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Gallium Melting and Entropy

  1. Jul 17, 2014 #1
    1. The problem statement, all variables and given/known data

    If 25.0 g of gallium melts in your hand, what is the change in entropy of the gallium?

    What about the change of entropy in your hand? Is it positive or negative? Is its magnitude greater or less than that of the change in entropy of the gallium?

    The melting temperature of gallium is 29.8°C, and its heat of fusion is 8.04 * 104 J/kg.


    2. Relevant equations

    ΔS = Q/Temp


    3. The attempt at a solution

    ΔSgal = ((.025 kg)(8.04*104 J/kg))/(20+273 K) assuming air temp is 20C
    ΔSgal = 6.86 J/K

    ΔShand = -((.025 kg)(8.04 * 104 J/kg))/(29.8+273 K)
    ΔShand = -6.63 J/K Negative since heat is leaving hand, reducing entropy
     
  2. jcsd
  3. Jul 17, 2014 #2

    Astronuc

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    Staff: Mentor

    I'm not sure why air temperature is used.

    The Ga melts in the hand, not in the air.

    The Ga melts at the melting temperature, no?

    Heat is transferred from hotter to cooler. What is the source of the thermal energy to melt the Ga? What might be the temperature of the hand be?
     
  4. Jul 17, 2014 #3
    The source of the thermal is the hand, which must be equal to or greater than 29.8°C. So, the changes in entropy of the hand and the gallium are equal and opposite?

    ΔSgal = ((.025 kg)(8.04*104 J/kg))/(29.8+273 K)
    ΔSgal = 6.64 J/K

    ΔShand = -((.025 kg)(8.04 * 104 J/kg))/(29.8+273 K)
    ΔShand = -6.64 J/K
     
  5. Jul 17, 2014 #4
    In the case of the gallium, the answer is pretty clearcut, because it takes place reversibly. In the case of the hand, the process is not reversible (because the temperature at the boundary (29.8) is less than the bulk temperature of the hand (37)). The initial and final temperatures of the hand are both very close to 37C, but a finite amount of heat has been transferred. You need to dream up a reversible path for this heat transfer , and calculate the entropy change from this path. One such path is where the interface is held only slightly below 37, and the same amount is transferred over a longer amount of time. For this reversible path, what is the entropy change?
     
  6. Jul 17, 2014 #5
    Would I then divide the QHand by 310 K? I'm not sure how to apply reversibility and irreversibility.
     
    Last edited: Jul 17, 2014
  7. Jul 17, 2014 #6

    ΔSgal = ((.025 kg)(8.04*104 J/kg))/(29.8+273 K)
    ΔSgal = 6.64 J/K

    ΔShand = -((.025 kg)(8.04 * 104 J/kg))/(37+273 K)
    ΔShand = -6.48 J/K
     
  8. Jul 18, 2014 #7
    Yes. Please understand that getting the entropy change for the hand by this approach is not very accurate because of a number of complexities that the problem statement glosses over. The hand is attached to a body (unless it's been chopped off), and there is interchange of heat with the rest of the body. There is also heat generated by chemical reactions in the body. And, there is heat transfer from the hand to the surrounding air, which may be affected by the cooling provided by the gallium to the hand. The approach we have taken here is to simply approximate the hand as a constant temperature reservoir at 37C. Getting the actual entropy change of the hand as a result of contact with the gallium is a much more complicated analysis which, IMHO, is not worth the effort.

    Chet
     
  9. Jul 18, 2014 #8



    Thank you very much! I assume it is supposed to be an ideal hand, possibly Thing from The Addam's Family. :)
     
  10. Jul 18, 2014 #9
    Ha!! Great joke. Incidentally, Art, please give my regards to Jerry, Elaine, and Kramer.

    Chet
     
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