, willya?
     
    Last edited: Aug 6, 2013
  8. Aug 6, 2013 #7

    CAF123

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    Does this not answer your below question?

     
  9. Aug 6, 2013 #8

    rude man

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    I just told you it did! That post was not directed at you.

    Actually, the work is done externally on the system by spinning the water. This is analogous to boring out a cannon (if the cannon was insulated from the environment).
     
    Last edited: Aug 6, 2013
  10. Aug 6, 2013 #9

    TSny

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    CAF123 has answered this.

    It's like a system where you have a closed box with adiabatic walls and fixed volume. Inside the box is a block of wood that is initially sliding along the floor of the box (call this the initial state). The block is brought to rest by friction between the block and the floor of the box (call this the final state).

    We define the system to be the block + the box. No energy is transferred to the system as heat or work during the time that the system passes from the initial state to the final state. Q = W = 0. The first law applied to this system implies that the total energy E of the system remains constant. You can still have energy transformations taking place inside the system. The internal energy U of the block and walls will increase as the KE of the block decreases.
     
  11. Aug 6, 2013 #10

    rude man

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    There is work done external to the system here also. It's the work done by gravity on the block as it slides.
     
  12. Aug 6, 2013 #11

    TSny

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    I am assuming the block is sliding horizontally along the floor of the box. The work done by gravity will then be zero.
     
  13. Aug 7, 2013 #12

    rude man

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    And what impelled the block to move horizontally? It had to be accelerated via an applied force, and W = ∫F ds.

    EDIT: it's occurred to me that this little exchange could go on forever. Perhaps I can truncate it by a quote from Mark W. Zemansky, Heat and Thermodynamics, 4th edition, p. 178:

    "Processes Exhibiting External Mechnical Irreversibility:

    Those involving the adiabatic dissipation of work into internal energy of a system, such as:

    2. Coming to rest of a rotating or vibrating thermally insulated liquid.


    Note the word "work" in that penultimate sentence.
     
    Last edited: Aug 7, 2013
  14. Aug 7, 2013 #13

    CAF123

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    The answer in the back of my book is that Q = W = 0. It doesn't give any reasons for these answers though in case you wondered.
     
  15. Aug 7, 2013 #14

    CAF123

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    Hi TSny,
    I am wondering how this conforms with what I wrote in (a). There I argued that the stirrer did mechanical work but the volume of the system was not changing. In (b), I suppose the rotating device can be regarded as a stirrer also, but you say there is no work done?

    Thanks.
     
  16. Aug 7, 2013 #15

    ehild

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    Thermodynamics deals with systems in steady state. It means that all its parts are in equilibrium with the surrounding parts in the system. Only in that case are the state variables like pressure, temperature, defined. Internal energy is function of the state variables. And it is the sum of the energies of the random motion of its particles. Internal energy is changed by work done on the system or heat transfer to the system.
    When stirred, external work is done on the fluid. The fluid is set into motion as a whole, and different parts move with respect to each other. It is some kind of regular motion, not random. The fluid as whole has got both momentum (angular momentum) and kinetic energy.
    The system is not in equilibrium. If you divide the fluid to small parts, these parts move with respect to each other, change their shapes, and exert force to each other, exchanging momentum and energy . After the stirring stopped the energy of the whirling parts of the fluid is transformed to that of the random motion of the molecules and thermal equilibrium is established. The liquid reaches a new state, with higher internal energy as before. The change of internal energy was caused by the work of the stirrer. At the same time, the adiabatic walls prevented heat exchange.

    In case b, there was work done on the system when the water was set into rotation. Till it moves on, it has angular momentum and kinetic energy as a whole, its thermodynamic state can not be specified. Isolated from the surroundings, the energy transforms to that of random motion of the constituents, thermal equilibrium is set up. You can not apply the laws of thermodynamics during the process leading to equilibrium, when the parts of the fluid interact with the other parts and with the wall, exchanging momentum and energy.

    ehild
     
    Last edited: Aug 7, 2013
  17. Aug 7, 2013 #16
    Hi Rude man.

    I agree with the analysis of part b. With regard to part a, it seems to me that the main thing that's happening is that stirrer is doing work to deform the fluid (viscous deformation), and the viscous heat generation within the system translates into an increase in internal energy. I regard the stirrer as part of the surroundings. So Q = 0 and ΔU = -W. I don't regard the interaction of the stirrer with the walls of the container as comprising the bulk of the work done by the stirrer.

    Chet
     
  18. Aug 7, 2013 #17

    rude man

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    In the final analysis I suppose one could argue that this is a semantic issue.

    Certainly, the moving liquid is doing work on the system, and when it stops, the internal energy has increased by that amount of work. And obviously it took work to set the liquid in motion. But i suppose one can ignore those work quantities, say that the rotating liquid represents initial kinetic energy, and that there is no work done on the system externally once the liquid is set in motion.

    To me, and to famed thermodynamicist Zemansky, the k.e. is doing work on the system consisting of the liquid and the container. I will leave it at that.
     
  19. Aug 7, 2013 #18

    ehild

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    PΔV = W, work by the system is valid only in quasi-static processes, so slow ones, that the system can be considered in equilibrium. Otherwise, the pressure is not defined.
    You can do work on a system in very different ways. Mechanical work, by moving a piston, but stirring a liquid is also mechanical work. In an electric field, the liquid or gas is polarized, it needs electric work done. A magnet can change the magnetic state of the system, doing magnetic work. Putting the liquid into the microwave oven, the electromagnetic field does work.
    In most cases the external work is easier to calculate than the work done by the system. The process can be fast, and the state variables of the system are well defined only in equilibrium - in quasi-static processes.
    Of course, the water stirred does work on the blades of the stirrer. There is force of interaction between the water and the blades, and the blades displace. There is force and displacement. But how do you know the force the water acts on the blades? At the same time, you can measure the electric power consumed by the stirrer, and you can also estimate its efficiency. It is not difficult to find the work done by the stirrer on the water.
    In case b, the work had been done by setting the liquid into rotation. After that, the parts of liquid do work on each other till equilibrium is set up. There is no external work done. The energy of the rotational motion will transform into energy of random motion of molecules, that is, into internal energy.

    ehild
     
  20. Aug 7, 2013 #19
    From the implied problem description, I agree with your interpretation in bold.

    Chet
     
  21. Aug 7, 2013 #20

    rude man

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    @Chet, thanks for chiming in.
    rudy
     
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