# Amount of water spilled when the temperature is changed

## 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?

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dV/dT = B V

Does not evaluate to logarithms
LOL, thanks, I don't know why I saw a T in the equation!

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)?

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.

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.

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?

haruspex
Homework Helper
Gold Member
ln(50) - ln(20)
How does integrating dT produce a ln function?

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
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.
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.