A point source of heat of power P is placed at the centre of a thin spherical shell of a mean radius R. The material of the shell has thermal conductivity K. Calculate the thickness of the shell if the temperature difference between the outer and inner surfaces of the shell in steady - state is T.

Options

A. `(4piR^(2)KT)/(2P)`
B. `(3piR^(4)KT)/(P)`
C. `(4piR^(2)KT)/(P)`
D. `(2piR^(2)KT)/(3P)`

Answer

To solve the problem, we need to calculate the thickness of the spherical shell given the power of the heat source, the thermal conductivity of the shell material, and the temperature difference between the inner and outer surfaces.

### Step-by-Step Solution:

1. **Understanding the Heat Transfer Equation**:
The rate of heat transfer (dq/dt) through a material can be described by Fourier's law of heat conduction:
\[ \frac{dq}{dt} = \frac{K \cdot A \cdot \Delta T}{x} \]
where:
- \( K \) is the thermal conductivity,
- \( A \) is the surface area through which heat is conducted,
- \( \Delta T \) is the temperature difference across the material,
- \( x \) is the thickness of the material.

2. **Identifying the Variables**:
From the problem:
- The power of the heat source \( P = \frac{dq}{dt} \).
- The mean radius of the shell \( R \).
- The thermal conductivity \( K \).
- The temperature difference \( \Delta T = T \).
- The area \( A \) of the inner surface of the shell is given by the formula for the surface area of a sphere:
\[ A = 4 \pi R^2 \]

3. **Substituting the Values into the Equation**:
Substitute \( P \), \( A \), \( \Delta T \), and \( x \) into the heat transfer equation:
\[ P = \frac{K \cdot (4 \pi R^2) \cdot T}{x} \]

4. **Rearranging to Solve for Thickness \( x \)**:
Rearranging the equation to solve for \( x \):
\[ x = \frac{K \cdot (4 \pi R^2) \cdot T}{P} \]

5. **Final Expression for Thickness**:
Thus, the thickness \( x \) of the shell is given by:
\[ x = \frac{4 \pi K R^2 T}{P} \]

### Final Answer:
The thickness of the shell is:
\[ x = \frac{4 \pi K R^2 T}{P} \]