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Thermodynamics-work, heat, & internal energy

  1. Feb 1, 2007 #1
    Thermodynamics--work, heat, & internal energy

    1. The problem statement, all variables and given/known data

    When a system is taken from state i to state f along path iaf in Figure 18-41, Q = 60 cal and W = 20 cal. Along path ibf, Q = 51 cal.

    Fig. 18-41 (see attatched)

    (a) What is W along path ibf?
    (b) If W = -13 cal for the return path fi, what is Q for this path?
    (c) If E_int,i = 7 cal, what is Eint,f?
    (d) If E_int,b = 18 cal what is Q for path ib?
    (e) For the same value of E_int,b, what is Q for path bf?

    2. Relevant equations

    [tex]\Delta E = Q - W[/tex] and is path independant

    3. The attempt at a solution

    (a) [tex]\Delta E_{iaf} = \Delta E_{ibf}[/tex]
    hereon, iaf will be abbreviated a, and ibf will be abbreviated b.
    [tex] Q_a - W_a = Q_b - W_b[/tex]
    [tex] Q_a - W_b + Q_b = -W_b [/tex]
    [tex] 60-20+51=-W_b=91[/tex]
    So W_b = -91 cal. Wrong. And yes the problem wants the answer in calories.
    On the chance that they're giving work performed on the system rather than work performed by the system, I changed Q-W to Q+W, and got that W_b=29 cal, which is still wrong.

    I used the same exact reasoning for part b (and got it wrong).

    I got part C correct [47 cal] using the fact that DeltaE is path independent, and Delta E = final energy - initial energy.

    Parts d and e I just have no idea where to begin. So I guessed 18 cal for both. Surprise! Both wrong...

    Attached Files:

  2. jcsd
  3. Feb 2, 2007 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    a) First of all, I would recommend that you use U instead of E. W and U are both forms of energy, as is heat flow: Q). Singling out internal energy as E suggests that Q and W are something other than energy, which is not true.

    dQ = dU + dW (where dW is the work done by the gas: PdV)

    We know from the data given that the area under the path from a to f (dW, the work done by the gas in following the iaf path) is 20. We also know that this required a heat flow, dQ, into the gas of 60. So the change in internal energy, dU from i to f is 40, as you appear to have found.

    Since, as you correctly stated, internal energy path independent, we know that following ibf, the change in U is the same: 40. So work done is just dQ - dU, where dQ is now the heat flow into the gas following ibf (51): dW = dQ - dU = 51-40 = 11 Cal.

    I think you will see that in your algebra you just got a sign wrong.

    b) the same reasoning as above applies to b).

    c) just add the change in U to Ui

    d) what is the change in U from i to b? You already know the work done (area under the graph from i to b). So you can get dQ.

    e) Hint: Is there any work done in going from b to f (what is the area under the graph from b to f)?

    Last edited: Feb 2, 2007
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