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How to solve for the spillover for a spherical container?o

  1. Nov 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A spherical brass shell has an interior volume of 1.60 x 10-3m^3. Within this interior volume is a solid steel ball that has a volume of 0.70 x 10-3m^3. The space between the steel ball and the inner surface of the brass shell is filled completely with mercury. A small hole is drilled through the brass, and the temperature of the arrangement is increased by 12 celsius degree. What is the volume of the mercury that spills out of the hole?

    2. Relevant equations
    Delta V = B*Vo*Delta t

    3. The attempt at a solution
    Delta V(Mercury) = 3.49e-6 m^3
    Delta V(Brass) = 1.09e-6 m^3
    Delta V(Steel) = 3.02e-7 m^3

    I've already computed for all of the changes in volume for the two spheres and the mercury. I'm having a problem on how to solve for the spillover because from our previous examples there are no spherical containers.
     
  2. jcsd
  3. Nov 19, 2015 #2

    RUber

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    My first thought is that the interior volume of the brass (if your equation is appropriate for hollow containers as well) is the limiting factor.
    If the original arrangement was full...then any change in the total volume of the contents (mercury and steel) exceeding the change in volume of the container (brass) will be the volume that spills out.
     
  4. Nov 19, 2015 #3
    So what would be the equation now if that is the case?
     
  5. Nov 19, 2015 #4
    Side note: I'm using the equations that our professor gave us about thermal expansion and the equation for a regular spillover is:

    Spillover = Delta V(Liquid) - Delta V(Container)

    I also don't have an idea how to solve for the spillover with another material inside the container.
     
  6. Nov 19, 2015 #5

    gneill

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    Staff: Mentor

    How did you determine the initial volume of mercury?
     
  7. Nov 19, 2015 #6
    It is stated in the problem that the sphere brass was filled with mercury so that means it is the same volume.
     
  8. Nov 19, 2015 #7

    gneill

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    What about the steel sphere that's also taking up space inside the brass shell?
     
  9. Nov 19, 2015 #8
    So that means I have to subtract the volume of the steel sphere to the volume of the brass sphere in order to get the volume of the mercury. But still I have the problem on how to get the spillover because it is a whole different equation since the container is spherical.
     
  10. Nov 19, 2015 #9

    gneill

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    The volumes will change according to the expansion formula in the same way regardless of their shapes, so you can determine the initial volume available for mercury inside the structure and then the volume that's available after the materials expand. You'll have a new mercury volume and the amount space available inside the container... what doesn't fit is what spills out.
     
  11. Nov 19, 2015 #10
    So I'll still be using the same formula that was given to me when solving for a regular spillover but with the addition of the steel sphere. So the equation will be:
    Spillover = Delta V(Mercury) + Delta V(Steel) - Delta V(Container)
    Is this right?
     
  12. Nov 19, 2015 #11

    gneill

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    That should work.

    Personally I find it easier to think in terms of before and after total volumes, but it reduces to the "Delta" approach if you lay out the math.
     
  13. Nov 19, 2015 #12
    Hey thanks for helping me! I was having a really hard time solving this problem.
     
  14. Nov 19, 2015 #13

    gneill

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    Staff: Mentor

    You're welcome. Glad to help.
     
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