Heat flows radially outwards through a spherical shell of outside radius `R_2` and inner radius `R_1`. The temperature of inner surface of shell is `theta_1` and that of outer is `theta_2`. At what radial distance from centre of shell the temperature is just half way between `theta_1 and theta_2`?

Explanation

To find the radial distance from the center of a spherical shell where the temperature is halfway between the inner temperature \( \theta_1 \) and the outer temperature \( \theta_2 \), we can follow these steps:

### Step 1: Understand the Problem
We have a spherical shell with inner radius \( R_1 \) and outer radius \( R_2 \). The inner surface temperature is \( \theta_1 \) and the outer surface temperature is \( \theta_2 \). We need to find the radius \( R \) at which the temperature is \( \frac{\theta_1 + \theta_2}{2} \).

### Step 2: Apply the Heat Transfer Equation
The heat current \( I \) through the spherical shell can be expressed using Fourier's law of heat conduction:
\[ I = -k A \frac{d\theta}{dr} \]
where \( k \) is the thermal conductivity, \( A \) is the surface area, and \( \frac{d\theta}{dr} \) is the temperature gradient.

For a spherical shell, the area \( A \) at a distance \( r \) from the center is:
\[ A = 4\pi r^2 \]

### Step 3: Set Up the Equation
The heat current can be expressed as:
\[ I = -k (4\pi r^2) \frac{d\theta}{dr} \]
Rearranging gives:
\[ \frac{d\theta}{dr} = -\frac{I}{4\pi k r^2} \]

### Step 4: Integrate the Equation
Integrate both sides from \( R_1 \) to \( R \) (where \( \theta = \frac{\theta_1 + \theta_2}{2} \)) and from \( R \) to \( R_2 \) (where \( \theta = \theta_2 \)):
\[ \int_{\theta_1}^{\frac{\theta_1 + \theta_2}{2}} d\theta = -\int_{R_1}^{R} \frac{I}{4\pi k r^2} dr \]

This results in:
\[ \frac{\theta_1 + \theta_2}{2} - \theta_1 = -\frac{I}{4\pi k} \left( -\frac{1}{R} + \frac{1}{R_1} \right) \]

### Step 5: Solve for \( R \)
After integrating, we can express \( I \) in terms of \( \theta_1 \) and \( \theta_2 \):
\[ I = \frac{4\pi k (\theta_2 - \theta_1)}{\frac{1}{R_1} - \frac{1}{R_2}} \]
Substituting this back into the equation allows us to isolate \( R \):
\[ \frac{2R_2 R_1}{R_2 - R_1} = \left( \frac{1}{R_1} - \frac{1}{R} \right) \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ R = \frac{2R_1 R_2}{3R_2 - R_1} \]

### Final Answer
The radial distance \( R \) from the center of the shell where the temperature is halfway between \( \theta_1 \) and \( \theta_2 \) is:
\[ R = \frac{2R_1 R_2}{3R_2 - R_1} \]

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