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Thermodynamic cycle related problem

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data

    The problem is as following:

    2. Relevant equations

    [itex]\Delta[/itex] E = [itex]\Delta[/itex] Q-[itex]\Delta[/itex] W

    3. The attempt at a solution
    Now I know that if work is done by the system then work sign is negative and that if volume remains unchanged then no work is done.

    Am I right to say that Internal energy @ C is 50 joules ?
    How does that answer my question ? I know over all energy is conserved in cycle.

    Any tips please.
    Edit i see that the screenshot hasnt loaded i will edir it once i get back home
    Last edited: Oct 9, 2011
  2. jcsd
  3. Oct 9, 2011 #2
    I have updated the main post.

    I think i have worked out solution to this problem.
    Given function are:

    [itex]\Delta[/itex] Q ABC = 80J.
    During this cycle 30j of work is done by the system. Therefore Internal energy at this cyle will be : [itex]\Delta E[/itex] = Q -W = 80-30 = 50j. Ec >EA by 50 joules ?

    Now to answer part b)
    It's given that work done by the system from A- > D = 20j. Heat entering at cycle DC = 20k aka internal energy. From previous part we know qabc = qab +qbc =70j so we can re arrange this to give us Qbc = 80-10 = 70j.
    From A->D 70 joules of heat enters.

    For the last answer where it asks for the comparison of internal energies I have got 50joules again.

    Can someone please criticize my working and shed some light over this.

    Thank you.
    Last edited: Oct 9, 2011
  4. Oct 9, 2011 #3
    Part (a) is Ok
    Part (b).... why is heat entering during DC be 20k when problem says it is !0?
  5. Oct 10, 2011 #4
    Sorry I mean't 10j. I don't know why I keep stating in wrong numbers which even the question doesn't mention. So what am I doing wrong for part b ? :/
  6. Oct 10, 2011 #5
    Sorry for my uprupt disappearance last time due a total power failure in our area.

    Note that if the transfer of heat energy from D to C is 10J , this DOES NOT mean that the transfer of h.e. from A to B is the same. This is because though the processes DC and AB are both at constant volume, yet they occur under different conditions of P and T.

    You may try using the 1st law of thermodynamics for AD and DC.
    Last edited: Oct 10, 2011
  7. Oct 10, 2011 #6
    thanks for your reply. As you have pointed out it's senseless of me to assume that heat at DC is similar to AB.
    Am I right to say that Qabc - 80j (given) then QADC= 80j too ? [ I get 70 joules as the answer of heat entering into phase AD. Since QADC = AD+ DC =80J ,where DC =10.

    What am I missing ?
    Last edited: Oct 10, 2011
  8. Oct 10, 2011 #7
    if QABC = 80J then QADC MAY NOT be equal to it because process ABC occurs under different P, V values from process ADC.

    But, since the internal energy U of an IDEAL gas depends only on T and the initial and final temperatures of processes ABC and ADC are the same, then the change in INTERNAL ENERGY U is the same for ABC and ADC.

    As I said before try using the !st law for AD and for DC.
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