Calculating Depth of Mercury Column at Higher Temperature

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The discussion focuses on calculating the depth of a mercury column in a glass tube at varying temperatures, specifically from 120°C to 1000°C. The key point of contention is the calculated change in volume of mercury, with one participant consistently arriving at a value of 2.072263 cm³, while the provided solution states it should be 0.0207 cm³. Participants express confusion over the discrepancy, questioning the accuracy of the solution and the assumptions made regarding the expansion coefficients of mercury and glass. The calculations involve determining the initial volume, volume expansion coefficients, and net change in volume to find the new depth of the mercury column. Ultimately, there is skepticism about the correctness of the provided answer, leading to a re-evaluation of the calculations.
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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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