To solve this problem, we need to understand how thermal expansion works and how it affects the dimensions of the copper ring and the aluminum sphere. The key here is that the sphere passes through the ring at thermal equilibrium, meaning their temperatures equalize, and we can use the coefficients of linear expansion for both materials to find the mass of the aluminum sphere.

Understanding Thermal Expansion

When materials are heated, they expand. This expansion can be quantified using the formula:

ΔL = α × L₀ × ΔT

Where:

• ΔL is the change in length (or diameter in this case).
• α is the coefficient of linear expansion for the material.
• L₀ is the original length (or diameter).
• ΔT is the change in temperature.

Coefficients of Linear Expansion

For our materials, the coefficients are approximately:

• Copper (Cu): α ≈ 16.5 × 10^-6 °C^-1
• Aluminum (Al): α ≈ 23.1 × 10^-6 °C^-1

Calculating the Change in Diameter

First, we need to calculate the change in diameter for both the copper ring and the aluminum sphere as they reach thermal equilibrium.

For the Copper Ring

The original diameter of the copper ring at 0ºC is 2.54000 cm. The temperature change when it reaches equilibrium with the aluminum sphere (at 100ºC) is:

ΔT (Cu) = 100ºC - 0ºC = 100ºC

Now, applying the formula for thermal expansion:

ΔL (Cu) = α_Cu × L₀_Cu × ΔT

ΔL (Cu) = (16.5 × 10^-6 °C^-1) × (2.54000 cm) × (100ºC)

ΔL (Cu) ≈ 0.0004191 cm

The new diameter of the copper ring at equilibrium:

D_Cu = 2.54000 cm + 0.0004191 cm ≈ 2.54042 cm

For the Aluminum Sphere

The original diameter of the aluminum sphere at 100ºC is 2.54533 cm. Since it cools down to the equilibrium temperature, we need to calculate the change in diameter:

ΔT (Al) = T_final - 100ºC

At equilibrium, the sphere just passes through the ring, so:

D_Cu = D_Al

Thus, we can set up the equation:

2.54042 cm = 2.54533 cm - ΔL (Al)

Solving for ΔL (Al):

ΔL (Al) = 2.54533 cm - 2.54042 cm ≈ 0.00491 cm

Now, applying the thermal expansion formula for the aluminum sphere:

ΔL (Al) = α_Al × L₀_Al × ΔT

We can rearrange this to find the change in temperature:

ΔT = ΔL (Al) / (α_Al × L₀_Al)

Substituting the known values:

ΔT = 0.00491 cm / (23.1 × 10^-6 °C^-1 × 2.54533 cm)

ΔT ≈ 78.5ºC

Thus, the final temperature of the aluminum sphere is:

T_final = 100ºC - 78.5ºC ≈ 21.5ºC

Finding the Mass of the Sphere

To find the mass of the aluminum sphere, we can use the volume formula for a sphere:

V = (4/3)πr³

First, we need to find the radius of the aluminum sphere:

r_Al = D_Al / 2 = 2.54533 cm / 2 ≈ 1.272665 cm

Now, calculating the volume:

V_Al = (4/3)π(1.272665 cm)³ ≈ 8.586 cm³

The density of aluminum is approximately 2.70 g/cm³. Therefore, the mass of the sphere can be calculated as:

m_Al = density × volume

m_Al = 2.70 g/cm³ × 8.586 cm³ ≈ 23.2 g

In summary, the mass of the aluminum sphere is approximately 23.2 grams. This problem illustrates the fascinating interplay between thermal expansion and material properties, showcasing how temperature changes can affect physical dimensions and lead to practical applications in engineering and design.