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Thermal expansion coefficent

  1. Mar 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that the volume thermal expansion coefficent of a solid is equal to the sum of its linear expansion coefficients in the three directions: B = Ax + Ay + Az

    2. Relevant equations

    B = (dV/V)/dT
    A = (dL/L)/dT

    3. The attempt at a solution

    My thought was using dV/V = ((dLx*dLy*dLz)/(Lx*Ly*Lz)) but when you use this it is clear that dL^3/L^3 does not equal 3* dL/L

    What am I doing wrong?

    Thanks for the help
  2. jcsd
  3. Mar 31, 2009 #2


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    dV= ( Lx +dLx)( Ly +dLy)( Lz +dLz) - Lx*Ly*Lz
    Simplify this and proceed.
  4. Apr 1, 2009 #3
    Can you explain mathematically how you got there?

    Also, the problem statement says at the end (So for an isotropic solid, which expands the same in all directions, B = 3A) if that makes the problem simpler.
  5. Apr 1, 2009 #4


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    Lx, Ly and Lz are the lengths of the block. When the temperature is raised through 1 degree C, the new lengths will be Lx + Ax, Ly + Ay and Lz+ Az.
    New volume will be (Lx + Ax)( Ly + Ay)( Lz+ Az). Original volume is Lx*Ly*Lz.
    Sp dV = (Lx + Ax)( Ly + Ay)( Lz+ Az) - Lx*Ly*Lz
    Find dV/V. Neglect the terms like Ax*Ay and so on because they are very small quantities.
  6. Apr 1, 2009 #5


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    dV = (volume after expansion) - (initial volume)​

    After expansion, the box dimensions have increased by dLx, dLy, and dLz, from their initial lengths Lx, Ly, and Lz.

    rl responded faster than I. :smile:
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