The thermal conductivity of copper is four times that of brass. Two rods of copper and brass of same length and cross-section are joined end to end. The free end of copper rod is at `0^(@)C` and that of brass rod at `100^(@)C`. Calculate the temperature of junction at equilibrium. neglect radiation losses.

Answer

To solve the problem of finding the temperature at the junction of two rods (copper and brass) joined end to end, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Let the thermal conductivity of brass be \( k \).
- Therefore, the thermal conductivity of copper is \( 4k \) (as given).
- Both rods have the same length \( L \) and cross-sectional area \( A \).
- The temperature at the free end of the copper rod is \( 0^\circ C \) and at the free end of the brass rod is \( 100^\circ C \).

2. **Define the Temperature at the Junction:**
- Let the temperature at the junction be \( T \).

3. **Apply the Heat Flow Equation:**
- The rate of heat flow through the copper rod can be expressed as:
\[ \frac{dQ}{dt} = \frac{k_{Cu} \cdot A}{L} (T - 0) = \frac{4k \cdot A}{L} T \]
- The rate of heat flow through the brass rod can be expressed as:
\[ \frac{dQ}{dt} = \frac{k_{Br} \cdot A}{L} (100 - T) = \frac{k \cdot A}{L} (100 - T) \]

4. **Set the Heat Flow Rates Equal:**
- Since the system is in equilibrium, the rate of heat flow through both rods must be equal:
\[ \frac{4k \cdot A}{L} T = \frac{k \cdot A}{L} (100 - T) \]

5. **Cancel Common Terms:**
- We can cancel \( k \), \( A \), and \( L \) from both sides:
\[ 4T = 100 - T \]

6. **Solve for \( T \):**
- Rearranging the equation gives:
\[ 4T + T = 100 \]
\[ 5T = 100 \]
\[ T = \frac{100}{5} = 20^\circ C \]

7. **Conclusion:**
- The temperature at the junction at equilibrium is \( T = 20^\circ C \).