Doubt in Thermal Expansion.

  • Thread starter Puneeth423
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  • #1
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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:

Answers and Replies

  • #2
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Your equation b is wrong.
 
  • #3
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Your equation b is wrong.
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.
 
  • #4
haruspex
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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.
 
  • #5
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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.
Since, [itex]\alpha[/itex] is small,[itex]\alpha[/itex]2 can be neglected and can be considered zero.
 
  • #6
K^2
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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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