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First Law of Thermodynamics (Eint = Q-W) question. I am only stuck on part (d)

  1. Sep 9, 2012 #1
    When a system is taken from state i to state f along path iaf in the figure, Q = 60 cal and W = 10 cal. Along path ibf, Q = 65 cal.
    hrw7_18-41.gif
    (a) What is W along path ibf? (b) If W = -19 cal for the return path fi, what is Q for this path? (c) If Eint,i = 11 cal, what is Eint,f?(d) If Eint,b = 27 cal what is Q for path ib?(e) For the same value of Eint,b, what is Q for path bf?
    2. Relevant equations
    Eint= internal energy of the system; Q= heat; W= work
    I have used Eint= Q - W to solve most of the problem.

    3. The attempt at a solution
    I only am marked wrong on part d which is very frustrating. So far I have:
    (a)Wibf = 15 cal because W= Q-E (b) Qfi= -69 cal because Q= -(E)+W (c) Eint,f= 61 cal because Eint,f= Eint+Ei (d) Qib= ? (e) Qbf= 34 because Eint,f =E-Eint,b
    Any thoughts?
     
    Last edited: Sep 9, 2012
  2. jcsd
  3. Sep 9, 2012 #2

    TSny

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    Hi, zag4life. Welcome to PF.

    How does the work for i [itex]\rightarrow[/itex]b compare to the work for i[itex]\rightarrow[/itex]b[itex]\rightarrow[/itex]f?
     
  4. Sep 9, 2012 #3
    Well, from b to f, because the volume does not change, work is equal to zero. The work for i,b,f would be the same as the work for i,b I think.
     
  5. Sep 9, 2012 #4

    TSny

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    Good. So, use that to get Q for ib.
     
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