Amount of water spilled when the temperature is changed

[Phys12]

Homework Statement

A glass bottle of nominal capacity 250 cm³ is filled brim full of water at 20^oC. If the bottle and content are heated to 50^oC, 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 v_o = 250 cm³, T_o=20^oC, T₁ = 50^oC, to find: v₁-v₀
From the equation of coefficient of volume of expansion,

dv/v = βdT
Integrating both sides, we get: ln(v₁) - ln(v_o) = 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?

 

[Phys12]

dV/dT = B V

Does not evaluate to logarithms

LOL, thanks, I don't know why I saw a T in the equation!

 

[Phys12]

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)?

 

[Cutter Ketch]

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.

 

[Phys12]

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.

 

[Phys12]

So I did the integral! β = 1/v(dv/dT)
=> βdT = dv/v => βΔT = ln(v₁/v₀) => v₁ = v₀e^(βΔT). 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?

 

[haruspex]

ln(50) - ln(20)

How does integrating dT produce a ln function?

 

[Phys12]

How does integrating dT produce a ln function?

Yeah, I did that incorrectly. I've fixed that in my post right before this one

 

[kuruman]

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.

 

[Phys12]

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!