Temperature of hot end and cold end of a rod, which is in steady state are `100^(@)C` and `40^(@)C` respectively. The area of cross section of rod and its thermal conductivity are uniform. The temperature of rod at its mid point is (Assume no heat loss thorugh lateral surface)

Options

A. `70^(@)C`
B. `60^(@)C`
C. `80^(@)C`
D. None of these

Answer

To find the temperature at the midpoint of the rod, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Temperature at the hot end (A) = \(100^\circ C\)
- Temperature at the cold end (B) = \(40^\circ C\)
- Length of the rod = \(L\)
- Area of cross-section = \(A\)
- Thermal conductivity = \(K\)
- We need to find the temperature at the midpoint (C).

2. **Assume the Temperature at the Midpoint**:
- Let the temperature at the midpoint (C) be \(T\).

3. **Apply the Heat Transfer Equation**:
- Since the rod is in steady state, the heat transfer rate (\( \frac{dQ}{dt} \)) through the left half (from A to C) must equal the heat transfer rate through the right half (from C to B).
- The heat transfer rate can be expressed as:
\[ \frac{dQ}{dt} = \frac{K \cdot A \cdot (T - 40)}{\frac{L}{2}} \quad \text{(from B to C)} \]
\[ \frac{dQ}{dt} = \frac{K \cdot A \cdot (100 - T)}{\frac{L}{2}} \quad \text{(from A to C)} \]

4. **Set the Heat Transfer Rates Equal**:
- Since there is no accumulation of heat at point C, we can set the two expressions for heat transfer equal to each other:
\[ \frac{K \cdot A \cdot (T - 40)}{\frac{L}{2}} = \frac{K \cdot A \cdot (100 - T)}{\frac{L}{2}} \]

5. **Cancel Common Terms**:
- The terms \(K\), \(A\), and \(\frac{L}{2}\) are common on both sides and can be canceled out:
\[ T - 40 = 100 - T \]

6. **Solve for T**:
- Rearranging the equation gives:
\[ T + T = 100 + 40 \]
\[ 2T = 140 \]
\[ T = 70^\circ C \]

7. **Conclusion**:
- The temperature at the midpoint of the rod is \(70^\circ C\).