Thermal Elongation of Alumina: 0.00036

[Karol]

Homework Statement

The pendulum of a clock is made of alumina. what is the relative elongation when cooled from 75⁰ fahrenheit to 45⁰?

Homework Equations

$$L=L_0(1-\alpha \Delta t)$$
From Fahrenheit to Celsius:
$$t_f=\frac{9}{5}t_c+32^0$$
Coefficient of thermal elongation of alumina: 24E-6^0 celsius

The Attempt at a Solution

30⁰ fahrenheit are: $\Delta t_c=\Delta t_f \frac{5}{9} \Rightarrow \Delta t_c=16.67^0$
$$\frac{L}{L_0}=\frac{L_0(1-\alpha \Delta t)}{L_0}$$
$$24E-6^0 \cdot 16.67=0.0004$$
Exactly, and it It should be 0.00036

 

[NascentOxygen]

It should be 0.00036

Is there a negative sign for its "elongation" when cooled?

 

[Karol]

In the book there isn't a negative sign, but i think it's clear that it shortens in the amount 1-0.00036

 

[nasu]

Is the coefficient given in the problem or you found it in a table?
Here is given as 22.2x10^(-6).
http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html

And is aluminum, right? Not alumina, which is aluminum oxide.

 

[Karol]

It is given in the book and it's aluminium