Calculating Depth of Mercury Column at Higher Temperature

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In summary, the question involves finding the depth of a mercury column at a higher temperature, given the radius and initial depth of the column. The solution involves calculating the change in volume of both the glass and mercury due to their respective expansion coefficients, and then finding the net change in volume and using it to calculate the change in length of the column. The given answer of 0.0207 cm^3 seems incorrect and 2.07 cm^3 is a more plausible answer.
  • #1
Beva@371
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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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  • #2
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
 
  • #3
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?
 
  • #4
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.
 
  • #5
Thanks Rickz, I redid the question trusting the 2.07 and got the answer 64.862. I think this is more likely.
 

Related to Calculating Depth of Mercury Column at Higher Temperature

1. What is a decimal point?

A decimal point is a punctuation mark (.) that is used to separate the whole number from the fractional part in a decimal number. It is also referred to as a decimal mark or decimal separator.

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Some common issues related to decimal points include confusion with commas in different countries, incorrect placement of the decimal point, and rounding errors.

3. How do I know when to use a decimal point?

A decimal point is used when representing numbers that are not whole, such as fractions or percentages. It is also used in mathematical operations involving numbers with a fractional part.

4. What are some tips for avoiding trouble with a decimal point?

To avoid trouble with a decimal point, it is important to understand its purpose and correct placement. It is also helpful to double check calculations involving decimal numbers and to use decimal point-friendly tools, such as calculators or spreadsheets.

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