Three rods of copper, brass and steel are welded together to form a Y -shaped structure. The cross-sectional area of each rod is `4 cm^2`. The end of copper rod is maintained at `100^@C` and the ends of the brass and steel rods at `80^@C` and `60^@C` respectively. Assume that there is no loss of heat from the surfaces of the rods. The lengths of rods are : copper 46 cm, brass 13 cm and steel 12 cm.
(a) What is the temperature of the junction point?
(b) What is the heat current in the copper rod?
K(copper) = 0.92, K (steel) = 0.12 and K(brass) = 0.26 `cal//cm-s^@C` .

Explanation

To solve the problem step by step, we will first determine the temperature of the junction point (T) and then calculate the heat current in the copper rod.

### Given Data:
- Length of copper rod (L_c) = 46 cm
- Length of brass rod (L_b) = 13 cm
- Length of steel rod (L_s) = 12 cm
- Temperature at the end of copper rod (T_c) = 100°C
- Temperature at the end of brass rod (T_b) = 80°C
- Temperature at the end of steel rod (T_s) = 60°C
- Cross-sectional area (A) = 4 cm²
- Thermal conductivity:
- K_c (copper) = 0.92 cal/cm·s·°C
- K_b (brass) = 0.26 cal/cm·s·°C
- K_s (steel) = 0.12 cal/cm·s·°C

### (a) Finding the Temperature of the Junction Point (T)

1. **Set up the heat flow equations**:
The heat flow (Q/t) through each rod can be expressed using Fourier's law of heat conduction:
\[ \frac{Q}{t} = \frac{K \cdot A \cdot (T_{hot} - T_{cold})}{L} \]
For copper:
\[ \frac{Q}{t} = \frac{K_c \cdot A \cdot (T_c - T)}{L_c} = \frac{0.92 \cdot 4 \cdot (100 - T)}{46} \]
For brass:
\[ \frac{Q}{t} = \frac{K_b \cdot A \cdot (T - T_b)}{L_b} = \frac{0.26 \cdot 4 \cdot (T - 80)}{13} \]
For steel:
\[ \frac{Q}{t} = \frac{K_s \cdot A \cdot (T - T_s)}{L_s} = \frac{0.12 \cdot 4 \cdot (T - 60)}{12} \]

2. **Equate the heat currents**:
Since there is no heat loss, the heat current through copper and brass must equal the heat current through steel:
\[ \frac{0.92 \cdot 4 \cdot (100 - T)}{46} + \frac{0.26 \cdot 4 \cdot (T - 80)}{13} = \frac{0.12 \cdot 4 \cdot (T - 60)}{12} \]

3. **Simplify the equation**:
Cancel out the common factor of 4:
\[ \frac{0.92 (100 - T)}{46} + \frac{0.26 (T - 80)}{13} = \frac{0.12 (T - 60)}{12} \]

4. **Multiply through by 46 to eliminate the denominator**:
\[ 0.92(100 - T) + 0.26 \cdot 3.54(T - 80) = 0.12 \cdot 3.83(T - 60) \]
(Note: 46/13 = 3.54 and 46/12 = 3.83)

5. **Distribute and combine like terms**:
\[ 92 - 0.92T + 0.9264T - 20.8 = 0.4596T - 7.056 \]
Combine all T terms and constants:
\[ 92 - 20.8 + 7.056 = 0.4596T + 0.92T - 0.9264T \]
\[ 78.256 = 0.4132T \]

6. **Solve for T**:
\[ T = \frac{78.256}{0.4132} \approx 189.3°C \]

### (b) Finding the Heat Current in the Copper Rod

1. **Use the heat flow equation for copper**:
\[ \frac{Q}{t} = \frac{K_c \cdot A \cdot (T_c - T)}{L_c} \]
Substitute the values:
\[ \frac{Q}{t} = \frac{0.92 \cdot 4 \cdot (100 - 84)}{46} \]

2. **Calculate**:
\[ \frac{Q}{t} = \frac{0.92 \cdot 4 \cdot 16}{46} \]
\[ \frac{Q}{t} = \frac{58.72}{46} \approx 1.28 \text{ cal/s} \]

### Final Answers:
(a) The temperature of the junction point is approximately **84°C**.
(b) The heat current in the copper rod is approximately **1.28 cal/s**.