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Doubt in Thermal Expansion.

  1. Dec 10, 2012 #1
    Consider a rod of length 'l'. When heated, its temperature got increased by 't'.
    Increase in its length = l[itex]\alpha[/itex]t.
    total length be (l1) = l + l[itex]\alpha[/itex]t.
    Again the rod is heated and the temperature further got increased by 't'.
    Total length be l2.
    l2 can be calculated in two ways,
    l2 = l1 + l1[itex]\alpha[/itex] t......(a)
    0r l2 = l + l[itex]\alpha[/itex](2t)........(b)
    Equating (a) and (b).
    l1 + l1[itex]\alpha[/itex] t = l+l[itex]\alpha[/itex](2t)
    {l + l[itex]\alpha[/itex]t} + {l + l[itex]\alpha[/itex]t}{[itex]\alpha[/itex] t}
    = l+l[itex]\alpha[/itex](2t).
    Solving above equation we get l([itex]\alpha[/itex]t)2 = 0.
    Where am i going wrong?
    Last edited: Dec 10, 2012
  2. jcsd
  3. Dec 10, 2012 #2
    Your equation b is wrong.
  4. Dec 10, 2012 #3
    What is wrong?
    Initial length of rod is 'l' and total rise in temperature is '2t'.
    l2 = l + l[itex]\alpha[/itex](2t).

    Its perfect.
  5. Dec 11, 2012 #4


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    The coefficient of expansion varies with temperature. For most substances and practical temperature ranges the variation is small, as is the quadratic term in your calculations.
  6. Dec 11, 2012 #5
    Since, [itex]\alpha[/itex] is small,[itex]\alpha[/itex]2 can be neglected and can be considered zero.
  7. Dec 11, 2012 #6


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    The formula l1 = l + l[itex]\alpha[/itex]t works only for very small t. Correct way to look after sweep over finite temperature range would be as follows.

    [tex]l+dl = l + l \alpha dt[/tex]

    Where dl is a small change in length, and dt is a small change in temperature. You can rearrange that into a differential equation.

    [tex]dl/dt = l \alpha[/tex]

    And that's easily solved.

    [tex]l_1 = l_0 e^{\alpha t}[/tex]

    Now if you substitute this into your formula, it works either way.

    [tex]l_2 = l_1 e^{\alpha t} = l_0 e^{\alpha t + \alpha t} = l_0 e^{\alpha 2t}[/tex]
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