Interactive Solution 12.29 presents a model for solving problems of this type. A thin spherical shell of silver has an inner radius of 5.64 x 10-2 m when the temperature is 24.1 °C. The shell is heated to 133 °C. Find the change in the interior volume of the shell.
To find the change in the interior volume of the shell, we can use the formula provided in Solution 12.29. According to the solution, the change in the interior volume (∆V) of the shell is given by the formula:
∆V = (4/3)π(∆R)(R^2 + R'^2 + RR')
Where:
- ∆V represents the change in the interior volume
- π represents the mathematical constant pi (approximately 3.14159)
- ∆R represents the change in radius (∆R = R' - R)
- R represents the initial radius of the shell
- R' represents the final radius of the shell
In this problem, we are given:
- The initial radius (R) of the shell as 5.64 x 10^(-2) m
- The initial temperature of the shell as 24.1 °C
- The final temperature of the shell as 133 °C
To find the final radius (R'), we need to use the concept of thermal expansion. The equation for thermal expansion is given by:
∆L = αL∆T
Where:
- ∆L represents the change in length
- α represents the coefficient of linear expansion
- L represents the initial length
- ∆T represents the change in temperature
In this problem, since we have a thin spherical shell, we can consider it to be a 2D object. Therefore, we can use the coefficient of superficial expansion (β) instead of α. The equation becomes:
∆A = βA∆T
Where:
- ∆A represents the change in area
- β represents the coefficient of superficial expansion
- A represents the initial area
- ∆T represents the change in temperature
Since the problem involves spherical shells, the initial and final areas can be given by the formula:
A = 4πR^2
Therefore, the change in area (∆A) is given by:
∆A = A' - A = 4πR'^2 - 4πR^2
Thus, the formula for finding the final radius (R') is:
R' = √(R^2 + ∆A / (4π))
Once we have found the final radius (R'), we can calculate the change in the interior volume (∆V) using the previously mentioned formula:
∆V = (4/3)π(∆R)(R^2 + R'^2 + RR')
To find the change in the interior volume of the shell, we need to use the formula for the change in volume due to temperature change for a spherical shell:
ΔV = α * V0 * ΔT
Where:
ΔV is the change in volume
α is the coefficient of linear expansion of the material
V0 is the initial volume of the shell
ΔT is the change in temperature
First, let's find the initial volume V0 of the shell:
V0 = (4/3) * π * (r1^3 - r0^3)
Where:
r1 is the outer radius
r0 is the inner radius
Given information:
r0 = 5.64 x 10^(-2) m
Temperature when r0 is measured = 24.1 °C
Let's convert the temperature to Kelvin since we will need absolute temperature in the calculations:
T0 = 24.1 °C + 273.15 = 297.25 K
Next, we need to find the final outer radius r1 when the shell is heated to 133 °C.
To find the change in radius, we can use the formula:
Δr = α * r0 * ΔT
Where:
Δr is the change in radius
α is the coefficient of linear expansion of the material
r0 is the initial radius
ΔT is the change in temperature
Given information:
ΔT = 133 °C - 24.1 °C = 108.9 °C
Now, we need to find the coefficient of linear expansion for silver. The coefficient of linear expansion for silver is α = 0.000019 / °C.
Now, let's calculate the change in radius Δr:
Δr = α * r0 * ΔT
= 0.000019 / °C * (5.64 x 10^(-2) m) * (108.9 °C)
Substituting the values:
Δr = 0.000019 / °C * 0.0564 m * 108.9 °C
Δr ≈ 0.00011689 m
Now, let's find the final outer radius r1:
r1 = r0 + Δr
= 5.64 x 10^(-2) m + 0.00011689 m
Substituting the values:
r1 ≈ 0.05611689 m
Now, let's calculate the initial volume V0:
V0 = (4/3) * π * (r1^3 - r0^3)
= (4/3) * π * [(0.05611689 m)^3 - (5.64 x 10^(-2) m)^3]
Substituting the values:
V0 ≈ (4/3) * π * (0.00017858 m^3 - 0.000179194 m^3)
V0 ≈ 0.000518 m^3
Finally, let's calculate the change in volume ΔV:
ΔV = α * V0 * ΔT
= 0.000019 / °C * 0.000518 m^3 * 108.9 °C
Substituting the values:
ΔV ≈ 0.000019 / °C * 0.000518 m^3 * 108.9 °C
ΔV ≈ 0.00001039 m^3
Therefore, the change in the interior volume of the shell is approximately 0.00001039 cubic meters.