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Amount of water spilled when the temperature is changed

  1. Sep 15, 2018 #1
    1. The problem statement, all variables and given/known data
    A glass bottle of nominal capacity 250 cm3 is filled brim full of water at 20oC. If the bottle and content are heated to 50oC, how much water spills over? (For water, β=0.21X10-3 K-1. Assume that the expansion of the glass is negligible.)

    2. Relevant equations
    (dv/dT)/v = β

    3. The attempt at a solution
    Let vo = 250 cm3, To=20oC, T1 = 50oC, to find: v1-v0
    From the equation of coefficient of volume of expansion,

    dv/v = βdT
    Integrating both sides, we get: ln(v1) - ln(vo) = 0.21x10-3 *ln(50) - ln(20)

    => ln(v1) = 5.521 + 0.21x10-3 x ln(5/2)
    => v1 = 250.048
    => Δv = 0.048cm3

    However, the answer is 1.6 cm3, what am I doing wrong?
  2. jcsd
  3. Sep 15, 2018 #2
    LOL, thanks, I don't know why I saw a T in the equation!
  4. Sep 15, 2018 #3
    Wait, if I move the volume to the other side of the equation, wouldn't I get log for volume? dv/V = BdT => ln(V1/V0) = B (T1 - T0)?
  5. Sep 15, 2018 #4
    That’s why I immediately deleted the post. I hit post too fast. On my phone I type very slowly, so it is embarrassing that my thumbs nevertheless out run my brain. Sorry.
  6. Sep 15, 2018 #5
    Ah, I see, no problem.
  7. Sep 16, 2018 #6
    So I did the integral! β = 1/v(dv/dT)
    => βdT = dv/v => βΔT = ln(v1/v0) => v1 = v0eΔT). When I plug in the values of v0, β and ΔT, I get the value of v1 as 251.58, hence, Δv = 1.58. Which is what the answer is. However, in a solution manual I found online, they use the equation Δv = v0βΔT. And they get the correct answer. My question is, are these two equivalent?

    From my derivation, v1 = v0e(βΔT)
    From the solutions manual, v1 = v0 + v0βΔT

    How are v0e(βΔT) and v0 + v0βΔT the same?
  8. Sep 16, 2018 #7


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    How does integrating dT produce a ln function?
  9. Sep 16, 2018 #8
    Yeah, I did that incorrectly. I've fixed that in my post right before this one
  10. Sep 16, 2018 #9


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    It's a series expansion. For small values of ##\beta \Delta T## to first order, ##e^{\beta \Delta t} \approx1+\beta \Delta T.## Calculate it both ways and see what you get.
  11. Sep 16, 2018 #10
    Yeah, I did something similar and I see that behavior. I plotted the two functions: https://www.desmos.com/calculator/vrhjyq5tk0. So,e^(βΔt) is the correct expression while the other one is just an approximation for small Δts. Awesome, thank you!
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