Problem with Thermal expansion: volume

• mimi83
In summary, the temperature of a block of lead is raised from 0°C to 100°C, causing a change in its density. The equation ΔV = Vi βΔT may be used to solve this problem, but it must be noted that the wrong answer was obtained when attempting to use it. It is possible that the wrong coefficient, β, was used, as it is typically used for compressibility rather than thermal expansion. If using the thermal expansion coefficient α, it may be necessary to account for expansion in 3-D, which would require a correction factor of 3.
mimi83

Homework Statement

The temperature of a block of lead is raised from 0°C to 100°C. What is the percentage change in its density? (The density of lead at 0°C is 11,300 kg/m3.)

ΔV = Vi βΔT

The Attempt at a Solution

it gave the wrong answer when i tried to use the above equation

What did you use for β?

Is that a given in the problem?

I'm used to seeing β used for compressibility.

If you are using α the thermal expansion coefficient, you may need to account for expansion in 3-D, which corrects the linear α to volume α with the factor 3.

αv ≅ 3*αL

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thexp2.html#c3

.

There may be a problem with the equation you are using. The correct equation for thermal expansion is ΔV = V0αΔT, where V0 is the initial volume, α is the coefficient of thermal expansion, and ΔT is the change in temperature. The equation you used, ΔV = Vi βΔT, is for volumetric thermal expansion, where β is the volumetric coefficient of thermal expansion. Since we are only concerned with density, we should use the linear coefficient of thermal expansion, α, which is equal to 3β. Additionally, you will need to convert the density of lead from kg/m3 to g/cm3 in order to get a percentage change in density. The correct equation to use would be Δρ = ρ0αΔT, where ρ0 is the initial density. Therefore, the percentage change in density would be (Δρ/ρ0) * 100%. Plugging in the values, we get a percentage change in density of approximately 0.03%.

1. What is thermal expansion and how does it affect volume?

Thermal expansion refers to the increase in size or volume of a substance when its temperature increases. This is due to the molecules within the substance moving faster and taking up more space. The amount of expansion depends on the material's coefficient of thermal expansion.

2. How does thermal expansion cause problems with volume?

When a substance undergoes thermal expansion, it can cause changes in its volume. This can lead to issues such as warping, distortion, or even structural damage in materials. In some cases, this can also affect the accuracy of measurements and cause problems in machinery or devices.

3. What are some common materials that are affected by thermal expansion?

Most materials are affected by thermal expansion to some degree, but some are more susceptible than others. Metals, plastics, and glass are commonly affected by thermal expansion, while materials like rubber and concrete have a lower coefficient of thermal expansion and are less prone to changes in volume.

4. How can thermal expansion be controlled or minimized?

There are several ways to control or minimize the effects of thermal expansion. One method is to use materials with a lower coefficient of thermal expansion, such as ceramics or composites. Another approach is to design structures or devices with built-in flexibility to accommodate for changes in volume. Additionally, thermal insulation can help regulate temperature and reduce expansion.

5. Are there any practical applications of thermal expansion?

Thermal expansion has numerous practical applications, such as in thermometers, thermostats, and bimetallic strips. It is also utilized in the construction of bridges and buildings, where materials need to be able to expand and contract without causing structural damage. In some cases, thermal expansion can also be harnessed to generate energy in devices like steam turbines.

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