Expansion of ruler and rod help

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Homework Help Overview

The discussion revolves around the thermal expansion of a metal rod and a steel ruler, specifically how to calculate the coefficient of expansion given temperature changes and measurements. The problem involves understanding the effects of temperature on both the rod and the measuring instrument.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the expansion of the rod and the ruler, questioning how to account for the ruler's expansion when measuring the rod's length. There are attempts to derive the coefficient of expansion using the provided measurements and temperature changes.

Discussion Status

Some participants have provided hints and suggestions for approaching the problem, such as finding the actual length of the expanded rod and simplifying expressions before calculations. There is an ongoing exploration of the calculations needed to determine the coefficient of expansion, with some participants expressing confusion over their results.

Contextual Notes

Participants note the importance of considering both the rod and ruler's expansions, as well as the specific temperature values given in the problem. There is mention of the known coefficient of expansion for steel, which may influence the calculations.

rdn98
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The length of a metal rod is measured to be 20.08 cm using a steel ruler when both the rod and the ruler are at 22oC. Both the rod and the ruler are raised to a temperature of 253oC. When the rod is measured at this higher temperature, its length is found to be 20.32 cm.

****
a) What is the coefficient of expansion of the metal?
b) You now make a rod of the same material but with twice the length. What will it coefficient of expansion be?
c) You now make a rod of the same material but with twice the diameter. What will it coefficient of expansion be?
//////

So far I am stuck on part a. If I can get this part, then I should be able to get the rest.

Now I know that the equation is delta L=L*(coefficient of expansion)*(delta T)

Now it appears that I have all the informatin given in the problem, so I plug it into the equation, but it doesn't work. Then I realize that the ruler is stretching also, which means the number scale on the ruler also changes. Now I have to somehow relate that number scale to the coefficient of expansion, but I'm not sure how.

I know from the book at the coefficient of expansion for steel is 11*10^-6 /degree C.
 
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Originally posted by rdn98
Then I realize that the ruler is stretching also, which means the number scale on the ruler also changes. Now I have to somehow relate that number scale to the coefficient of expansion, but I'm not sure how.
First find the actual length of the expanded rod. You know the length of the expanded rod as measured by the expanded steel ruler. So, knowing the expansion of the steel, find the actual length of each "1 cm" marking on the ruler. Then you can convert to real units.

Hint: Write down the final expression (for the coefficient of expansion of the rod) and simplify it before you start doing any arithmetic.
 
Thread necromancy!

I've got slightly different numbers: The length of a metal rod is measured to be 20.08 cm using a steel (α = 1.1e-005 (oC)-1) ruler when both the rod and the ruler are at 22oC. Both the rod and the ruler are raised to a temperature of 253oC. When the rod is measured at this higher temperature, its length is found to be 20.32 cm.

When heated, the rod expands. However, the ruler also expands. So the rod doesn't actually expand from 20.08 to 20.32. I know the rod expands more than the ruler, because after heating, the rod is measured to be longer. If they expanded the same the length wouldn't change and if the rod expanded less the final length would be less than the original length.


I used 1 cm because I need to find the actual length of each '1 cm marking' after heating.
ΔLs = L [itex]\alpha[/itex] Δt
ΔLs = (1 cm) (1.1e-5 °C^-1) (253°C - 20°C)
ΔLs = .002563
1 hcm (heated cm) = 1.002563 cm (real cm)

If the rod now measures 20.32 hcm, so I convert my units:

20.32 hcm * (1.002563 cm/1 hcm) = 20.37208 cm

So now I have an actual length and can calculate the [itex]\alpha[/itex].

[itex]\alpha[/itex] = ΔL/(L*Δt)
[itex]\alpha[/itex] = .00435

And it tells me NO!
 
Doh!

If I actually use a ΔL instead of L it works out.

/alpha = 6.243e-5

Thanks for being here so I can think this through! :)
 

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