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Entropy Of Mixing and Specific Heat

  1. May 18, 2013 #1

    morrobay

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    1. The problem statement, all variables and given/known data
    100g of lead, specific heat .0345 cal/g/C at 100 C is mixed with 200 g of water at 20 C.
    Find the difference in entropy of system at end from value before mixing.
    Note: I think the book has wrong answer of 1.42 cal/K



    2. Relevant equations
    Before mixing entropy for water = 4000 c/293 K = 13.65 c/K
    Entropy for lead = 100g * .0345 c/g/C * 100C = 345 c/373K = .925 c/K
    Total before mixing = 14.57c/K


    3. The attempt at a solution
    The lead loses 276 calories going from 100 C to 20 C
    So now the water temp is 4276 c/ 200g * 1c/g/C = 294.38K entropy =
    4276 c/ 294.38K=14.52c/K
    For the mixed in lead = 69 c/.0345 c/g /C= 2000gC /100g= 293K S = 69c/293k = .23c/K
    So total after mixing = 14.75c/K
    ΔS = 14.75c/K - 14.57c/K = .18c/K
    I think the book got the wrong answer with : 4345 c/293 K = 14.82 c/K
    And 345 c/293K = 1.177 c/K for total 15.99 c/K - 14.57c/K = 1.42 c/K
     
    Last edited: May 18, 2013
  2. jcsd
  3. May 18, 2013 #2

    haruspex

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    Will the lead go to 20C?
     
  4. May 18, 2013 #3

    morrobay

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    No it will go to the water temp. 294.38K, 21.38 C losing 271.2 cal, Did the lead first lose 276c then gain back 4.8 cal ? - Some confusion on this mixing of specific heats since Im reading that the lead is in powder forum.
    78.62 C * .0345 c/g/C * 100 g = 271.2 cal. For entropy value of 73.8 c/294.38K =.25c/k This still does not get close to the book answer of ΔS = 1.42 c/K for after mix - before mix. By the way the book only gave answer. I just plugged in those values in trying to determine how they got that by error. But it gave the after mix value 15.99 c/K - before mix 14.57c/K = 1.42c/K (book)
    But it looks wrong since the 345 calories are applied twice
     
    Last edited: May 18, 2013
  5. May 18, 2013 #4

    haruspex

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    I get a different answer again. I don't believe this is valid:
    You cannot say there is some absolute quantity of entropy in a body. You can only discuss changes in entropy. If a body with heat capacity H cal/C drops in temperature from T to T-δT (absolute) then it loses HδT of heat and HδT/T of entropy. Integrating, ΔS = H ln(T1/T0). If I apply that formula to this problem I get a net increase in entropy of 0.107 cal/K.
     
  6. May 19, 2013 #5

    morrobay

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    It was my understanding that in a case like this for water and lead that S = Joules/K
    So I dont understand why 4000c/293K & 345c/373 K are not valid before mix.
    After the mix there is a total of 4345 cal in 300 gr. mixture.
    With 200 g having Cp =1 c/g/C and 100 gr with Cp = .0345 c/g/C
    This problem and answer of S2 - S1 = 1.42 c/K are from my Halliday Resnick textbook.
    Maybe the problem is on what the temperature is in S2 from above.
    Could you work that ? thanks
     
  7. May 19, 2013 #6

    rude man

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    Luckily I happen to have the Resnick & Halliday text. I looked up the problem and its answer. The OP correctly described the problem but the answer is +0.11 cal/K.

    I computed +0.13 cal/K but that is undoubtedly due to my approximate temperature calculations. So all is well.
     
  8. May 19, 2013 #7

    haruspex

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    So the OP quotes the wrong book answer? Different edition perhaps?
    0.11 fits with my 0.107, so are we agreed that the ln() equation I derived is the right one to use?
     
  9. May 19, 2013 #8

    rude man

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    Yes.

    I checked the answer carefully, and it's 0.11. I assume the OP just picked the wrong problem answer although none of the answers in the relevant chapter (Chapt. 25 in my edition) were anything near what the OP quoted (1.42).
     
  10. May 19, 2013 #9

    morrobay

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    Yes this is chapter 25 problem #15 in a very early combined 1st & 2nd edition 1960.
    I got in a used bookstore in Thailand. Since Im out of the U.S. The correct answer in book .1cal/K. was in error in the #17 answer place.
    With T2 = 294.36 K I got .11c/K for S2-S1 = mCp ln T2/T1

    Problem # 17 : A brass rod is in contact thermally with a heat reservoir at 127°C at one end
    and 27°C at the other end. What is ΔS for the conduction of 1200 calories through the rod,
    end to end ? Is this correct ? ΔS = (1200 cal)( ln 400K/300K) = 345c/K
    Not close to 1.42 c/K ?
     
  11. May 20, 2013 #10

    haruspex

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    No, this is different. The entropy change is in general ∫dQ/T. In the OP, T varied with Q, so you got the ln() formula. This time T is constant.
     
  12. May 20, 2013 #11

    rude man

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    What is delta S for the cold reservoir?
    What is delta S for the hot reservoir?
    What is delta S for the brass bar?
    Add them up!

    You need to think about why this is not a dQ/T integration! More importantly, you need to think about why the other problem was an integration. What were you physically doing in the other problem? Remember, the integral has to be along a reversible path.

    PS I did not get 1.42 cal/K for this problem either. @Haruspex?
     
    Last edited: May 20, 2013
  13. May 20, 2013 #12

    haruspex

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    No, but at least it's the right order of magnitude this time.
     
  14. May 20, 2013 #13

    rude man

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    Yay team - sort of.
     
  15. May 20, 2013 #14

    morrobay

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    1200 calories = Q
    T1 = 400K
    T2 = 300K

    Q/T2- Q/T1 = 1c/K ΔS

    If this is correct then book has #15 & #17 in error
     
  16. May 20, 2013 #15

    haruspex

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    That's what I get.
     
  17. May 21, 2013 #16

    rude man

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    Bravo!
     
  18. May 21, 2013 #17

    morrobay

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    The first problem was integration since ΔS = ∫12 = dQ/T with T varing is based on ∫12 1/x dx = ln x2/x1
    The second problem with T constant = 1/T ∫12 dQ = Q/T

    Question : In the first mixing problem with a plot of Q on y axis and T on x axis
    and ΔS the area under curve. Would this plot look like the inverse y=1/x curve ?
    By the way I did not mean to imply that the 1960 text was my text in 1960
     
  19. May 21, 2013 #18

    haruspex

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    Since S = ∫dQ/T, for S to be the area under the curve the natural arrangement would be y = 1/T, x = Q. Since, in the first question, Q = kT then yes, that would give a curve like y = 1/x.
     
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