In the above example, if temperature of inner surface P is kept constant at `theta_1` and of the outer surface Q at `theta_2(lttheta_1)`. Then,
Find.
(a) rate of heat flow or heat current from inner surface to outer surface.
(b) temperature `theta` at a distance `r (altrltb)` from centre.

Explanation

To solve the problem step by step, we will break it down into two parts: (a) calculating the rate of heat flow (heat current) from the inner surface to the outer surface, and (b) determining the temperature at a distance \( r \) from the center.

### Part (a): Rate of Heat Flow (Heat Current)

1. **Identify the temperatures**: We have the inner surface temperature \( \theta_1 \) and the outer surface temperature \( \theta_2 \) where \( \theta_1 > \theta_2 \).

2. **Calculate the temperature difference**: The temperature difference \( \Delta T \) is given by:
\[ \Delta T = \theta_1 - \theta_2 \]

3. **Determine the thermal resistance**: The thermal resistance \( R \) for a spherical shell can be calculated using the formula:
\[ R = \frac{1}{4 \pi k} \left( \frac{1}{a} - \frac{1}{b} \right) \]
where \( k \) is the thermal conductivity, \( a \) is the radius of the inner surface, and \( b \) is the radius of the outer surface.

4. **Use the formula for heat current**: The rate of heat flow (heat current \( H \)) can be calculated using:
\[ H = \frac{\Delta T}{R} \]
Substituting the expression for \( R \):
\[ H = \Delta T \cdot 4 \pi k \left( \frac{1}{a} - \frac{1}{b} \right) \]

5. **Substitute \( \Delta T \)**:
\[ H = (\theta_1 - \theta_2) \cdot 4 \pi k \left( \frac{1}{a} - \frac{1}{b} \right) \]

### Part (b): Temperature \( \theta \) at a distance \( r \) from the center

1. **Assume a temperature at distance \( r \)**: Let the temperature at a distance \( r \) from the center be \( \theta \).

2. **Write the heat current equations**: The heat current flowing into point \( A \) (at distance \( r \)) and out of point \( A \) can be expressed as:
\[ H_1 = \frac{\theta - \theta_2}{R_1} \quad \text{and} \quad H_2 = \frac{\theta_1 - \theta}{R_2} \]
where \( R_1 = \frac{1}{4 \pi k} \left( \frac{1}{r} - \frac{1}{a} \right) \) and \( R_2 = \frac{1}{4 \pi k} \left( \frac{1}{b} - \frac{1}{r} \right) \).

3. **Set the net heat current to zero**: Since there is no heat source at point \( A \), the heat currents must balance:
\[ H_1 - H_2 = 0 \quad \Rightarrow \quad H_1 = H_2 \]

4. **Substituting the expressions for \( H_1 \) and \( H_2 \)**:
\[ \frac{\theta - \theta_2}{\frac{1}{4 \pi k} \left( \frac{1}{r} - \frac{1}{a} \right)} = \frac{\theta_1 - \theta}{\frac{1}{4 \pi k} \left( \frac{1}{b} - \frac{1}{r} \right)} \]

5. **Cross-multiply and simplify**:
\[ (\theta - \theta_2) \left( \frac{1}{b} - \frac{1}{r} \right) = (\theta_1 - \theta) \left( \frac{1}{r} - \frac{1}{a} \right) \]

6. **Rearranging terms**: After simplifying, you can express \( \theta \) in terms of \( \theta_1 \), \( \theta_2 \), \( a \), \( b \), and \( r \).

### Final Result

- The rate of heat flow from the inner surface to the outer surface is:
\[ H = (\theta_1 - \theta_2) \cdot 4 \pi k \left( \frac{1}{a} - \frac{1}{b} \right) \]

- The temperature \( \theta \) at a distance \( r \) from the center can be expressed as:
\[ \theta = \frac{B \cdot A - R \cdot \theta_2 + A \cdot (B - R) \cdot \theta_1}{AB - BR + AB - AR} \]