Why Does Turpentine Not Overflow When Heated in an Aluminum Cylinder?

AI Thread Summary
The discussion centers on the thermal expansion of turpentine in a hollow aluminum cylinder when heated from 20°C to 80°C. The key confusion arises from the application of expansion formulas, where the user initially believes both the aluminum and turpentine should be treated the same. However, it is clarified that the solutions manual used the linear expansion coefficient for aluminum and the volumetric expansion for turpentine, which explains the differing calculations. The user realizes that the volumetric expansion should have been applied consistently for both substances. Ultimately, understanding the distinction between linear and volumetric expansion coefficients is crucial for solving the problem correctly.
Feodalherren
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Homework Statement


A hollow aluminum cylinder 2cm deep has an internal capacity of 2L at 20C. It is completely filled with turpentine and then slowly warmed to 80C.
a) How much turpentine overflows?
b)If the cylinder is then cooled back to 20 C how far below the cylinder's rim does the turpentine recede?


Homework Equations


Vnew=Vold(1+αΔT)^3


The Attempt at a Solution


I'm so confused. I just plugged and chugged into the formula.

It seems that the solutions manual uses the formula for the Al container, getting 2.0087L, which I follow. Then for the turpentine, they just do
2L(1+αΔT)=2.108L
So they do NOT cube it. What am I missing here? Why aren't they cubing the turpentine?
 
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Did they use the volume or the linear expansion coefficient?
 
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Ahh they used linear for the Al and volume for the turpentine. Thanks a bunch!
 
Feodalherren said:
Ahh they used linear for the Al and volume for the turpentine. Thanks a bunch!
They should have cubed for both, unless the used the volumetric coefficient of expansion for aluminum.

Chet
 

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