What is the Correct Temperature Yesterday?

In summary, the problem involves a survey crew measuring the distance between two points on two separate days using a steel chain without correcting for temperature. The temperature is given as 28C for the second day, and the question asks for the temperature on the first day. The problem also provides a correction equation for temperature, but the four possible answers given are not correct. The conversation between two individuals shows their approach to solving the problem and their confusion over the given information. One individual suggests using the equation for thermal linear expansion to solve the problem, while the other disagrees and explains their reasoning using a hypothetical example. The conversation ends with the individual asking for help in understanding the problem.
  • #1
jrs
3
0
While studying for the Fundamentals of Engineering exam, I came across the following problem that has me puzzled:

Yesterday a survey crew measured the distance between two points as 805.420m. They used a steel chain, but no correction was made for the temperature. Today, they measured the distance as 805.316m, also without correction. If a measured length is Lm and the temperature is Tc in degrees C, then the length is corrected for temperature by adding "(11.66)(10^-6)(Tc-20)Lm" to it. If today's temperature is 28C, what was the temperature yesterday?


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:
 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:
 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)

please help. thanks in advance.

-- Jonathan
 
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  • #2
Well, I got one of the answers listed. This is my approach:

I used the equation for thermal linear expansion

[tex] \Delta L=\alpha L_o \Delta T [/tex]

so [tex]L-L_o=\alpha L_o(T-T_o)[/tex]

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

As the temperature increases, the chain will expand. The measured distance will get smaller as this happens.
Yes, the chain will expand as temperature increases, but I can't see how you conclude the measured distance gets smaller. :confused:
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.
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:
  • #3
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?)
 

What is thermal expansion confusion?

Thermal expansion confusion refers to the misunderstanding or lack of knowledge about the phenomenon of thermal expansion, which is the tendency of matter to increase in size when heated and decrease in size when cooled.

Why is thermal expansion important to understand?

Thermal expansion is important to understand because it affects the properties and behavior of materials, and can have practical applications in various fields such as engineering, construction, and manufacturing. It can also cause damage or failure in structures if not accounted for.

What causes thermal expansion?

Thermal expansion is caused by the increase in the kinetic energy of molecules when heated, leading to an increase in the distance between them. This results in an overall increase in the size or volume of the material.

How does thermal expansion affect different materials?

Different materials have different coefficients of thermal expansion, which is a measure of how much they expand or contract with a change in temperature. Materials with higher coefficients of thermal expansion will expand or contract more significantly than those with lower coefficients.

How can thermal expansion be managed or controlled?

Thermal expansion can be managed or controlled by using materials with low coefficients of thermal expansion, incorporating joints or gaps to allow for expansion and contraction, and using thermal expansion calculations to design structures or objects that can withstand the effects of thermal expansion.

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