Two identical metal rods A and B are joined end to end. The free end of A is kept at 27°C and the free end of B at 37°C. Calculate the temperature of the interface. Thermal conductivity of `A = 385 "Wm"^(-1) "K"^(-1)`, that of B `110 "Wm"^(-1) "K"^(-1)`.

Explanation

To solve the problem, we need to find the temperature at the interface of two identical metal rods A and B that are joined end to end. The free end of rod A is at 27°C and the free end of rod B is at 37°C. The thermal conductivities of rods A and B are given as 385 W/m·K and 110 W/m·K, respectively.

### Step-by-Step Solution:

1. **Identify the Known Values:**
- Temperature at the free end of rod A (T1) = 27°C = 300 K (27 + 273)
- Temperature at the free end of rod B (T2) = 37°C = 310 K (37 + 273)
- Thermal conductivity of rod A (k1) = 385 W/m·K
- Thermal conductivity of rod B (k2) = 110 W/m·K

2. **Set Up the Heat Flow Equation:**
Since the rods are in thermal contact and in steady state, the rate of heat flow through both rods must be equal. We can express this as:
\[ \frac{k_1 A (T - T_1)}{L} = \frac{k_2 A (T_2 - T)}{L} \]
Here, \(T\) is the temperature at the interface, \(A\) is the cross-sectional area (which cancels out), and \(L\) is the length of the rods (which also cancels out).

3. **Simplify the Equation:**
Cancelling \(A\) and \(L\) from both sides gives us:
\[ k_1 (T - T_1) = k_2 (T_2 - T) \]

4. **Substitute the Known Values:**
Substitute \(k_1\), \(k_2\), \(T_1\), and \(T_2\) into the equation:
\[ 385 (T - 300) = 110 (310 - T) \]

5. **Expand and Rearrange the Equation:**
Expanding both sides:
\[ 385T - 115500 = 34100 - 110T \]
Rearranging gives:
\[ 385T + 110T = 34100 + 115500 \]
\[ 495T = 149600 \]

6. **Solve for T:**
Dividing both sides by 495:
\[ T = \frac{149600}{495} \approx 302.02 \text{ K} \]

7. **Convert Back to Celsius:**
To convert the temperature back to Celsius:
\[ T = 302.02 - 273 \approx 29.02°C \]

### Final Answer:
The temperature at the interface is approximately **29.02°C**.