# Doubt in Thermal Expansion.

1. Dec 10, 2012

### Puneeth423

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?

Last edited: Dec 10, 2012
2. Dec 10, 2012

### xAxis

3. Dec 10, 2012

### Puneeth423

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

Its perfect.

4. Dec 11, 2012

### haruspex

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. Dec 11, 2012

### Puneeth423

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

6. Dec 11, 2012

### K^2

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.

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