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A little entropy

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A boater dumps 8.6 kg of ice from his cooler into the lake at the end of a fishing trip. The ice is at -6°C, and the lake is at 20.9°C. What is the change in entropy of the new lake-ice cube system after the ice cubes come to equilibrium with the lake?

    2. Relevant equations

    [tex]\Delta[/tex]S = [tex]\frac{\Delta Q}{T}[/tex]

    3. The attempt at a solution

    Not sure where I'm going wrong, but my answer is around 13 times too large.

    I calculated 3 different dQ's corresponding to bringing the ice up to 0 C, melting the ice, then bringing the water up to 20.9 C. I then summed each dQ and got 3840.8 kJ, divided by 294.05K and end up with 13.061 kJ/K. (the actual "right" answer is 0.873 kJ/K, I just cannot get to it!)
     
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  3. Mar 2, 2009 #2

    mgb_phys

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    That's the enthalpy change for the ice - what about the change for the lake?
    Which direction is the lake's enthalpy change in?
     
  4. Mar 2, 2009 #3
    Well, the lake is transferring heat into the ice, and since the dQ1 + dQ2 = 0, the lake is contributing -3840.8 kJ.

    If I understand this correctly, my above answer is actually the loss of entropy of the lake itself. I then need to find the gain in entropy of the ice. (that can't be the gain in entropy of the ice since that isn't at a constant temperature, right?)
     
  5. Mar 2, 2009 #4

    mgb_phys

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    For each of your dQ/T into the ice there is a corresponding dQ/T from the lake, but the temperature is different (the T of the lake is higher) so they don't exactly cancel.
    The 0.873 is the difference between these sets of values.
     
  6. Mar 2, 2009 #5
    After your first reply, thats what I was thinking.

    However, how/when do I determine the T for each dQ?

    ie, bringing the ice from -6 C to 0 C isn't a constant temperature, so how do I decide which value to use?
     
  7. Mar 2, 2009 #6

    mgb_phys

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    You are really summing the entropy change of each infinitesimal bit of energy - so there is a fraction of a Joule going into the ice at -6.0C, then another at -5.99999C etc.
    It's easy to see this is equivalent to just taking the midpoint temperature of each step.

    So T1 for ice warming to zero is (-6-0)/2C, T2 while melting doesn't change, T3 for the ice water warming is (20.9-0)C/2. The entropy change of the ice is going to be dominated by the melting step anyway because it uses so much energy.

    The important point is that the temperature of the heat reservoir (ie the lake) doesn't change for the entropy flow in the other direction.
     
  8. Mar 2, 2009 #7

    Andrew Mason

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    The ice temperature changes as the heat is absorbed by the ice, so you have to do some calculus.

    [tex]\Delta S = \Delta S_{ice} + \Delta S_{lake}[/tex]


    Since dQ is a function of temperature of the ice/water + heat of fusion , you have to break it out into three different sections and calculate the change in entropy of the ice in each:

    [tex]\Delta S_{ice} = \int_{T_i}^0 dQ/T + \frac{\Delta Q_{fusion}}{T_{fusion}} + \int_{0}^{T_f} dQ/T[/tex]

    AM
     
  9. Mar 2, 2009 #8
    First off, I really do appreciate the help!

    However, a couple questions:

    Is it really clear why we can use the midpoint? Sure, I get close to the right answer, but by using the midpoint we are assuming that T is changing at a constant rate, which isn't the case. While this implies the need for calculus, where can I determine a function to use for T?


    Another note: maybe I'm just rounding incorrectly somewhere too.

    Again, thanks for the help!
     
  10. Mar 2, 2009 #9
    Found my error, found the right answer, but still wondering why it's clear why we can use the midpoint of dT.

    Specific heat of ice [tex]\neq[/tex] specific heat water. Whoops.


    EDIT : Nevermind on that too. As long as we're not interested in the time it takes for the temperature to change, we don't care about the rate of heat flow.

    Thanks for your help!
     
  11. Mar 3, 2009 #10
    Just thought I'd bump this quick.

    Just finished a test involving topics ranging from the 2nd Law of Thermodynamics and Entropy to light/electromagnetic waves to refraction/reflection and mirrors/lenses.

    With all those topics, fully 1/3 of the test involved an entropy problem similar to my HW posted. So thanks to you I've now got ~100% on the test rather than ~70%.

    Just thought I'd share ! :biggrin:
     
  12. Mar 3, 2009 #11

    Andrew Mason

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    I am not sure what you mean by the midpoint.

    In calculatng dS = dQ/T, you can use the midpoint of each process (eg. from 267-273 use 270 and from 273-293 use 283) to get a reasonably close figure. But it is a rough approximation.

    I get total dQ = 2050*8.6*6 + 334000*8.6 + 4181*8.6*20 =
    = 105780 + 2872400 + 719132 = 3,697,312 J

    So:

    [tex]\Delta S_{lake} = -3,697,312/293 = -12619 J/K[/tex]

    And:

    [tex]\Delta S_{ice} = C_{ice}m ln(273/267) + L_{fusion}m + C_{water}m ln(293/273)[/tex]

    [tex]= 2050 * 8.6 * .022 + 334000*8.6 + 4181*8.6*.0707 = 387.9 + 10521.6 + 2542.1 = 13452 J/K[/tex]

    So the total entropy change is: 13452 - 12619 = 833 J/K


    I don't see how the correct answer would be 863 J/K.

    AM
     
  13. Mar 4, 2009 #12
    Maybe a typo on my part? Whatever I put into the homework site was accepted (i think the range is +/- 1% or so).

    But can you explain why using the midpoint is only a rough approximation? Isn't it a linear graph when we don't consider the time involved?
     
    Last edited: Mar 4, 2009
  14. Mar 4, 2009 #13
    Okay, here's my full answer.
    given specific heat of ice = .5 cal/g*K and specific heat water = 1 cal/g*K.
    Latent heat of fusion for water is 334 J/g
    1 calorie = 4.184 J

    dQ_1 = Heat to bring ice to 0ºC
    = 2.092 (J/g*K) * 8600 (g) * (0--6) (K) = 107,947 J

    dQ_2 = Heat to melt ice.
    = 334 (J/g) * 8600 (g) = 2,872,400 J

    dQ_3 = Heat to bring melted ice to 20.9ºC.
    = 4.183 (J/g*K) * 8600 (g) * (20.9-0) (K) = 751,547 J

    dQ_net = 3,731,894 J

    So, for the entropy change of the lake, dS_lake = -3,731,894 (J) / (273.15+ 20.9) (K) = -12,691.4 J/K

    dS_1 = 107,947 J / (273.15-3) (K) = 399.582 J/K
    ds_2 = 2,872,400 J / 273.15 K = 10515.8 J/K
    dS_3 = 751,457 J / (273.15+10.45) K = 2650.02 J/K

    dS_net = dS_1 + dS_2 + dS_3 + dS_lake = ~874.002 J/K
     
  15. Mar 4, 2009 #14

    Andrew Mason

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    I see that I misread the temperature of the lake as 20 C instead of 20.9.

    The reason you can use the midpoint to get an approximation is because the area under the graph f(T) = k/T for a small interval of T in a region where the graph is not steep (ie.where T is not small) is approximated by taking the area of the rectangle at the height of the midpoint in the interval.

    AM
     
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