# Thermal expansion of liquid in a tube

• Soren4
In summary, the conversation discusses calculating the new height of liquid in a cylindrical glass tube with temperature changes. The absolute change in height of the liquid can be found using a formula that takes into account the change in volume and area of the tube. The relative change in height, or the change in height relative to the tube, can also be calculated by considering the change in volume and area of the tube. However, there is some uncertainty about whether this approach is correct. The conversation also touches on the concept of thermal expansion and how equations for linear, planar, and volumetric expansion are derived.
Soren4

## Homework Statement

A cylindrical glass tube (linear thermal expansion coefficient ##\alpha##) contains liquid (volume thermal expansion coefficient ##\beta##). The height of the tube is ##h_{t,0}## and the height of the liquid inside of it is ##h_{l,0}##. If the temperature changes of an amount ##\Delta T## what is the new height of the liquid? If the cylindrical tube is provided of a measuring scale, what is the new height of liquid measured from the scale? Do not neglet any thermal expansion.

## Homework Equations

Thermal expansion coefficients

## The Attempt at a Solution

I get a bit confuse in considering all the expansions. The relation I would use is $$\frac{\Delta V}{V}\approx\frac{\Delta h}{h} +\frac{\Delta A}{A}$$
• To find the new "absolute" height of the liquid I would simply consider the change in volume ##\Delta V_{l}=V_{l,0} \beta \Delta T##, and then the change in the area of the cylinder ##\Delta A_{t}=A_{t,0} 2 \alpha \Delta T##. Then I would write
$$\frac{\Delta h_{l}}{h_{l,0}} =\frac{\Delta V_{l}}{V_{l,0}}+ \frac{\Delta A_{t}}{A_{t,0}}=(\beta-2\alpha) \Delta T$$
So actually in this case I did not consider the change in height of the tube, since, what is asked is simply the (absolute) change in height of the liquid.​
• While, to get the new height of liquid "relative to the tube" I would consider the "relative change in volume" $$\Delta V_{l,relative}=\Delta V_{l}-\Delta V_{t}=(V_{l,0} \beta- V_{t,0} 3\alpha)\Delta T$$
Here is my main doubt: does this "relative" change already takes into account the fact that the area of the tube changes and that the height of the tube changes? If so, considering this "relative change" I can write​
$$\frac{\Delta h_{l,relative}}{h_{l,0}}= \frac{\Delta V_{l,relative}}{V_{l,0}}$$
And get the new height relative to the tube, but I'm not convinced about this last equation I wrote.Are these two processes correct or are there any mistakes (conceptual or of other kind) ?

Hey Soren4,
The absolute looks OK.

Still thinking about the relative equation.
But as a hint, you have the absolute height of the liquid.
You can determine the absolute height of the marking on the tube.
Does your equation confirm the relative change in height tube/liquid?By the way, do you have an understanding about how
say, the equation of ΔV/V that you have listed comes about

ie from assuming small changes in temperature
linear ( 1 + α )
planar ( 1 + α ) ( 1 + α )
volumetric ( 1 + α )( 1 + α )( 1 + α )

Soren4,
For the volumes of liquid before and after changing T we have $$V_0 = h_0A_0$$ and $$V_1=h_1A_1$$ The second equation can be rewritten as
$$(V_0+\Delta V)=(h_0+\Delta h)(A_0+\Delta A)$$
$$V_0\left ( 1+\frac{\Delta V}{V_0} \right )=h_0A_0\left ( 1+\frac{\Delta h}{h_0} \right )\left ( 1+\frac{\Delta A}{A_0} \right )$$
$$1+\frac{\Delta h}{h_0}=\frac{1+\frac{\Delta V}{V_0}}{1+\frac{\Delta A}{A_0}}=\frac{1+\beta \Delta T}{1+2 \alpha \Delta T}$$
The new height
$$h_1=\frac{1+\beta \Delta T}{1+2 \alpha \Delta T}h_0$$

You can use the same approach in the second part of your problem.

## 1. How does thermal expansion affect the volume of liquid in a tube?

Thermal expansion is the tendency of matter to expand or contract in response to changes in temperature. When a liquid is heated, its molecules gain energy and move around more, causing the liquid to expand. This expansion can cause the volume of liquid in a tube to increase, potentially leading to overflow if the tube is not designed to accommodate the expansion.

## 2. What factors affect the degree of thermal expansion in a liquid?

The degree of thermal expansion in a liquid is affected by several factors, including the type of liquid, the temperature change, and the material of the tube. Different liquids have different coefficients of thermal expansion, meaning they expand at different rates. Additionally, the greater the temperature change, the greater the expansion will be. The material of the tube can also play a role, as some materials have a higher coefficient of thermal expansion than others.

## 3. Can thermal expansion cause a liquid to change state?

In most cases, thermal expansion alone is not enough to cause a liquid to change state. However, it can play a role in the phase transition of a substance when combined with other factors, such as pressure. For example, water will expand when heated, but it will not turn into steam until it reaches its boiling point and the pressure is low enough.

## 4. How is thermal expansion used in practical applications?

Thermal expansion is used in various practical applications, such as thermometers, thermostats, and thermocouples. These devices use the expansion and contraction of materials to measure changes in temperature. Additionally, thermal expansion is also taken into account in the design of structures, such as bridges and pipelines, to ensure they can withstand temperature changes without breaking or leaking.

## 5. Can thermal expansion be a problem in industrial processes?

Yes, thermal expansion can be a problem in industrial processes, particularly in systems that involve liquids or gases. If the expansion of a liquid is not accounted for, it can cause pipes or containers to burst, leading to costly repairs and potential safety hazards. It is important for engineers and designers to consider the effects of thermal expansion in industrial processes and implement measures to prevent any potential issues.

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