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Homework Help: Heat Thermodynamics Questions

  1. Jan 25, 2010 #1
    Alright so my friends and I have been trying really hard on our new assignment, and there are some problems that we still can't do. It's like no matter how hard we try the book doesn't explain it well enough for us to be confident in our answers. So Pleeease help!

    1. A mixture of hydrogen and oxygen is enclosed in a rigid insulating container and exploded by a spark. The temperature and pressure both increase. Neglect the small amount of energy provided by the spark itself.
    A) Has there been a flow of heat into the system.
    B)Has any work been done by the system.
    C)Has there been any change in internal energy U of the system?

    2.The water in a rigid. insulated cylindrical tank is set in rotation and left to itself. It is eventually brought to rest by viscous forces. The tank and the water constitute the system.
    A)Is any work done during the process in which the water is brought to rest?
    B)Is there a flow of heat?
    C)Is there any change in the internal energy U?

    3.Compressing the system represented in the figure along the adiabatic path a-c requires 1000J. Compressing the system along b-c requires 1500J but 600J of heat flow out of the system.
    A)Calculate the work done, the heat absorbed, and the internal energy change of the system in each process and in the total cycle a-b-c-a.
    B)Sketch this cycle on a P-V diagram.
    C)What are the limitations on the values that could be specified for process b-c given that 1000J are required to compress the system along a-c

    1.A)There is no heat flow because the system is insulated.
    B)Yes adiabatic work is done. (I'm not sure why though)
    C)Yes the change in internal energy is equal to the adiabatic work done (I think it's an increase in energy.

    2.A)Yes there is dissipative work caused by the viscous forces.
    B)There is no flow of heat because the system is insulated.
    C)Yes the change in internal energy equals the dissipative work (decrease in energy?)

    3.A)For a-c I wrote there is 1000J work done, no heat flow, and 1000J change in energy.
    For b-c there is 1500J work done, 600J heat flow, 2100J change in energy.
    For a-b-c-a there is no heat absorbed, 600J work done, no change in internal energy.
    I haven't yet sketched the diagram but I am not sure what to do for the limitations on the values that could be specified for b-c given that 1000J are required to compress the system along a-c/

    Diagram for 3: nn4f9e.jpg


    Thanks for the help :D
     
  2. jcsd
  3. Jan 25, 2010 #2
    Anyone?
     
  4. Jan 26, 2010 #3
    My old mechanics teacher always used to tell us "draw a box round it" , look at your problems draw a boundry round them and ask "does anything cross the boundry?"
     
  5. Jan 26, 2010 #4
    Can anyone help clarify these?
     
  6. Jan 26, 2010 #5
    Take your first question.
    Draw a boundry round the the container, does anything pass across the boundry?
    If nothing crosses the boundry going in no heat can have entered the system.
    If nothing crosses the boundry going out the system has not done any work.
    If the system has performed no work and no heat has entered the system can the internal energy have changed?
     
  7. Jan 26, 2010 #6
    Ok got it thanks a lot!. I figured out the third question, but in the case of the "boundary", that would make my second answer correct right?
     
  8. Jan 26, 2010 #7

    fluidistic

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    I'm confused about this. Because the problem states that there is an increase of temperature. Plus, the internal energy of an ideal gas is directly proportional to the temperature. Hence an increase of temperature should mean an increase of internal energy. But as you implied, according to the first law of thermodynamics, it cannot be. So what's going on?
     
  9. Jan 26, 2010 #8
    fuidistic
    Heat isn't the only form of energy.
     
  10. Jan 26, 2010 #9

    fluidistic

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    Hmm I know (for instance work), but how does it matter here?
    I don't want to hijack the thread, so if you want to reply me by PM, it's ok.
     
  11. Jan 26, 2010 #10
    fluidistic
    If you accept that the thermal energy of the system has increased (the gas has got hotter) and that there has been no energy imported into the system, what else might have happened.
    Work is heat and heat is work is a true statement but NOT excusivly true.
     
  12. Jan 27, 2010 #11

    fluidistic

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    I accept that the thermal energy (temperature) of the system has increased and that no energy has been imported into the system. It is explicitly said that
    I also agree that there is no work done since the volume is constant. Hence one might say that the variation of internal energy of the system is 0, according to the first law.
    My problem is looking at the definition of the internal energy of an ideal gas: it depends on temperature. So a hotter gas means a gas with a greater internal energy.
    But how can that be? It seems to be in contradiction with the first law of thermodynamics here.
    Do you understand where I'm misunderstanding?
     
  13. Jan 27, 2010 #12
    fluiodistic
    What happens when hydrogen and oxygen are mixed and a spark applied?
     
  14. Jan 27, 2010 #13

    fluidistic

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    It forms water if I remember well. (from high school since I've no chemistry in my physics degree).
    Hmm ok... But to be rigorous, don't you think we'd have to show that it is under the liquid form? Because if it would be a gas, it seems the first law of thermodynamics couldn't be applied. Is it sufficient to affirm that it is under the liquid form otherwise the first law wouldn't be right?
     
  15. Jan 27, 2010 #14
    It forms water but what is given off during the reaction?
     
  16. Jan 27, 2010 #15

    fluidistic

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    I don't remember. Heat I guess.
     
  17. Jan 27, 2010 #16
    Hydrogen and oxygen form water and release energy. Having said that I hope my second answer is right...
     
  18. Jan 27, 2010 #17
    OK lets wrap this up. The hydrogen and oxygen turn into water in its gaseous phase (steam). The energy released by the chemical reaction provides the heat to raise the temperature and pressure, nothing needs to enter the system as energy is changed from one state to another.
     
  19. Jan 28, 2010 #18

    fluidistic

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    Ok thank you very much. So as in this (https://www.physicsforums.com/showthread.php?p=2552092#post2552092) thread says, there is the chemical energy that Resnick-Halliday never mentioned.
     
  20. Jan 28, 2010 #19

    Andrew Mason

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    A) There is heat flow into the system. The post-explosion system is definitely at a higher temperature than the pre-explosion system (higher internal energy) and there is no work done (volume remains constant). Apply the first law to see that Q cannot be 0: [itex]\Delta Q = \Delta U + W[/itex]

    B) If there is no change in volume, how can there be any work done?\\

    C) The change in energy is equal to the heat flow.


    AM
     
  21. Jan 28, 2010 #20

    fluidistic

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    Dear Andrew Mason,
    I think you are wrong by saying that there was a heat flow. I don't know if you have read the previous posts, in case of yes, please correct Jobrag and I. And I also suggest you to read https://www.physicsforums.com/showthread.php?p=2552092#post2552092 where Matterrave suggest to use a more generalized form of the firs law to explain the increase of temperature of the gas when both the work done by the gas and the heat flow are null.
     
  22. Jan 28, 2010 #21

    Andrew Mason

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    The heat flow from chemical reaction is referred to as the [itex]\Delta H[/itex] or change in Enthalpy of the reaction and is equal to the change in internal energy + work done by the thermodynamic system. This is just an application of the first law of thermodynamics to chemical reactions. The law is unchanged. There is no more general statement of the first law. The thread you referred to is incorrect if it suggests that [itex]\Delta Q = \Delta U + W[/itex] is not sufficient. For chemical reactions, one has to take into account the number of molecules before and after and the enthalpy of the reaction when determining the internal energy, of course. But one is still measuring [itex]\Delta U \text{ and } \Delta Q[/itex]. See: http://en.wikipedia.org/wiki/Enthalpy and http://en.wikipedia.org/wiki/Chemical_thermodynamics

    AM
     
    Last edited: Jan 28, 2010
  23. Jan 28, 2010 #22

    fluidistic

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    Ok thanks a lot for the information (I think it was necessary, since we were all wrong before you came it seems!). I didn't realize the number gas molecules would go down.
    Also, I'd like to understand better how it is possible that there's a heat flow into the system if the system is insulated.
    What about if I imagine a system with nothing out of it, i.e. the system is the whole universe?
     
    Last edited by a moderator: Apr 24, 2017
  24. Jan 28, 2010 #23

    Andrew Mason

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    The system is only insulated from outside heat flows. It is not insulated from internal heat sources. The internal heat sources are the molecules of gas, which generate heat from a chemical reaction. The additional internal (kinetic) molecular energy did not exist before the chemical reaction occurred. It resulted from the conversion of electromagnetic potential energy into molecular kinetic energy. This could be thermodynamically equivalent to sticking a battery and a switched heating element inside a container filled with an inert gas and then closing the switch. The fact that the container is insulated from the outside does not mean there is no heat flow into the gas.

    AM
     
  25. Jan 28, 2010 #24

    fluidistic

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    Thank you very much. You've been extremely helpful.
     
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