• Support PF! Buy your school textbooks, materials and every day products Here!

Amount of water spilled when the temperature is changed

  • #1
301
30

Homework Statement


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.)

Homework Equations


(dv/dT)/v = β[/B]

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?
 

Answers and Replies

  • #2
301
30
dV/dT = B V

Does not evaluate to logarithms
LOL, thanks, I don't know why I saw a T in the equation!
 
  • #3
301
30
LOL, thanks, I don't know why I saw a T in the equation!
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)?
 
  • #4
901
385
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)?
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.
 
  • #5
301
30
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.
Ah, I see, no problem.
 
  • #6
301
30
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?
 
  • #7
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,780
5,049
ln(50) - ln(20)
How does integrating dT produce a ln function?
 
  • #8
301
30
How does integrating dT produce a ln function?
Yeah, I did that incorrectly. I've fixed that in my post right before this one
 
  • #9
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
8,727
2,136
How are v0e(βΔT) and v0 + v0βΔT the same?
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.
 
  • #10
301
30
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.
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!
 

Related Threads on Amount of water spilled when the temperature is changed

  • Last Post
Replies
3
Views
971
  • Last Post
Replies
4
Views
3K
  • Last Post
2
Replies
26
Views
2K
  • Last Post
Replies
4
Views
4K
Replies
7
Views
3K
Replies
2
Views
1K
Replies
2
Views
545
Replies
6
Views
5K
  • Last Post
Replies
6
Views
449
Replies
0
Views
2K
Top