Calculating Depth of Mercury Column at Higher Temperature

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Homework Help Overview

The discussion revolves around a problem involving the calculation of the depth of a mercury column in a glass tube at varying temperatures. The problem includes parameters such as the radius of the tube, initial depth of mercury, and coefficients of linear expansion for both glass and mercury.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are examining the calculations related to volume expansion and questioning the results obtained for the change in volume of mercury. There are discussions about discrepancies in numerical results and the interpretation of the coefficients of expansion.

Discussion Status

There is an ongoing exploration of the calculations, with participants expressing confusion over the results and questioning the validity of the provided answer. Some participants are attempting to verify their own calculations against the original poster's findings.

Contextual Notes

Participants note that the problem involves specific temperature values and coefficients of expansion, which are critical to the calculations. There is an acknowledgment of potential errors in the interpretation of these values, as well as a recognition of the original source of the problem being from a textbook without a provided solution.

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Homework Statement



Here is the question:

A glass tube of radius 0.8cm contains liquid mercury to a depth of 64.0cm at 120C. Find the depth of the mercury column at 1000C.Assume that the linear expansion coefficient of the glass is 10X10-6K-1 and the linear expansion coefficient of mercury is 0.61X10-4K-1


3. And here is the solution... But what I don't understand is how the area in bold below was achieved. Every time I do the calculation i arrive at 2.072263 and the answer tells me I should be getting 0.0207! where are the two extra decimal points coming from?

Radius r =0.8 cm
Area of cross section A = πr 2
= 2.0106 cm 2
Depth h = 64 cm
Initial volume V = Ah
= 128.679 cm 3
Initial temperature t = 12 o C
Final temepreature t ' = 100 o C
The linear expansion coefficient of the glass α = 10X10-6K-1
The linear expansion coefficient of mercury α ' = 0.61X10-4K-1
The volume expansion coefficient of the glass γ = 3α
= 30X10-6K-1
The volume expansion coefficient of mercury γ ' = 3α '
= 1.83X10-4K-1
Change in volume of glass dV = Vγ ( t ' - t )
= 0.33971256 cm 3
Change in volume of mercury dV ' = Vγ' ( t ' - t )
= 0.020722466 cm 3

Net change in volume dV" = dV - dV '
= 0.318990094 cm 3
So, Change in length of the column dL = dV" / A
= 0.1586 cm
The depth of the mercury column at 1000C , h ' = h -dL
= 63.84 cm
 
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88K*1.83X10-4 K-1*128.679 cm 3

0.020722466 cm 3

the question you have written has T 120C and 1000C factor of 10 out from what the solution has

but not a factor of 100.. which is weird
 
The answer 0.0207 cm^3 doesn't look right to me. Mercury has a higher linear of expansion thus its expansion should be greater than that of the glass. For me 2.07 cm^3 is a more convincing answer. Where did you get that solution anyway?
 
It's one of the even numbered questions in the back of my text with no solution so I found that one online. But try as I might (and I'm very new to math and physics so could easily be missing something here) I can't get 1.83 x 10(^-4)(88) 128.56 to equal 0.0207... I get 2.07 and like you said, find the answer really odd.
 
Thanks Rickz, I redid the question trusting the 2.07 and got the answer 64.862. I think this is more likely.
 

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