# Thermodynamics - Thermal Expansion

1. Jan 18, 2005

### Kelvin

The following questions are extracted from "Physics" Halliday Resnick & Krane Chapter 21 problem

7.
(a) Show that if the lengths of two rods of different solids are inversely proportional to their respective coefficients of linear expansion at the same initial temperature, the difference in length between them will be constant at all temperature.​

(b) What should be the lengths of a steel and a brass rod at 0 degree celcius so that all temperatures their difference in length is 0.30 m ?​

My attempt to this question is:

The definition of the coefficient of linear expansion is
$$\begin{split} \alpha &= \frac{\Delta L_1}{\Delta T} \frac{1}{L_1}\\ \beta &= \frac{\Delta L_2}{\Delta T} \frac{1}{L_2}\\ \\ \frac{\alpha}{\beta} &= \frac{\Delta L_1}{\Delta L_2} \frac{L_2}{L_1}\\ \end{split} \begin{split} \frac{\alpha L_1}{\beta L_2} &= \frac{\Delta L_1}{\Delta L_2} \equiv K\\ \end{split}$$

Where $$K$$ is a dimensionless constant since the length is inversly proportional to the coefficient of linear expansion.

Suppose $$L_1'$$ and $$L_2'$$ are the length after the temperature change $$\Delta T$$. The difference in length is
$$\begin{split} \Delta L \equiv L_1' - L_2' &= L_1 + \Delta L_1 - \left( L_2 + \Delta L_2 \right) \\ &= L_1 - L_2 + \left( \Delta L_1 - \Delta L_2 \right) \end{split}$$​

What I think is that the given condition is not sufficient, but neccessary. One more condition required such that "the difference in length between them will be constant at all temperature" is
$$K \equiv 1$$​
If this condition is imposed,

$$\alpha L_1 = \beta L_2$$
$$\Delta L_1 = \Delta L_2$$
$$\Delta L = L_1 - L_2$$

which depends on the intial length of two rods only.

(b) We want
$$\Delta L = 0.30 \hbox{ m}$$​
for all temperature. We have the following equations:

$$\begin{split} \alpha L_1 = \beta L_2 \\ \Delta L = L_1 - L_2 = 0.30 \hbox{ m}\\ \alpha \left( \Delta L + L_2 \right) &= \beta L_2 \\ \\ L_2 &= \frac{\alpha \Delta L}{\beta - \alpha}\\ &= \frac{11 \times 10^{-6} K^{-1} \times 0.30 \hbox{ m}}{\left(19 - 11\right) \times 10^{-6} K^{-1}}\\ &= 41 \hbox{ cm}\\ L_1 &= \Delta L + L_2\\ &= 41 \hbox{ cm} + 30 \hbox{ cm}\\ &= 71 \hbox{ cm} \end{split}$$​

The part (b) result agree with the given answer. I wonder whether the part(a) question is wrong or I am wrong?

Wow, lateX is so difficult to use. It took me more than an hour to type out the equations nicely. Now I am typing the second question.

Last edited: Jan 18, 2005
2. Jan 18, 2005

### dextercioby

The first part is wrong.Check the division in the passing from the eq.(1) to eq.(2)

Daniel.

3. Jan 18, 2005

### Kelvin

Thank you
I have corrected that