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Entropy Question Help pleasee

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data
    An open insulated vessel is divided into two parts by a vertical non-conducting partition.On one side of this partition is 2.5 kg of water initially at 38°C, and on the other side is 5.0 kg of water initially at a temperature of 70°C. On removal of the partition and the subsequent mixing, uniform conditions in the vessel are finally attained. Assuming no external heat transfer, evaluate the increase in entropy of the system. Assume that Cp for water is constant at 4.187 kJ/kg K


    2. Relevant equations
    Δs= m x Cp x ln(Tfinal/Tinitial)


    3. The attempt at a solution

    This is a mixing process so it must be irreversible and no heat transfer so adiabatic. Δs must be bigger than zero.

    m1= 2.5 kg , m2= 5.0 kg , T1= 313.15 K, T2= 345.15 K

    m1CpΔT+m2CpΔT= 0 (internal energy is dependent on the temperature only, temp doesnt change like Joule's experiment)

    I assumed that they would have a final temp that would be the same for both m1 and m2.
    I solved the equation above and got Tfinal= 334.48 K

    For m1:

    Δs = 2.5 x 4.187 x ln( 334.48/313.15)= 0.6898 kJ/K

    For m2:

    Δs = 5.0 x 4.187 x ln( 334.48/345.15)= -0.13148 kJ/K

    Δs increase for the entire system =0.6898 + (-0.13148) =+0.0331 kJ/K

    This is the correct answer according to the solutions so I did the maths right. However I am confused with the entropy change for m2=5.0 kg . As it can be seen in the calculation above that we have obtained a negative change in entropy. Isn't this undefined in the sense it disobeys the second law of thermodynamics. 'Entropy of the universe is always increasing' So how come we are allowed to proceed with the calculations if it isn't possible to have this in the first place.. If someone can enlighten me on this it would be great because i'm finding the second law quite hard to understand.

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 3, 2011 #2

    Mapes

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    There's no problem with local entropy decreasing. It occurs any time a material cools down. The Second Law tells us that entropy must increase somewhere else, however, by at least an equal amount. And this is what occurred, by your calculations. Does this make sense?
     
  4. Nov 3, 2011 #3
    Can you expand it a bit more, I haven't heard the term local entropy before so not entirely sure still
     
  5. Nov 3, 2011 #4

    Mapes

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    "Local entropy" just means the entropy within any well-defined region, not necessary open or closed. If the region is closed, though, the entropy inside cannot decrease. If it's open, then entropy can decrease but entropy in another region must increase by a value at least as large.
     
  6. Nov 3, 2011 #5
    Thanks very much for the helpful explanation
     
  7. Nov 4, 2011 #6
    I forgot to add this last section to the question..

    At the end of the mixing process, the container now transfers energy to its surroundings, and the entropy of the mixture is restored to its value before mixing began. State whether the combination of the mixing and energy transfer processes is reversible or irreversible and explain how attaining the final state does not involve violation of the second law of thermodynamics.

    The mixing process is irreversible because Δs (net) bigger than zero. In order to have a reversible process the Δs=0. It doesn't satisfy this requirement there is irreversible.

    Not sure how to answer the last section about the non-violation of the second law. Is this somehow related to the local entropy in any way?
     
  8. Nov 4, 2011 #7

    rude man

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    Think about what Mapes said about local vs. universal entropy change. The answer's staring you in the face!
     
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