How does temperature affect the height of a liquid column in a glass tube?

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The discussion focuses on how temperature changes affect the height of a liquid column in a glass tube. A specific example involves a 1.55 m long glass tube filled with liquid, initially at 23.0°C, and the calculations for height change when heated to 40.0°C. The initial calculations incorrectly assumed constant cross-section, which led to an inaccurate height change. The correct approach involves considering the change in volume of the liquid and the expansion of the tube's cross-section. Ultimately, the cross-sectional area simplifies in the calculations, allowing for an accurate determination of the height change without needing the tube's radius.
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A 1.55 m long vertical glass tube is half-filled with a liquid at 23.0°C. How much will the height of the liquid column change when the tube is heated to 40.0°C? Take αglass = 1.0 × 10-5 °C-1 and βliquid = 4.0 × 10-5 °C-1.

First I found how much the height of the glass changed by using
\DeltaV=\alphaglass x 1.55m x 17^{o}C.
I added that to the original height of the cylinder, and took the ratio of that value to the original, 1.5502635m/1.55m = 1.00017. (Sorry for not using sig. figs. I want to be as exact as possible.)
The textbook we have says that the V_{o} = L^{3}_{o}. Using that relationship, I found what the change in volume was, given the coefficient of the volume expansion, then converted that back into height by taking the cube root. I used the ratio found earlier to find the actual height of the liquid, then took the difference between the two giving me .7753074067m - .775 m = 3.07E-4m.
This answer is apparently incorrect, so can anyone shed some light on this situation? Thanks!
 
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The change in the cross section of the tube has an influence on the final height. You cannot assume that the cross section stays the same and just subtract the lengths.

$\Delta h =\Delta V_{liquid}/A_{tube}$
where
A_{tube}=A_0(1+2\alpha \Delta T)$
\Delta V_{liquid}=V_0 \beta \Delta T$
and
$ V_0 =A_0 L_0, L_0=1.55m/2 $
A_0 will simplify in the end.
 
How do i find the Surface Area of the tube without knowing its radius?
 
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It's not the surface area but the cross section area.
And you don't need it. It simplifies in the end, as I already mentioned.
 
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