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Aluminium bar

  1. May 21, 2008 #1
    1. The problem statement, all variables and given/known data

    A 40mm diameter bar of aluminium is 2.5m long the bar is has its temperature raised from 20°c to 40°c. Determine the increase in length of the bar in mm. Take the Al = 24 × 10-6 °C-1
    How much heat has been transferred into the bar if the SCH = 0.9kj/kg °C


    2. Relevant equations



    3. The attempt at a solution

    Am stuck on where to start could any one help please
     
  2. jcsd
  3. May 21, 2008 #2

    Hootenanny

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    Saying that you are stuck on where to start isn't an attempt at a solution. You much have some ideas as to where to start. Have you read your class notes or course text?
     
  4. May 21, 2008 #3
    increase in length = linear expansion coefficent × lenght of material (t2 - t1)

    I do not understand where the t2 - t1 come from.

    Is it the temperture increase - the temperture of the bar before it is rasied.
     
  5. May 21, 2008 #4

    tiny-tim

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    Hi speedy46! :smile:

    Yes, it's the temperature increase - final temperature minus initial temperature.

    Increase in length is more if the original length is more,

    and it's more if the temperature increase is more.

    So it's proportional to both of them (in other words, to their product). :smile:
     
  6. May 21, 2008 #5
    would this also work for decreassing temperture on a material?
     
  7. May 21, 2008 #6

    tiny-tim

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    Yes … the effect is reversible.

    If a particular temperature increase makes it longer, then the same decrease will return it to its original length. :smile:

    And cooling it further will make it even shorter than its normal length.

    Some armies used to keep iron cannonballs inside brass "barrels". If the temperature fell below freezing point overnight, the iron and the brass would both decrease in size, but the brass would decrease faster, so the cannonballs would get closer to the top, and sometimes one of them would even be forced out of the brass barrel! :biggrin:
     
  8. May 21, 2008 #7
    please could you tell me if this right or not

    increase in length = 24 × 10-6 × 40 mm (40 - 20) 24 ×10-6 × 40 × 20 = 0.0192
     
  9. May 21, 2008 #8

    tiny-tim

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    Looks good! :smile:
     
  10. May 21, 2008 #9
    thank you now i have to work out how much heat has been transfered into the bar if the SCH= 0.9KJ/ KG degrees Celsius.

    What does the SHC stand for please
     
  11. May 21, 2008 #10

    tiny-tim

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  12. May 21, 2008 #11
    Thank you

    To work out heat transfer

    Q= (t1-t2)/(l÷Ka)

    so 20 - 40 = 20

    20 divide by 2.5m divide by ?

    I do not know what the Ka stands for
     
  13. May 24, 2008 #12

    tiny-tim

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    Hi speedy46! :smile:
    Sorry, I've no idea what Ka is … where did you get it from? :confused:

    All you need can be found in the units of the SCH in the question:
    /kg °C means per kilogram, and per degree.

    So you find the number of kJ by multiplying 0.9 by the weight and also by the change in temperature … and nothing else! :smile:

    btw, wikipedia gives the SCH as 0.9 J /g K … can you see why that's the same?
     
  14. May 27, 2008 #13
    so the change in temperture is 20 degress 20-40=20

    but the question does not give the weight or can i work this out.
     
  15. May 27, 2008 #14

    tiny-tim

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    hmm … that's very true! :rolleyes:

    they only give you the volume, and you would need to know the density also. :cry:
     
  16. May 27, 2008 #15
    density of aluminum is 2.7g/cm³

    volume of the bar is πr^2*h π * 20 2 * 2500 = 3141592.654

    I changed meters in to millmeters to keep them in the same units 1000*2.5= 2500mm
     
    Last edited: May 27, 2008
  17. May 28, 2008 #16
    so now what do I need to do knowing the volume and density
     
  18. May 28, 2008 #17

    tiny-tim

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    mass = volume times density,

    then SCH = 0.9kj per kilogram, and per degree. :smile:
     
  19. May 28, 2008 #18
    2.7 3 times 33141592.654 = 61835968.21

    is this correct
     
  20. May 28, 2008 #19

    tiny-tim

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    Sorry, you've lost me. :confused:

    Write it out in full … and remember you need the mass in kg.

    (and don't work to 10 significant figures! :rolleyes: )
     
  21. May 28, 2008 #20
    volume of the bar is pie times 20 squared times 2500 = 3141592
    The density of aluminum is 2.7g/cm3

    Mass = volume times mass

    mass = 3141592 times 2.7g/cm3

    mass = 61835955.34

    6.1kg

    is this correct am unsure
     
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