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Homework Help: Analyze a Thermodynamic Cycle

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data


    I have filled in a couple of spots as seen below, but I am a little confused as to how to proceed. If I get a good hint on one, I am sure the rest will follow. I think I am overlooking something obvious here.

    I know that in a cyclic process, all E (internal) will sum to zero. Does anything else sum to zero?
    2. Relevant equations Energy Balance equations

    3. The attempt at a solution

  2. jcsd
  3. Sep 21, 2008 #2
  4. Sep 21, 2008 #3
  5. Sep 22, 2008 #4

    Andrew Mason

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    You have to apply the first law:

    [tex]\Delta Q = \Delta U + W[/tex] where W is the work done BY the system.

    I am not sure what [itex]\Delta E[/itex] in the tables refers to. Is it the total change in mechanical energy of the system (eg. a piston)?

    For 1-2 and 2-3, [itex]W = \Delta Q - \Delta U[/itex] = ?
    For 3-4, [itex]\Delta Q = \Delta U + W[/itex] = ?

    For 4-1, since we know that this is a cycle, what is the sum of all the [itex]\Delta U[/itex]s from 1-2, 2-3, 3-4 and 4-1? That will tell you what [itex]\Delta U[/itex] from 4-1 is.

  6. Sep 22, 2008 #5
    It's the total change in energy (kE+Pe+U). Thanks for the hint! I completely overlooked that! The First law says that the change in

    internal Energy=Q-W

    So that would mean that my work for 1-2 was found incorrectly eh?
    Last edited: Sep 22, 2008
  7. Sep 22, 2008 #6

    Andrew Mason

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    Right. I am still not sure what is meant by PE and KE and how it differs from [itex]\Delta U[/itex] and W. If the system is using a piston, say, to lift a weight, the gain in PE is the increase in gravitational potential energy. But this is also a measure of the work done by the system.

  8. Sep 22, 2008 #7


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    The internal energy (U) is the energy associated with the molecular structure of a system and the degree of molecular activity. The kinetic energy (KE) exists as a result of the system’s motion relative to an external reference frame. The energy that a system possesses as a result of its elevation in a gravitational field relative to the external reference frame is the potential energy (PE). The sum of these is the total energy (E) of the system.

    Most closed systems remain stationary during a process, thus, experience no change in their kinetic and potential energies. In this case the change in the stored energy is identical to the change in internal energy for stationary systems.

    Heat (Q) and work (W) differ in the fact that they are transport mechanisms for energy across a system boundary (i.e. between the system and its surroundings). Systems possess energy, but not heat or work.

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