# A problem that i spotted about expansion;

1. Nov 3, 2009

### Cryphonus

1. The problem statement, all variables and given/known data

So this question was on my mind for about 2 years and since i discovered such a good forum i wanted to ask to get an answer;I asked this question to various instructors but couldnt get an answer unfortunatly, so here it is

The formula to express the expansion for 1 dimension (for a metal stick for example) is:

L=L(initial).λ.Δt

So my problem is the formula itself;

Lets imagine a stick with 10 cm and the temperature of the stick is 10 celcius lets say; After that i decided to increase the temperature for 10 celcius more, so our temperature is 20 celcius and eventually our stick is going to expand a bit,

L=L(initial). λ.Δt

L initial(Lets take it 10 meters) So:

L(for 20 celcius)=10.λ(which is a constant doesn't matter in this case).10 (change in the temperature)

We calculated the new length of the stick no problem so far;

The temperature of the environment is 20 celcius at them moment; so i want to cool the air a little by bringing the temperature to 10 celcius again.If we calculate the new situation now;

L=L(the value for 20 celcius which is bigger than 10 meters for sure, lets take it 12 in that case).λ(constant doesnt matter).Δt (which is again 10)

so the value that we find is actually much bigger than the one for 10 celcius,since our L is bigger at 20 celcius ,our stick should be much more small then 10 meters which was our initial value,

If you try this you will also see that the stick will go to 0 if we keep changing the temperature to 10-20.If we do this infinite times our stick should be vanished according to the mathematical equation.

2. Relevant equations

3. The attempt at a solution

2. Nov 3, 2009

### Andrew Mason

Your premise is wrong. This is not the formula for thermal expansion.

The change in length (not the full length) is a linear function of temperature change and initial length:

$$\Delta L = L_0\lambda\Delta t$$

where $\lambda$ is the co-efficient of thermal expansion.

$$L = L_0(1 +\lambda\Delta t)$$

AM

3. Nov 3, 2009

### Staff: Mentor

I think the OP is pointing out a problem with the linear expansion formula for large values of λ. For example, say λ = 0.05 and the rod starts out with a length L0 of 10 m.

For ΔT = +10 degrees, we have L1 = 10 + .05*10*10 = 15 m.

If we then cool it down by 10 degrees (to its original temperature), we get:
L'0 = 15 - .05*15*10 = 7.5m ... which is smaller than it started out!

The answer is that the expansion formula only makes sense when λ*ΔT is small enough. (Usually, λ << 1.)

Last edited: Nov 3, 2009