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Homework Help: Density change due to thermal linear expansion

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    A certain metal has a coefficient of linear expansion of 2.00 × 10-5 K-1. It has been kept in a laboratory oven at 325°C for a long time. It is now removed from the oven and placed in a freezer at -145°C. After it has reached freezer temperature, the percent change in its density during this process is closest to

    2. Relevant equations

    ΔL=[itex]\alpha[/itex]L_oΔT, where alpha is the coefficient of linear expansion


    ρ = m/V

    3. The attempt at a solution

    Because density is related to volume, I figured that although we are given the coefficient of linear expansion, it should be converted to the coefficient of volumetric expansion, so that we can work with the second equation listed. But since we're not given any values for volume or length, I get an answer of ΔV=.0282V_0. Plugging this into ρ=m/V, doesn't really get me anywhere.

    I'm sure my approach is not quite right, but there aren't any other solutions coming to mind.

    Any help would be greatly appreciated. Thank you!
  2. jcsd
  3. Feb 7, 2012 #2


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    Often when original dimensions are not given, it is because the actual dimension do not affect the answer you seek.
    That means you can use L, W and H for the dimensions, and in the end, those variables will cancel out.

    One way to test for that, if you don't like using algebra, would be to assume some number values and calculate an answer, then assume a different set of starting numbers and see if the final answer is the same.

    A lot then depends on whether you were after [or required to give] just an answer or also how you got that answer.

    For assumed values, it is useful to use prime numbers like 7, 11, 13, 17, 19, 23 etc as it is less likely for answers to be coincidentally the same. [never use a 2, since 2+2 and 2*2 give the same answer - and you might accidentally cloud an error].
  4. Feb 7, 2012 #3
    I'm sorry, I'm still not understanding.

    I tried assuming a value of 7 for V_0, finding a chance in volume of -.1974, meaning that the volume decreased by 2.86 percent, if I did that correctly. If the volume decreases, then density ought to increase. The correct answer turns out to be +2.90%, but given that the other options are -2.9%, -2.74%, and +2.74%, I'm wondering if it's just coincidence that I got a value of 2.86, and that in fact I should be doing something to give me an exact value of 2.9.

    Thanks for the help, by the way.
  5. Feb 7, 2012 #4


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    I was suggesting you assume values for length, and width and height of the block, since you had the co-efficient of linear expansion. and work from first principles.
    Not sure how you got your figures of -.1974 and then 2.86%. I would like to see you basic calculations.
  6. Feb 7, 2012 #5
    Sure thing.

    So, first thing I did was obtain a value for beta (3)(2*10^-5)=6*10^-5

    Using ΔV=βV_0ΔT with V_0=7, and ΔT=-470, I got ΔV=-.1975.

    Then I found the difference in V, by subtracting .1975 from 7, which gave 6.8, and then I divided 6.8 by 7 to find the percentage that 6.8 is out of 7, which gave me 97.14%, but since we're looking at the amount it changed, .1975, I did 100-97.14 to find 2.86.
  7. Feb 7, 2012 #6


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    When you subtract .1975 form 7 you don't get 6.8 ?
  8. Feb 7, 2012 #7


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    btw: did you notice that to get ΔV you included a 7 in your multiplications, and then when you found your percentages, you divided by7 - so it would not have mattered what value you chose, and may just as well have used Vo.
  9. Feb 7, 2012 #8
    7-.1975 = 6.8025

    6.8025/7 = .9718*100 = 97.17

    100-97.17 = 2.83

    Seems I'm still not arriving at the value I need...
  10. Feb 7, 2012 #9


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    I am not exactly happy with this conversion β = 3α.

    It seems to me that if the linear expansion coefficient, and an appropriate temperature change resulted in a 10% reduction in each linear dimension, then the Length, Width and Height would all be 0.9 times the original.

    Vo = Lo * Wo * Ho

    The final Volume would be

    Vf = 0.9*Lo * 0.9*Wo * 0.9*Ho

    Vf = 0.729*LoWoHo

    so the volume is reduced by 27.1% , not 30%

    Perhaps I am not seeing it correctly?
  11. Feb 7, 2012 #10
    That makes sense, but I still don't really understand how to approach the problem as a whole...
  12. Feb 8, 2012 #11


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    By my calculation, a 470 degree change brings in a change (reduction) of (.0094) or 0.94% on each dimension, so each becomes 99.06% of their original value.
    When you cube that [L x W x H} that means the final volume is 97.206% of the original [or 0.97206 times the original

    Now, density is Mass / Volume and mass is fixed, so the new density is (1 / 0.97206) times the original

    1/0.97206 = 1.028743; which rounds to 1.029 or a 2.9% increase.

    If you trace the figures through carefully - without rounding too much { if I don't round and just carry through the 10 or so places my calculator works to, my final displayed figure is 1.028738585 (I think the calculator is actually working to a couple more places than that, and there is a technique to see the extra figures - but I have forgotten what it is.)
  13. Feb 8, 2012 #12


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    This technique follows on from the binomial expansion.

    (1+x)3 = 1 + 3x + 3x2 + x3

    Now if x is quite small, x2 is very small and x3 really tiny

    so (1+x)3 ≈ 1 + 3x

    Thus the idea we use with percentage errors.

    If the side length of a cube has a 2% uncertainty, then the Volume has a 6% uncertainty.
    This works fine because the 2% figure we started with is approximate already, and was possibly only 1.97% anyway [or 2.1%]

    [Try evaluating (1.01)3 and see how different it is from 1.03]

    In this example, a greater degree of accuracy was needed, so those extra terms (3x2 + x3) had to be considered as well. That is how you 2.83% got even closer to the 2.9% on offer.
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