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Entropy problem

  1. Oct 26, 2014 #1
    1. The problem statement, all variables and given/known data
    An aluminum can, with negligible heat capacity, is filled with 150g of water at 0 ∘C and then is brought into thermal contact with a similar can filled with 150g of water at 53∘C.
    Find the change in entropy of the system if no heat is allowed to exchange with the surroundings. Use ΔS=∫dQ/T.

    2. Relevant equations
    ΔS=∫dQ/T.
    dQ = mcTdT

    3. The attempt at a solution
    cup1 T = 273 K cup2: T = 53 + 273 = 326 K
    first I need to find the final temp they will both be at

    [itex] Q_{tot} = 0 [/itex]

    [itex] \int dQ_1 + \int dQ_2 = 0 [/itex]


    [itex] dQ_1 = mcTdT [/itex]

    [itex] \int dQ_1 = mc \int_{273}^{T_f} TdT [/itex]

    [itex] \int dQ_1 = mc ( \frac{1}{2} \left.T^2 \right|_{273}^{T_f} ) [/itex]

    [itex] \int dQ_1 = \frac{1}{2}mc (T_f^2 - 273^2) [/itex]

    [itex] dQ_2 = mcTdT [/itex]

    [itex] \int dQ_2 = mc \int_{326}^{T_f} TdT [/itex]

    [itex] \int dQ_2 = mc ( \frac{1}{2} \left.T^2 \right|_{326}^{T_f} ) [/itex]

    [itex] \int dQ_2 = \frac{1}{2}mc (T_f^2 - 326^2) [/itex]

    plugging back in

    [itex] \int dQ_1 + \int dQ_2 = 0 [/itex]

    [itex] \frac{1}{2}mc (T_f^2 - 273^2) + \frac{1}{2}mc (T_f^2 - 326^2) = 0 [/itex]

    [itex] (T_f^2 - 273^2) + (T_f^2 - 326^2) = 0 [/itex]

    [itex] T_f^2 - 74529 + T_f^2 - 106276 = 0 [/itex]

    [itex] 2T_f^2 = 180805 [/itex]

    [itex] T_f^2 = 90402.5 [/itex]

    [itex] T_f = 301 [/itex]

    now i am stuck when it comes to finding the actual entropy. I tried using

    [itex]ds = \frac{dQ}{T} [/itex]

    [itex]ds = \frac{mcTdT}{T} [/itex]

    [itex]ds = mcdT [/itex]

    [itex]s_1 = \int_{273}^{301} mcdT [/itex]

    and

    [itex]s_2 = \int_{326}^{301} mcdT [/itex]

    and then

    [itex] s_1 + s_2 = s_{sys} [/itex]

    but it didnt give me the right answer which makes me question my entire method. where did i go wrong?
     
  2. jcsd
  3. Oct 26, 2014 #2

    rude man

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    whence this???
     
  4. Oct 26, 2014 #3
    its specific heat. my book gives it as Q = mc(deltaT) and the differential form as
    dQ = mc(T)dT
     
  5. Oct 26, 2014 #4

    rude man

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    dQ = mc(T) dT is not the same as dQ = mcT dT!. And then you carried this mistake all the way thru your computations.

    It should be apparent that, assuming constant c between the two temperatures, that the mix temperature will be the average of the two.

    Strangely enough, your mix temperature wasn't off by much.

    EDIT: that goes for your entropy change calculations also, for which you again used the wrong formula.
     
  6. Oct 26, 2014 #5
    I am confused... I guess the difference between dQ = mcTdT and dQ = mc(T)dT is that (T) should be replaced by a function of temperature but they dont tell me how the temp is changing as a function so I am not sure what to plug in there
     
  7. Oct 26, 2014 #6

    rude man

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    "mc(T)" says "mass times specific heat as a function of temperature". The units are J/s or W.
    "mcT" says "mass times specific heat times temperature". Units are J-K/s or W-K.
    The two are completely different entities.
    In your case there is no difference between c and c(T) since c is assumed constant between your two temperatures. So forget c(T) and use c only:
    dQ = mc dT
    dS = dQ/T = mc dT/T.
    c is specific heat in J/s/kg.
     
  8. Oct 26, 2014 #7
    yeah the whole problem makes a lot more sense now. thank you!
     
  9. Oct 27, 2014 #8

    rude man

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  10. Jul 30, 2015 #9
    please solve
    Two vessels A and B each of volume 3m^3 are connected by tube if negligible volume. Vessel A contains air at 0.7 MPa , 95C while vessel B contains air at 0.35MPa , 205C. Find the change in entropy of system assuming the mixing to be complete and adiabatic .
     
  11. Jul 30, 2015 #10
    J/(kgK)
     
  12. Jul 30, 2015 #11
    Please do not hijack someone else's thread to seek help with your own problem. Please start a separate thread using the required template, and show some effort.

    Chet
     
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