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Thermodynamics Entropy problem

  1. Feb 17, 2013 #1
    1. The problem statement, all variables and given/known data
    http://postimage.org/image/9artvc1h5/ see attached


    2. Relevant equations
    ΔS=Q/T


    3. The attempt at a solution

    The real issue I'm having is how to go about solving this, I think I'm going to need to use kinematics to find the time the gold is exposed to the air correct? but then I'm not sure how to use that time value or anything. My book has no problems like this. Any and all guidance will be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Feb 18, 2013 #2

    Andrew Mason

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    You do not have to worry about the time the gold is exposed to the air. You just need to know the total heat flow into the surroundings.

    When everything reaches equilibrium, has the temperature of the air or ice changed? So if you know the amount of heat flow into the surroundings, what is its change in entropy?

    What is the heat flow into the surroundings generated by the cooling of the gold? (hint: you need an expression for ΔQ as a function of ΔT).

    What is the change in entropy of the gold? (hint: ΔS = ∫dQrev/T).

    What is the heat flow into the surroundings generated by the kinetic energy of the gold? Assume that energy is absorbed by the ice and air.

    What is the total amount of heat flow into surroundings? So what is the change in entropy (hint: that part is easy).

    Add the changes in entropy together.

    AM
     
    Last edited by a moderator: May 6, 2017
  4. Feb 18, 2013 #3
    So I calculated the Q for the gold to reach -20 degrees using the specific heat of gold (wasn't provided for practice test??), I'm not sure if that's the right move. I assume because the surroundings are at -20 that there is no real change in the temp of the air and ice overall, otherwise I wouldn't know what to do.

    Q=m(spH)ΔT

    Q=(1500 g)(.126 J/gK)(-20-1500)= -287280 J

    For the entropy change of the gold

    ΔS=∫dQ/T=c∫dT/T=(m(spH))ln(T2/T1)=(189)ln(253/1773)= -367.99 J/K

    Now, when you say kinetic energy of the gold, do you mean from its speed falling? If so I used the change in gravitational potential to calculate that part. I dunno if I can just say its Q2 but yeah.

    ΔUg=mghf-mghi=0-(1.5 kg)(9.8 m/s^2)(150 m)=-2205 J=Q2??

    total heat flow into surroundings 289485 J

    change in entropy of surroundings ΔS=Q/T=(289485 J)/(253 K)= 1144.21 J/K

    adding changes in entropy gives 776.219 J/K, is this change in entropy of the universe then?

    Not very confident in all of this so please let me know wherre I'm wrong. Thank you :)
     
  5. Feb 18, 2013 #4
    Does my work look alright? I don't want to let the thread die quite yet. If it goes down again I suppose I can take it to mean I'm all good.
     
  6. Feb 19, 2013 #5

    Andrew Mason

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    Everything looks good and is well explained. Well done.

    AM
     
  7. Feb 19, 2013 #6
    Thank you very much for your help.

    For reference I believe I forgot a piece, the latent heat of fusion for the gold which solidifies at 1064 °C

    Q_fusion=-m*Lh=-(1.5 kg)(63000 j/kg)= -94500 J
    ΔS=Q/T=-94500/(1064+273)= -70.68 J/K

    Making total change in entropy of the gold -438.67 J/K

    making heat flow into surroundings 383985 J and entropy change 1517.727 J/K

    change in entropy of universe becomes 1079.057 J/K
     
    Last edited: Feb 19, 2013
  8. Feb 19, 2013 #7

    Andrew Mason

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    You are thinking! Since the gold is a liquid at 1500C you might want to check the heat capacity for liquid gold as it may be higher than for solid gold - not that it will make a huge difference here.

    AM
     
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