A thin spherical shell of silver has an inner radius of 5.58 x 10-2 m when the temperature is 25.7 °C. The shell is heated to 133 °C. Find the change in the interior volume of the shell.

2.2 x 10^-7 m^3

To find the change in the interior volume of the shell, we can use the concept of thermal expansion. When an object is heated, its dimensions (length, area, or volume) change due to the expansion of its material.

In this case, we need to consider the thermal expansion of the silver shell. The formula for the change in volume due to thermal expansion is given by:

ΔV = V₀ * β * ΔT

Where:
ΔV is the change in volume
V₀ is the initial volume
β is the thermal expansion coefficient
ΔT is the change in temperature

First, let's calculate the initial volume (V₀) of the shell. The shell is a thin spherical shell, so its volume can be calculated using the formula:

V₀ = (4/3) * π * (R₂³ - R₁³)

Where:
R₁ is the inner radius (initial) of the shell
R₂ is the outer radius (initial) of the shell

Given:
R₁ = 5.58 x 10^(-2) m

We need to find R₂ before we can calculate V₀.

The outer radius (R₂) can be found by considering the fact that the shell is thin, which means the difference between R₂ and R₁ is very small compared to the radius itself. Hence, we can approximate R₂ as:

R₂ ≈ R₁

Now we can calculate V₀ using the formula:

V₀ = (4/3) * π * (R₂³ - R₁³)
≈ (4/3) * π * (R₁³ - R₁³)
≈ 0

Since the initial volume (V₀) is approximately zero, the change in the interior volume (ΔV) will also be zero. Therefore, there is no change in the interior volume of the shell.

Please note that this result is based on the assumption that the shell is a thin spherical shell and the difference between the initial and final radii is very small.