1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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)?

    AM
     
    Last edited: Feb 2, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Thermodynamics-work, heat, & internal energy
Loading...