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Thermal expansion olympiad problem

  1. Jun 17, 2012 #1
    1. The problem statement, all variables and given/known data

    I only know volumetric expansion and linear expansion which require the coefficient constants that are not given in the problem. I also read somewhere that the mass remains constant during thermal expansion or is this false?
    The problem is attached pls take a look at it.

    2. Relevant equations

    delta V=BVi Dleta T

    3. The attempt at a solution
    I tried finding the total energy but could not do anything after that.
     

    Attached Files:

  2. jcsd
  3. Jun 17, 2012 #2
    Your question has nothing to do with thermal expansion.
     
  4. Jun 17, 2012 #3

    cepheid

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    This problem has nothing to do with thermal expansion. Thermal expansion, as you've noted, results in an increase in volume, not mass. It's the tendency of things to get bigger with increasing temperature.

    So if not thermal expansion, then what is this question about? My hint to you would be: think "mass-energy equivalence" and Einstein's famous equation. You can easily compute from the given data how much light energy is absorbed by the surface of the Earth over the course of a day. Then you just have to convert that into an equivalent mass value.
     
  5. Jun 17, 2012 #4
    Wow it uses E=mc^2? I didnt think it like that! Thanks for the help!!
     
  6. Jun 17, 2012 #5
    I actually havent learnt how to use that equation but when i use it by Dividing E/c^2=m
    , I got the wrong answer, (E) the answer given was (D)
     
  7. Jun 17, 2012 #6
    show the details of your calculation.
     
  8. Jun 17, 2012 #7

    cepheid

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    Pay attention, in particular, to the geometry.
     
  9. Jun 17, 2012 #8
    Surface area of earth=4∏r2≈5.1×1014m2

    5.1×1014×1500W/m2=7.65×1017

    Watt=J/s, 7.65×1017×60×60×24≈6.61×1022=Total energy in one day.
    E=mc2, Thus m=E/c2=[itex]\frac{6.61×10^{22}}{3.0×10^{8}}


    [/itex]≈7.34×105
     
  10. Jun 17, 2012 #9
    Answers D and E differ by the same factor as the area of a great circle to the total area of a sphere.

    It is my opinion that answer D is the correct answer.
     
  11. Jun 18, 2012 #10
    Why do we take it be a circle instead of a sphere?
     
  12. Jun 18, 2012 #11
    Because the incident light delivers power equal to the product of the intensity to the perpendicular area. What is the shape perpendicular to Sun's rays (assuming they're parallel)?
     
  13. Jun 18, 2012 #12

    cepheid

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    A way to visualize what Dickfore said. Say you have a whole "wall" of parallel rays coming from the sun. What is the effective area of this "beam" that is blocked out by Earth? In other words, what's the shape of the Earth's shadow?


    Code (Text):

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    ------------>[SIZE="7"]O[/SIZE]
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    It's a poor diagram, but the idea is that the "beam" of photons that is intercepted and absorbed is a circular cylinder whose radius is equal to the radius of the Earth. Therefore the cross sectional area of this cylinder is the area of the circle: πR2, where R is the radius of the Earth. So the incident power absorbed has to be equal to the flux (in W/m2) multiplied by this perpendicular cross-sectional area over which the light is blocked out.

    Now, you might argue that the incident rays impact over a hemispherical area of 2πR2 on the surface of the Earth. That's true, but the rays are not perpendicular to the surface of the Earth at every point. The angle of incidence may be perpendicular at the equator, but closer to the poles it's a very oblique angle. This means that the flux received is not 1.5 kW/m2 everywhere on the surface of the Earth. There are some places at higher latitudes where the 1.5 kW gets spread out over an area larger than 1 m2 because of this oblique incidence. I think that if you worked out the power received over the entire hemisphere, (which might be mathematically tricky at your level and probably requires integration), you'd find the same answer as we gave using the above argument about just what portion of the incident light is ultimately blocked out or "removed" from the beam.

    In other words, effectively, the power absorbed from the incident light by this hemisphere, which is not perpendicular to the incident light at every point, is the same as the power that would have been absorbed by a flat circular disc of the same radius that WAS perpendicular to the incident light at every point, because from the point of view of the "stream" of photons, they have the same cross-section.
     
    Last edited: Jun 18, 2012
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