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Thermal expansion of a rod with variable alpha

  1. Feb 12, 2016 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    The coefficient of linear expansion of a rod of length 1 meter (at 300K) varies with temperature as ##\alpha=\frac{1}{T}##, where T is the temperature. Find the increment in length when the rod is heated from 300K to 600K

    2. Relevant equations
    $$\Delta L=L\alpha \Delta T$$

    3. The attempt at a solution
    I considered a small element of length ##dl## at a distance of ##l## from one end.
    Let the elements length increase by ##dx##.
    $$dx=dl\frac{2}{T}dT$$
    How will I integrate the above expression
     
  2. jcsd
  3. Feb 12, 2016 #2

    haruspex

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    That does not balance in terms of infinitesimals. You can't equate one to a product of two.
    There's no need to consider an element of length, the whole rod will expand in the same proportion.
    Also, I didn't understand where the 2 came from. Should that be an alpha?
     
  4. Feb 12, 2016 #3

    Titan97

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    I put the value of alpha. How can directly use ##\Delta L =L\alpha z
    \Delta T## when alpha varies with temperature? Shouldn't I integrate?

    Also, if I take a length ##l##, it will expand by ##dl=dl\frac{2}{T}dT##.
    Integrating, I got the answer as ##2\ln 2##. But answer given is ##3##.
     
  5. Feb 12, 2016 #4

    haruspex

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    According to your original post, alpha is 1/T, but you seem to have substituted 2/T. From the given answer, 2/T is correct.
    If you take L as the whole length, the increase in length, dL, equals LαdT, not dLαdT.
    Please post your steps from there.
     
  6. Feb 12, 2016 #5
    Titan97: In terms of differential changes in length and temperature, the correct equation should read ##dL=\alpha L dT##. In fact, this is the truly correct equation, and the equation ##\Delta L=\alpha L \Delta T## is only an approximation.

    By the way, did you mean in your original post that ##\alpha=1/T##, or did you mean ##\alpha = 2/T##?
     
  7. Feb 13, 2016 #6

    Titan97

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    OK. So $dL=L\alpha dT$
    $$\int_L^{L+\delta L}\frac{dL}{L}=\int_{T_1}^{T_2}\frac{2}{T}dT$$
    $$\ln\frac{L+\delta L}{L}=2\ln\frac{T_2}{T_1}$$
     
  8. Feb 13, 2016 #7
    This is the same thing as $$\frac{L+\delta L}{L}=\left(\frac{T_2}{T_1}\right)^2$$
    Can you see that?
     
  9. Feb 13, 2016 #8

    haruspex

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    Right. How to get rid of the logs?
     
  10. Feb 13, 2016 #9

    Titan97

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    @haruspex I got the same expression as @Chestermiller 's post #7.
    $$\ln(x)=2\ln(y)$$
    $$\ln(x)=\ln(y^2)$$
    $$x=y^2$$
    $$1+\frac{\delta L}{L}=(\frac{T_2}{T_1})^2$$
    ##T_2/T_1=2##
    $$1+\frac{\delta L}{L}=4$$
    $$\frac{\delta L}{L}=3$$
    Since ##L=1\text{m}##
    $$\delta L=3\text{m}$$
     
  11. Feb 13, 2016 #10

    haruspex

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    Great. Please mark as solved!
     
  12. Feb 16, 2016 #11

    Titan97

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    @haruspex I was going to mark it as solved today. But someone did that for me. I have also seen that some people can mark questions unsolved even if OP marked it as solved.
     
  13. Feb 16, 2016 #12
    Yes. I had marked it solved.
     
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