How to Calculate Thermal Expansion in a Rod of Length 'l'?

[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?

 

[xAxis]

Your equation b is wrong.

 

[Puneeth423]

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]

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.

 

[Puneeth423]

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]

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.

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