# Amount of water spilled when the temperature is changed

• Phys12
In summary, the amount of water spilled when the temperature is changed depends on the temperature difference, surface area, and container shape. When the temperature increases, the water molecules have more energy and collide with each other, causing the water to expand and potentially overflow. On the other hand, when the temperature decreases, the water molecules lose energy and become more compact, resulting in a decrease in volume and less water spilled. The surface area and shape of the container can also affect the amount of water spilled, as a larger surface area or uneven shape can cause the water to spill more easily. Overall, the change in temperature can greatly impact the amount of water spilled, and it is important to consider these factors when handling liquids.
Phys12

## 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?

Cutter Ketch said:
dV/dT = B V

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

Phys12 said:
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)?

Phys12 said:
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.

Cutter Ketch said:
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?

Phys12 said:
ln(50) - ln(20)
How does integrating dT produce a ln function?

haruspex said:
How does integrating dT produce a ln function?
Yeah, I did that incorrectly. I've fixed that in my post right before this one

Phys12 said:
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.

Chestermiller
kuruman said:
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!

## 1. How does temperature affect the amount of water spilled?

When water is heated, it expands and takes up more space, leading to an increase in volume. This means that as the temperature of water increases, the amount of water spilled will also increase.

## 2. Is it true that more water is spilled when the temperature is colder?

No, this is not necessarily true. While it is true that water contracts when cooled, causing a decrease in volume, the amount of water spilled may still vary depending on factors such as the shape and size of the container, the rate at which the water is poured, and the type of surface the water is spilled on.

## 3. How significant is the impact of temperature on the amount of water spilled?

The impact of temperature on the amount of water spilled can vary depending on the specific conditions. In some cases, the difference may be minimal and hardly noticeable, while in others, it can be more significant. Factors such as the temperature difference and the amount of water being spilled can also affect the impact.

## 4. Can the amount of water spilled be predicted based on temperature changes?

While temperature does play a role in the amount of water spilled, it is not the only factor that affects it. Other variables, such as the surface tension of the water and the force of gravity, also play a role in determining the amount of water spilled. Therefore, it is not possible to accurately predict the amount of water spilled based on temperature changes alone.

## 5. How can the amount of water spilled be controlled when the temperature is changed?

To control the amount of water spilled when the temperature is changed, one can use containers with a larger volume or pour the water at a slower rate. Additionally, using surfaces with higher surface tension, such as waxed paper, can help reduce the amount of water spilled as the temperature changes.

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