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Homework Help: Trouble with a decimal point

  1. Sep 9, 2011 #1
    1. The problem statement, all variables and given/known data

    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
     
  2. jcsd
  3. Sep 10, 2011 #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
     
  4. Sep 10, 2011 #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?
     
  5. Sep 10, 2011 #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.
     
  6. Sep 10, 2011 #5
    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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