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Conduction rate problem

  1. Dec 16, 2011 #1
    1. The problem statement, all variables and given/known data
    The average rate at which energy is conducted outward through the ground surface in North America is 54 mW/m^2, and the average thermal conductivity of the near-surface rocks is 2.5 W/mK. Assuming a surface temperature of 10 C, find the temperature at a depth of 35 km. Ignore the heat generated by the presence of radioactive elements.


    2. Relevant equations
    P = kA(Th-Tc)/L

    where P is the conduction rate
    k = thermal conductivity
    A = face area
    L = thickness
    Th = temperature of hot reservoir
    Tc = temperature of cold reservoir

    3. The attempt at a solution

    I can plug all values into the equation for conduction rate except for A. Is there something I'm missing in terms of the surface area for this problem? In the answers I've seen online, people seem to assume A is 1m^2. However, I can't figure out why that assumption is made. (The answer seems to be 766 C, which is found using A=1m^2).

    Thanks for your help!
     
  2. jcsd
  3. Dec 16, 2011 #2

    Delphi51

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    Homework Helper

    You could apply the formula to any area you like.
    But if you are going to use the 54 mW/m^2 you must express A in m^2 as well.
    In the old days we used to put in the units for all quantities and check to make sure they worked out the same on both sides of a formula. That took too much time, but when in doubt, it is still worth doing occasionally.
     
  4. Dec 16, 2011 #3
    But if we had arbitrarily chosen an A of 2 m^2, wouldn't the answer to this question be different? The only quantity that would change in the equation would be A, no? I'm confused because the problem doesn't seem to tell you how large of an area we're considering, whereas the solution seems to automatically use a value of 1 m^2.
     
  5. Dec 16, 2011 #4

    Delphi51

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    Homework Helper

    Oops, I gave you a dumb answer!
    When you work out the units on the right side, taking k to be in W/mK, you get Watts.
    That formula is giving you the power, not the power per square meter as the wording of the question implies. It is only natural that if you double the area, you get double the power flowing through.

    Mix up in formulas? The formula is given as ΔQ/ΔT = kA*ΔT here:
    http://en.wikipedia.org/wiki/Thermal_conductivity#Equations
     
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