A 20.0 g copper ring at 0.000°C has an inner diameter of D = 2.54000 cm. An aluminum sphere at 108.0°C has a diameter of d = 2.54508 cm. The sphere is placed on top of the ring, and the two are allowed to come to thermal equilibrium, with no heat lost to the surroundings. The sphere just passes through the ring at the equilibrium temperature, and the coefficients of linear expansion of aluminum and copper are αAl = 23*10^-6/C° and αCu = 17*10^-6/C°, respectively. . What is the mass of the sphere?
The mass of the sphere can be calculated using the equation m = (4/3)πr^3ρ, where r is the radius of the sphere and ρ is the density of aluminum.
The radius of the sphere can be calculated using the equation r = (d - D)/2, where d is the diameter of the sphere and D is the diameter of the ring.
The density of aluminum can be calculated using the equation ρ = m/V, where m is the mass of the aluminum and V is the volume of the aluminum.
The volume of the aluminum can be calculated using the equation V = (4/3)πr^3, where r is the radius of the sphere.
Substituting the values into the equations, we get:
m = (4/3)π((2.54508 cm - 2.54000 cm)/2)^3 * (20.0 g / ((4/3)π((2.54508 cm - 2.54000 cm)/2)^3))
m = 20.0 g * (2.54508 cm - 2.54000 cm)^3 / ((4/3)π(2.54000 cm)^3)
m = 0.0014 g
Therefore, the mass of the sphere is 0.0014 g.
To find the mass of the sphere, we need to use the concept of thermal expansion and thermal equilibrium.
Step 1: Calculate the change in diameter of the ring and sphere:
ΔD = d - D
Given:
D = 2.54000 cm
d = 2.54508 cm
ΔD = d - D = 2.54508 cm - 2.54000 cm = 0.00508 cm
Step 2: Calculate the change in temperature:
ΔT = T_f - T_i
Given:
Initial temperature of the ring, T_i = 0.000°C
Final temperature of the system (thermal equilibrium), T_f = ?
Step 3: Use the coefficient of linear expansion to relate the change in temperature to the change in length:
ΔT = ΔL * α
For copper (ring):
ΔL_cu = D * αCu
For aluminum (sphere):
ΔL_al = d * αAl
Step 4: Set the change in length of copper equal to the change in length of aluminum:
ΔL_cu = ΔL_al
D * αCu = d * αAl
Step 5: Solve for the final temperature, T_f:
T_f = (ΔL_al * T_i) / ΔL_cu + T_i
Substituting in the given values:
T_f = (d * αAl * T_i) / (D * αCu) + T_i
Step 6: Substitute the values and calculate the final temperature:
T_f = (2.54508 cm * 23 * 10^-6 /C° * 0.000°C) / (2.54000 cm * 17 * 10^-6 /C°) + 0.000°C
T_f ≈ 108.036°C
Step 7: Calculate the mass of the sphere:
The mass of the sphere can be calculated using the mass formula:
m = ρ * V
Given information:
ρ_aluminum = 2.70 g/cm^3 (density of aluminum)
We need to find the volume of the sphere, V.
The volume of a sphere can be calculated using its diameter:
V = (4/3) * π * (d/2)^3
Substituting the values:
V ≈ (4/3) * π * (2.54508 cm / 2)^3
V ≈ 13.20585 cm^3
Step 8: Calculate the mass of the sphere:
m = ρ_aluminum * V
Substituting the values:
m ≈ 2.70 g/cm^3 * 13.20585 cm^3
m ≈ 35.699 g
Therefore, the mass of the sphere is approximately 35.699 g.
To find the mass of the sphere, we need to use the principle of thermal expansion. The diameter of the copper ring increases as it is heated, while the diameter of the aluminum sphere also increases due to its own expansion. When the two objects reach thermal equilibrium, the sphere just passes through the ring.
Let's start by finding the change in diameter for both the copper ring and the aluminum sphere.
ΔDCu = αCu * DCu * ΔT
Where ΔDCu is the change in diameter of the copper ring, αCu is the coefficient of linear expansion for copper, DCu is the initial diameter of the copper ring, and ΔT is the change in temperature.
Substituting the given values:
ΔDCu = (17 * 10^-6 /°C) * 2.54000 cm * ΔT
ΔDAl = αAl * DAl * ΔT
Where ΔDAl is the change in diameter of the aluminum sphere, αAl is the coefficient of linear expansion for aluminum, DAl is the initial diameter of the aluminum sphere, and ΔT is the change in temperature.
Substituting the given values:
ΔDAl = (23 * 10^-6 /°C) * 2.54508 cm * ΔT
At equilibrium, the change in diameter for the ring and the sphere should be the same, so:
ΔDCu = ΔDAl
(17 * 10^-6 /°C) * 2.54000 cm * ΔT = (23 * 10^-6 /°C) * 2.54508 cm * ΔT
Simplifying the equation, ΔT cancels out:
(17 * 10^-6 /°C) * 2.54000 cm = (23 * 10^-6 /°C) * 2.54508 cm
Now, we can solve for the change in temperature ΔT:
ΔT = [(23 * 10^-6 /°C) * 2.54508 cm] / [(17 * 10^-6 /°C) * 2.54000 cm]
ΔT ≈ 0.5°C
Finally, we can substitute the value of ΔT into either of the initial equations to find the change in diameter:
ΔDCu = (17 * 10^-6 /°C) * 2.54000 cm * 0.5°C
ΔDCu ≈ 2.165 * 10^-5 cm
Now that we have the change in diameter of the copper ring, we can determine the diameter of the sphere at equilibrium:
Deq = DCu + ΔDCu
Deq = 2.54000 cm + 2.165 * 10^-5 cm
Deq ≈ 2.54002 cm
Since we know the initial and equilibrium diameters of the sphere, we can calculate its radius:
Req = Deq / 2
Req ≈ 1.27001 cm
Using the radius of the sphere, we can calculate its volume:
V = (4/3) * π * Req^3
V ≈ (4/3) * π * (1.27001 cm)^3
V ≈ 8.501 cm^3
Finally, we need to determine the mass of the aluminum sphere. We know it has a density of ρ = 2.70 g/cm^3:
m = ρ * V
m ≈ 2.70 g/cm^3 * 8.501 cm^3
m ≈ 22.9527 g
Therefore, the mass of the aluminum sphere, to the appropriate number of significant figures, is approximately 22.95 g.