Two walls of thickness `d_(1) and d_(2)` and thermal conductivites `K_(1) and K_(2)` are in contact. In the steady state, if the temperature at the outer surfaces are `T_(1) and T_(2)` the temperature at the common wall is

Options

A. `(K_(1)T_(1)d_(2) + K_(2)T_(2)d_(1))/(K_(1)d_(2) + K_(2)d_(1))`
B. `(K_(1)T_(1) + K_(2)T_(2))/(d_(2) + d_(1))`
C. `((K_(1)d_(1) + K_(2)d_(2))/(T_(1) + T_(2)))T_(1)T_(2)`
D. `(K_(1)T_(1)d_(1) + K_(2)T_(2)d_(2))/(K_(1)d_(1) + K_(2)d_(2))`

Answer

To find the temperature at the common wall between two walls with given thicknesses and thermal conductivities, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let the thickness of the first wall be \( d_1 \) and its thermal conductivity be \( K_1 \).
- Let the thickness of the second wall be \( d_2 \) and its thermal conductivity be \( K_2 \).
- Let the outer temperatures be \( T_1 \) (for the first wall) and \( T_2 \) (for the second wall).
- We need to find the temperature at the common wall, denoted as \( T \).

2. **Understand the Heat Transfer**:
- The heat transfer through both walls is in steady state, meaning the rate of heat flow (thermal current \( I \)) is constant across both walls.
- According to Fourier's law of thermal conduction, the heat transfer \( I \) can be expressed as:
\[ I = \frac{K \cdot A \cdot (T_{hot} - T_{cold})}{L} \]
- For the first wall:
\[ I = \frac{K_1 \cdot A \cdot (T_1 - T)}{d_1} \]
- For the second wall:
\[ I = \frac{K_2 \cdot A \cdot (T - T_2)}{d_2} \]

3. **Set the Heat Transfer Equations Equal**:
- Since the heat transfer is the same through both walls, we set the two equations for \( I \) equal to each other:
\[ \frac{K_1 \cdot A \cdot (T_1 - T)}{d_1} = \frac{K_2 \cdot A \cdot (T - T_2)}{d_2} \]
- The area \( A \) cancels out from both sides.

4. **Cross Multiply to Solve for \( T \)**:
- Rearranging gives:
\[ K_1 \cdot d_2 \cdot (T_1 - T) = K_2 \cdot d_1 \cdot (T - T_2) \]
- Expanding both sides:
\[ K_1 \cdot d_2 \cdot T_1 - K_1 \cdot d_2 \cdot T = K_2 \cdot d_1 \cdot T - K_2 \cdot d_1 \cdot T_2 \]

5. **Combine Like Terms**:
- Rearranging gives:
\[ K_1 \cdot d_2 \cdot T_1 + K_2 \cdot d_1 \cdot T_2 = K_1 \cdot d_2 \cdot T + K_2 \cdot d_1 \cdot T \]
- Factor out \( T \) on the right side:
\[ K_1 \cdot d_2 \cdot T_1 + K_2 \cdot d_1 \cdot T_2 = T(K_1 \cdot d_2 + K_2 \cdot d_1) \]

6. **Solve for \( T \)**:
- Finally, solve for \( T \):
\[ T = \frac{K_1 \cdot d_2 \cdot T_1 + K_2 \cdot d_1 \cdot T_2}{K_1 \cdot d_2 + K_2 \cdot d_1} \]

### Final Answer:
The temperature at the common wall is given by:
\[ T = \frac{K_1 \cdot d_2 \cdot T_1 + K_2 \cdot d_1 \cdot T_2}{K_1 \cdot d_2 + K_2 \cdot d_1} \]