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Linear and Volume Expansion problem

  1. May 5, 2015 #1
    1. The problem statement, all variables and given/known data
    An aluminium container (cylinder) is filled with glycerine. Glycerine has a volume expansion coefficient of 4.85 × 10^-4 (°C)-1. Aluminium has a linear expansion coefficient of 2.40 x 10^-5 (°C)-1. At a temperature of 20.00°C the scale on the side of the aluminium container reads that there are 462.7 mL of glycerine in the container. At 105.2°C what volume of glycerine will the scale read?

    2. Relevant equations
    [tex]\Delta L = \alpha L \Delta T[/tex]
    [tex]\Delta V = \beta V \Delta T[/tex]

    3. The attempt at a solution
    When the aluminium container is 20 degree, 1 ml is 1 ml in scale, when it is 105.2 degree, the scale increase by 2.40 x 10^-5 x 1 x (105.2-20) =0.0020448 ml. Then I am not sure what the next step is.
     
  2. jcsd
  3. May 5, 2015 #2
    The next step is to correct your previous step. How is β of a material related to α of the same material?

    Chet
     
  4. May 5, 2015 #3
    β=3α? is this also work for liquid, is this mean I need to divide β by 3 to get the linear expansion for glycerine?
     
  5. May 5, 2015 #4
    No. You need to multiply the α of aluminum by 3 to get the volume expansion of the aluminum.

    Chet
     
  6. May 5, 2015 #5
    I will try again, Thanks
     
  7. May 8, 2015 #6
    Could you tell if I am correct please?

    αa = 2.4*10^-5 (°C)-1, βa = 3*αa = 7.2*10^-5 (°C)-1. therefore 1 ml in 20.00°C will become 1 * 7.2*10^-5 * (105.2 - 20) + 1 = 1.00613 ml.
    which means if put 20.00°C,462.7 mL( in 20.00°C aluminium container) glycerineat into 105.2°C aluminium container, the reading will be smaller which is 462.7/1.00613 = 459.88 mL.
    Then if 20.00°C,459.88 mL glycerine (in 105.2°C aluminium container) increase temperature to 105.2°C the final reading would be = 459.88 * 4.85 × 10^-4 * (105.2-20) + 459.88 = 478.88 mL
     
  8. May 8, 2015 #7
    I don't think this is correct.

    If the original volume at 20 C is 462.7 mL , what will be the actual volume of the aluminum container at this marker after the container has been heated to 105.2 C? (The open volume has to increase, not decrease).

    The volume of glycerine at 105.2 C is calculated correctly.

    Chet
     
  9. May 8, 2015 #8
    Sorry, I think I didn't make myself clear enough, of course if the container increase temperature it's open volume will increase, but in the first paragraph, I actually try to say that after container increase temperature, if we put 462.7 mL glycerin in 20 C scale to 105.2 C scale the reading will be smaller, as temperature of the liquid is still 20 C, volume doesn't change and therefore if moving it to a bigger scale, the reading is smaller. Is this correct?

    If my calculation for glycerin at 105.2 C is correct, is this the final reading from the container at 105.2 C? Thanks
     
  10. May 9, 2015 #9
    I think what you did in post # 6 is correct. I did it a little differently. Here was my approach:

    I said that the volume of glycerine increased to 462.7(1+4.85 x 10-4 x 85.2) = 481.8 ml

    The volume of the container, up to the 462.7 mark on the scale, increases to 462.7(1+7.2 x 10-5 x 85.2) = 465.5 ml
    So, at 105.2 C, the ratio of volume indicated on the scale in the aluminum container to actual volume is (462.7)/(465.5) = 0.994
    So the final volume of glycerine indicated on the scale is 0.994 x 481.8 = 478.9 ml. This matches with your value.

    Chet
     
  11. May 9, 2015 #10
    Thank you so much, really appreciate you answered all my question with patience !
     
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