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Linear Expansion of Solid problem

  1. Jan 30, 2007 #1
    1. The problem statement, all variables and given/known data

    There is an aluminum ring, .07 m in diameter at 5 deg. C. There is also an aluminum shaft whose diameter is .07003 m in diameter at 5 deg. C. What temp. should the shaft be cooled such that the ring will fit over the shaft.



    2. Relevant equations

    /\L = Coefficient of Lin. Expansion (Original Length) /\T




    3. The attempt at a solution

    step 1) .07 - .07003 = 2.4 x 10^-5 (.07003) (5 - Tf)

    * The original length is .07003 since is what is being cooled to eventually reach .07 m. Also, the Coefficient of Lin. Expansion for Aluminum is 2.4 x 10^ -5

    step 2) -3.0 x 10^-5 = 1.681 x 10^-6 (5 - Tf)

    step 3) -3.0 x 10^-5 = 8.405 x 10^-6 - 1.681 x 10^-6 Tf


    now when i solve for Tf, it is not coming out to -12.85 deg. C, which the book states is the answer...so I'm assuming there is an error somewhere, or a few places.....any help would be appreciated , thanks
     
  2. jcsd
  3. Jan 30, 2007 #2

    AlephZero

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    Science Advisor
    Homework Helper

    Your step 3 is OK. Check you didn't make a mistake with the signs or the exponents. Multiplying everything by 10^6 might help.

    -30 = 8.405 - 1.681 Tf

    EDIT:

    Oops, P3X-018 is right. I made two mistakes which cancelled out and gave me the right answer. Need more coffee....
     
    Last edited: Jan 30, 2007
  4. Jan 30, 2007 #3
    I assume you get a reversed sign? When you calculate the temperature difference you use "initial temperature - final temperature" you need to use the same order about the difference in the diameter, that is "inital diameter - final diameter". You do it reversed for the length, what is the initial diameter?
     
  5. Jan 30, 2007 #4
    Linear expansion

    consider
    for variation in lenth Delta L for any variation in temp Delta T is

    Delt a L = Coeff of Linear expansion * Original Length * Delta T
    or
    Delta T = Delta L / ( C of L exp * Original Length)
    in the present case


    delta L = 0.07003 - 0.07000 = 0.00003
    then delta T = 0.00003 / ( 2.4 * 10 ^-5 *0.07003)
    = ( 3 * 10^5)/ (2.4 * 7003)
    = 300000/(1687.02)

    =17.849
    since delta L is negative, delta T also will be negative , which means
    delta T = - 17.849
    or -17.85 = final temp -original temp = Ft - 5 or

    FT= -12.85
    hope this is clear
     
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