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Thermal Expansion: Finding common temperature (Ring/Sphere)

  1. Mar 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the common temperature at which the inner diameter of the ring is 0.05% larger than the diameter of the sphere.

    Given:
    Lead sphere
    d = 5cm
    To = 70°C
    ∝ = 29x10-6/C°
    Steel ring
    d = 4.9975cm (the sphere is 0.05% larger than the inner diameter of the ring, hence this value)
    To = 70°C
    ∝ = 12x10-6/C°
    2. Relevant equations
    Af = Ao(1+2∝Δt)
    Δt = Tf - To
    3. The attempt at a solution
    Equate both final area of sphere (S) and ring (R)
    AFR = AFS (1.0005)
    Substitute
    Af = Ao(1+2∝Δt)
    AoR(1+2∝RΔt) = (1.0005)AoS(1+2∝SΔt)
    Arrange equation to find Δt
    Δt = (1.0005)(AoS) - (AoR) / 2(AoR)(∝R) - (1.0005)(2)(AoS)(∝L)
    Enter the values
    Δt = (1.0005)(π(5cm/2)2) - (π(4.9975cm/2)2) / 2(π(4.9975cm/2)2)(12x10-6/C°) - 2(1.0005)(π(5cm/2)2)(29x10-6/C°)
    The answer is
    Δt = -44.0416 C°
    Which is incorrect because the Final temperature is supposed to be 11.2766°C, using my answer and finding for the final temperature with 70°C as initial.
    Tf = 25.9584°C
    EDIT:
    I've found out that if I multiplied 1.0005 on the diameter of lead sphere (5cm) immediately before others, I'll get an answer of Δt = -58.6820 C° where the final temperature will be 11.3180°C which is pretty close but not the exact answer :(
     
  2. jcsd
  3. Mar 8, 2017 #2

    rl.bhat

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    Homework Helper

    In the case of ring, you have to consider linear expansion, not the areal expansion. Similarly in the care of sphere you have to consider volume expansion.
     
  4. Mar 8, 2017 #3
    Linear expansion for both.
     
  5. Mar 8, 2017 #4
    @rl.bhat I tried using volume and linear but I arrived with a bigger value.

    @mjc123 Tried to make both linear and I arrived with Δt = - 58.7521 C° which is pretty good, finding the final temperature I got 11.2473 °C although the given correct answer is 11.2766 °C. I'll try to recompute later.

    The problem is actually one out of four question, I used area of expansion on the three others and found the exact correct answer. This is the last question and I'm having trouble in getting the exact value (11.2766 °C).
     
  6. Mar 8, 2017 #5

    haruspex

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    Gold Member
    2016 Award

    The question statement violates the usual rules of specifying accuracy. E.g. it should give the lead diameter as 5.0000cm.
    Having extended all the data to that accuracy, you might think that the answer can be given to that accuracy, but not so. In this question, small differences are calculated between numbers of similar magnitude. E.g. 5.0025-4.9975=0.0050, so 5 digits of accuracy falls to only two.
    Thus, it does not really make any sense to quote the answer more precisely than 11°C.

    But for what it is worth, I get 11.256. If you care about that difference, please show exactly how you calculated it.
     
  7. Mar 8, 2017 #6
    That was what I was thinking but it bothered me that it was the only answer that didn't exactly gave the exact value from the answer sheet. Thanks for the help guys.
     
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