Volume Thermal Expansion: Videos & Practice Problems

Volumetric thermal expansion describes how three-dimensional objects, like cubes or spheres, increase in volume when temperature rises. The equation for volume change, ΔV = βV0ΔT, uses the volumetric expansion coefficient, β, which is three times the linear expansion coefficient, α. For example, a lead ball's volume decreases when cooled, calculated using the same principles of thermal expansion, demonstrating the relationship between temperature and volume in materials.

Volume Thermal Expansion

In the study of thermal expansion, we explore how materials change in size when subjected to temperature variations. While linear thermal expansion focuses on one-dimensional objects, volumetric thermal expansion applies to three-dimensional objects, such as cubes and spheres. When the temperature of a 3D object increases, its volume expands in all three dimensions, leading to a change in volume, denoted as ΔV.

The equation for volumetric thermal expansion is expressed as:

$$\Delta V = \beta \cdot V_0 \cdot \Delta T$$

Here, β represents the volumetric expansion coefficient, \(V_0\) is the initial volume, and ΔT is the change in temperature. This equation mirrors the linear expansion equation, which uses ΔL for length changes, highlighting the relationship between temperature change and volume change.

To find the final volume, the equation can be rearranged to:

$$V_f = V_0 \cdot (1 + \beta \cdot \Delta T)$$

It is important to note that the volumetric expansion coefficient (β) is related to the linear expansion coefficient (α) by the equation:

$$\beta = 3\alpha$$

This means that for the same material, the volumetric expansion coefficient is three times the linear expansion coefficient. For example, if the linear expansion coefficient of aluminum is \(2.4 \times 10^{-5} \, \text{K}^{-1}\), then its volumetric expansion coefficient would be:

$$\beta = 3 \cdot (2.4 \times 10^{-5}) = 7.2 \times 10^{-5} \, \text{K}^{-1}$$

To illustrate this concept, consider a lead ball with an initial temperature of 333 K and an initial volume of 50 cm³. If the temperature decreases to 303 K, we can calculate the change in volume (ΔV). First, we determine ΔT:

$$\Delta T = T_f - T_i = 303 \, \text{K} - 333 \, \text{K} = -30 \, \text{K}$$

Next, we calculate β using the linear expansion coefficient (α = \(2.9 \times 10^{-5} \, \text{K}^{-1}\)):

$$\beta = 3 \cdot (2.9 \times 10^{-5}) = 8.7 \times 10^{-5} \, \text{K}^{-1}$$

Now, substituting the values into the volumetric expansion equation:

$$\Delta V = 8.7 \times 10^{-5} \cdot 50 \, \text{cm}^3 \cdot (-30)$$

Calculating this gives:

$$\Delta V = -0.13 \, \text{cm}^3$$

This negative value indicates a decrease in volume as the temperature drops. Understanding these principles of volumetric thermal expansion is crucial for applications in engineering and materials science, where temperature fluctuations can significantly impact the performance and integrity of materials.

Expansion of a Hemispherical Dome

In this scenario, we explore the concept of volumetric thermal expansion using a geodesic hemispherical dome made of aluminum. A hemispherical dome can be visualized as half of a sphere, and in this case, the radius of the dome is measured at 25 meters when the temperature is -10 degrees Celsius. As the temperature rises to 30 degrees Celsius, the aluminum expands, increasing the radius and consequently the interior volume of the dome.

The change in volume, denoted as ΔV, is calculated using the formula for volumetric thermal expansion:

ΔV = β V₀ ΔT

Here, β represents the volumetric expansion coefficient for aluminum, which is approximately 7.2 x 10^-5 °C^-1. The initial volume V₀ of the hemispherical dome needs to be determined first. The volume of a hemisphere is given by:

V_hemisphere = (2/3) π r³

Substituting the radius of 25 meters into this formula, we find:

V₀ = (2/3) π (25)³ = 3.27 x 10⁴ m³

Next, we calculate the change in temperature ΔT, which is the difference between the final and initial temperatures:

ΔT = 30 - (-10) = 40 °C

Now, substituting the values into the volumetric expansion formula:

ΔV = (7.2 x 10^-5) (3.27 x 10⁴) (40)

Calculating this gives us a change in volume of:

ΔV = 94.2 m³

This indicates that the aluminum dome has expanded significantly, resulting in an increase of 94.2 cubic meters of interior space due to the temperature rise. Understanding this concept is crucial in fields such as engineering and architecture, where material properties and thermal expansion must be considered in design and construction.

Overflowing Mercury

In this scenario, we explore the thermal expansion of a glass flask filled with mercury when the temperature is increased from 0 degrees Celsius to 100 degrees Celsius. Both the glass and the mercury expand, but they do not expand equally due to their different coefficients of volumetric expansion, denoted as β (beta). The coefficient for mercury is greater than that for glass, which leads to a situation where the mercury overflows from the flask.

To quantify the overflow, we define the change in volume for both substances. The change in volume for the mercury, denoted as ΔV_mercury, can be expressed using the formula:

ΔV_mercury = β_mercury × V_initial × ΔT

Similarly, the change in volume for the glass is given by:

ΔV_glass = β_glass × V_initial × ΔT

Here, V_initial is the initial volume of both the glass and the mercury, which is 250 cm³, and ΔT is the change in temperature, which is 100 degrees Celsius.

The volume of mercury that spills out, denoted as V_spill, is calculated by subtracting the change in volume of the glass from that of the mercury:

V_spill = ΔV_mercury - ΔV_glass

Substituting the expressions for ΔV, we have:

V_spill = (β_mercury - β_glass) × V_initial × ΔT

Plugging in the values, we find:

V_spill = 250 cm³ × 100 × (1.8 × 10^-4 - 1.2 × 10^-5)

Calculating this gives us a result of V_spill = 4.2 cm³. This positive value indicates that indeed, some mercury will spill out of the flask when the temperature is raised, confirming our understanding of thermal expansion in different materials.

Here’s what students ask on this topic:

What is the equation for volumetric thermal expansion?

The equation for volumetric thermal expansion is given by:

$\mathrm{\Delta V}=\beta V$₀ΔT

Here, $\mathrm{\Delta V}$ is the change in volume, $\beta$ is the volumetric expansion coefficient, $V$₀ is the initial volume, and $\mathrm{\Delta T}$ is the change in temperature. This equation shows how the volume of a three-dimensional object changes with temperature.

How is the volumetric expansion coefficient (β) related to the linear expansion coefficient (α)?

The volumetric expansion coefficient $\beta$ is related to the linear expansion coefficient $\alpha$ by the equation:

$\beta =3\alpha$

This relationship holds because volumetric expansion considers changes in all three dimensions (length, width, and height), whereas linear expansion considers changes in only one dimension.

What is the difference between linear thermal expansion and volumetric thermal expansion?

Linear thermal expansion refers to the change in length of a one-dimensional object when its temperature changes. The equation used is:

$\mathrm{\Delta L}=\alpha L$₀ΔT

Volumetric thermal expansion, on the other hand, refers to the change in volume of a three-dimensional object when its temperature changes. The equation used is:

$\mathrm{\Delta V}=\beta V$₀ΔT

Here, $\beta$ is three times the linear expansion coefficient $\alpha$.

How do you calculate the final volume of an object after thermal expansion?

The final volume $V$_f of an object after thermal expansion can be calculated using the equation:

$V$_f=V₀(1+βΔT)

Here, $V$₀ is the initial volume, $\beta$ is the volumetric expansion coefficient, and $\mathrm{\Delta T}$ is the change in temperature. This equation accounts for the increase in volume due to the temperature change.

What happens to the volume of a lead ball when it is cooled from 333 K to 303 K?

When a lead ball is cooled from 333 K to 303 K, its volume decreases. To calculate the change in volume $\mathrm{\Delta V}$, we use the equation:

$\mathrm{\Delta V}=\beta V$₀ΔT

Given the initial volume $V$₀ is 50 cm³, the linear expansion coefficient $\alpha$ is 2.9 × 10^-5 K^-1, and $\beta =3\alpha$, we find:

$\beta =8.7\times 10⁻5$

With $\mathrm{\Delta T}=-30$ K, the change in volume is:

$\mathrm{\Delta V}=8.7\times 10⁻5\times 50\times -30=-0.13$ cm³

Thus, the volume decreases by 0.13 cm³.