# Doubt in Thermal Expansion.

Consider a rod of length 'l'. When heated, its temperature got increased by 't'.
Increase in its length = l$\alpha$t.
total length be (l1) = l + l$\alpha$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$\alpha$ t......(a)
0r l2 = l + l$\alpha$(2t)........(b)
Equating (a) and (b).
l1 + l1$\alpha$ t = l+l$\alpha$(2t)
{l + l$\alpha$t} + {l + l$\alpha$t}{$\alpha$ t}
= l+l$\alpha$(2t).
Solving above equation we get l($\alpha$t)2 = 0.
Where am i going wrong?

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## Answers and Replies

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Your equation b is wrong.

Your equation b is wrong.
What is wrong?
Initial length of rod is 'l' and total rise in temperature is '2t'.
l2 = l + l$\alpha$(2t).

Its perfect.

haruspex
Science Advisor
Homework Helper
Gold Member
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.

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, $\alpha$ is small,$\alpha$2 can be neglected and can be considered zero.

K^2
Science Advisor
The formula l1 = l + l$\alpha$t works only for very small t. Correct way to look after sweep over finite temperature range would be as follows.

$$l+dl = l + l \alpha dt$$

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

$$dl/dt = l \alpha$$

And that's easily solved.

$$l_1 = l_0 e^{\alpha t}$$

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

$$l_2 = l_1 e^{\alpha t} = l_0 e^{\alpha t + \alpha t} = l_0 e^{\alpha 2t}$$