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.(adsbygoogle = window.adsbygoogle || []).push({});

First I found how much the height of the glass changed by using

[tex]\Delta[/tex]V=[tex]\alpha[/tex]glass x 1.55m x 17[tex]^{o}[/tex]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[tex]_{o}[/tex] = L[tex]^{3}_{o}[/tex]. 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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# Homework Help: Volume Thermal Expansion

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