Thermal Expansion in Liquids and Steel

In summary, the conversation discusses a problem with finding the correct final temperature using linear and volume expansion equations. The mistake was made in equalizing the volume changes between the oil and pot, and it is suggested to add a difference of 0.1 to the equation to get the correct answer.
  • #1
4
1
Homework Statement
The coefficient of linear expansion of steel is 11 x 10-6 K-1 and the and the coefficient of volume expansion of a certain cooking oil is 0.92 x 10-3 K-1. A steel cooking pot is 90% filled with this cooking oil, both at 24 oC. If the pot and oil are both slowly and uniformly heated, the temperature at which the oil will start to spill out of the pot is closest to:

A 50 oC

B 75 oC

C 100 oC

D 150 oC

E 200 oC
Relevant Equations
∆L=αLi∆T
∆V=βVi∆T
β=3α
This is a repost as I didn't read the community guidelines the first time. Hopefully this is better!

First thing I did was write down both the linear and volume expansion formulas.
I then multiplied my alpha by 3 and used the volume expansion equation for both materials.
∆V oil > ∆V pot, at the point where the oil bubbles over.
I used volumes of 0.9 for the oil and 1 for the pot and made the equations equal to each other as the ∆V would be the same at the point where it bubbled over.
In doing this, I got a final temperature of 24ºC which is clearly incorrect as the final temperature must be higher than the initial.
I have a feeling that my mistake was made in equalising the ∆V but I'm not too sure. Have been attempting this problem in different ways but can't seem to get the right answer.
The correct answer is D
Thanks!
 
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  • #2
kgm01 said:
I have a feeling that my mistake was made in equalising the ∆V but I'm not too sure.
Right. If both the oil and the pot increase their volume by the same amount, then the oil will still have less volume than the pot.
 
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  • #3
Ohh true. So if I did ∆V oil > ∆V pot + 0.1 (difference) I would get the correct answer. Thank you!
 
  • #4
kgm01 said:
Ohh true. So if I did ∆V oil > ∆V pot + 0.1 (difference) I would get the correct answer. Thank you!
Yes. Good.
 

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