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

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Homework Help Overview

The discussion revolves around the effect of temperature on the height of a liquid column in a vertical glass tube. The original poster presents a scenario involving a glass tube that is half-filled with a liquid and seeks to understand how heating the tube affects the liquid's height.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the change in height of the liquid column based on the thermal expansion of both the glass and the liquid. They express uncertainty about their calculations and seek clarification on their approach.
  • Some participants question the assumption that the cross-sectional area of the tube remains constant during the temperature change and suggest that this should be considered in the calculations.
  • There are inquiries about how to determine the surface area of the tube without knowing its radius, indicating a need for further clarification on the geometric aspects involved.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the relationship between volume change and cross-sectional area, but there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are navigating the complexities of thermal expansion coefficients for both the glass and the liquid, as well as the implications of the tube's geometry on the height of the liquid column. There is an acknowledgment of potential simplifications in the calculations that may affect the outcome.

plfarrell
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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?
 
Last edited:
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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