# Homework Help: Thermal expansion confusion

1. May 4, 2007

### jrs

While studying for the Fundamentals of Engineering exam, I came across the following problem that has me puzzled:

The problem then goes on to give four possible answers (37.06 degrees C, 38.06 degrees C, 39.06 degrees C, and d. 40.06 degrees C) though I'm convinced that none of them can be correct.

Okay, I'll babble for a bit. Enough that you can see that I've got a few ideas, but also enough that you can see that I'm confused:

The second to last sentence in the problem is more confusing than it's worth, I think. From it I gather that the coefficient of thermal expansion of the chain is "11.66 x 10 ^ -6" with units of 1/degrees C. Also it seems as though they are trying to imply that at 20 degrees C, the change in temperature will be zero - my confusion level crosses into murk at about this point.

Back to the problem. Here are some things that I know:

Thermal Expansion is:

Code (Text):

DL/L = a DT

where:

DL   delta L - the change in length of the chain due to thermal
expansion.

L    - the "unchanged" length of the chain.  the length of
the chain at DT = 0, I suppose.

a    alpha - the coefficient of thermal expansion.

DT   delta T - the change in temperature that caused the chain to
expand from L to L + DL.

As the temperature increases, the chain will expand. The measured distance will get smaller as this happens. For example, if it was so hot that the chain expanded to 200% between measurements (unrealistic of course, but easy to think about) then the same distance would only measure 50% of its previous value.

From that little thought experiment it seems clear enough that according to the problem, the "temperature yesterday" should be less than "today's temperature". So none of the four given answers can be correct.

However I'm a bit stubborn and after I get started on a problem I like to grok it as best I can. So, what *is* the temperature today anyway?

It seems a bit more confusing to me because I'm only given 2 measurements - neither of which are "real" - they're only what the chain measured while it was expanded to some degree or another.

I've reasoned that:

Code (Text):

Lm = L / ( L + DL )

where:

Lm   length measured - the measured length.

L    - the length of the chain at DT = 0.
- another way of saying "what the chain would measure
if DT = 0"

DL   delta L - the change in length of the chain due to thermal
expansion.

Now I've got two equations to work with, but it all falls apart here - I'm just not having the insight to figure out what to think next.

I hope someone is sparked with a bit of curiosity and wants to help me to an understanding. This isn't a homework problem (I graduated with a physics degree in '92 - to say I'm rusty would be an understatement though as I've been programming computers for that last 15 years rather than doing physics)

-- Jonathan

2. May 5, 2007

### hage567

Well, I got one of the answers listed. This is my approach:

I used the equation for thermal linear expansion

$$\Delta L=\alpha L_o \Delta T$$

so $$L-L_o=\alpha L_o(T-T_o)$$

I used "yesterday" as Lo and To, and "today" as L and T. I just put in the numbers given, and solved for To.

Yes, the chain will expand as temperature increases, but I can't see how you conclude the measured distance gets smaller.
I see it the other way around. If the chain measured longer yesterday, then that means there must have been a higher temperature yesterday compared to today.

Last edited: May 5, 2007
3. May 5, 2007

### jrs

I'm thinking about it this way:

Say I've got a chain, and at some temperature T0 it measures the distance between two fixed points to be the full length of the chain - call that 1 unit of measure. The next day is very hot and the chain has expanded to twice it's original length. At that time (at temperature T) the two fixed points only measure up to half the length of the chain. So the measured distance is 1/2 units long.

I see what you've done, and you're probably doing what whoever wrote the question intended the solver to do. But the way I see it, in order for that solution to be valid the values for L and L0 must have been measured by another scale. The one that is "on the chain" so to speak, isn't going to work.

Think about it this way (continuing with my totally unrealistic example): on the first day I measure 1 unit long (the full length of the chain). So I'll use 1 unit for the value of L0 (as you said you did). On day two (when the chain has expanded to twice it's original length) I'll go measure a different distance so that I get L0 also to be one unit (of course you an I know that I'm measuring something that is twice as far (because the chain has expanded) but the chain doesn't know that). Now plugging L into the equation and solving for T I get... ta-da, no change in temperature - because my measured distances were the same.

Does that make any sense?

(completely off topic - how do you do the laTex equations?)