Two rods, one made of copper and the other steel of the same length and cross sectional area are joined together. (The thermal conductivity of copper is `385 J.s^(-1) .m^(-1). K^(-1)` and steel is `50 J.s^(-1).m-^(1).K^(-1)`.) If the copper end is held at `100^(@)C` and the steel end is held at `0^(@)C`, what is the junction temperature (assuming no other heat losses) ?

Options

A. 0.12
B. 0.5
C. 0.73
D. 0.88

Answer

To find the junction temperature of the two rods made of copper and steel, we can use the concept of thermal conductivity and heat flow. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have two rods:
- Rod 1 (Copper): Thermal conductivity \( K_1 = 385 \, \text{J.s}^{-1} \text{m}^{-1} \text{K}^{-1} \)
- Rod 2 (Steel): Thermal conductivity \( K_2 = 50 \, \text{J.s}^{-1} \text{m}^{-1} \text{K}^{-1} \)

Both rods have the same length \( L \) and cross-sectional area \( A \). The copper end is maintained at \( 100^\circ C \) and the steel end at \( 0^\circ C \).

### Step 2: Set Up the Heat Current Equation
The heat current \( Q \) flowing through both rods must be equal since they are in steady state (no heat loss). We can express the heat current through each rod as:

\[ Q = \frac{(T_1 - T_j)}{R_1} = \frac{(T_j - T_2)}{R_2} \]

Where:
- \( T_1 = 100^\circ C \) (temperature at the copper end)
- \( T_2 = 0^\circ C \) (temperature at the steel end)
- \( T_j \) = junction temperature
- \( R_1 \) and \( R_2 \) are the thermal resistances of copper and steel respectively.

### Step 3: Calculate the Thermal Resistance
The thermal resistance \( R \) for each rod can be calculated using the formula:

\[ R = \frac{L}{K \cdot A} \]

Thus, we have:
- For copper: \( R_1 = \frac{L}{K_1 \cdot A} \)
- For steel: \( R_2 = \frac{L}{K_2 \cdot A} \)

### Step 4: Express the Heat Current
Substituting the resistances into the heat current equations, we get:

\[ \frac{(100 - T_j)}{\frac{L}{K_1 \cdot A}} = \frac{(T_j - 0)}{\frac{L}{K_2 \cdot A}} \]

### Step 5: Simplify the Equation
Since \( L \) and \( A \) are common in both terms, they can be canceled out:

\[ \frac{(100 - T_j) \cdot K_1}{L} = \frac{T_j \cdot K_2}{L} \]

This simplifies to:

\[ (100 - T_j) \cdot K_1 = T_j \cdot K_2 \]

### Step 6: Rearrange the Equation
Rearranging gives us:

\[ 100K_1 = T_j(K_1 + K_2) \]

### Step 7: Solve for \( T_j \)
Now, we can solve for \( T_j \):

\[ T_j = \frac{100K_1}{K_1 + K_2} \]

### Step 8: Substitute the Values
Substituting the values of \( K_1 \) and \( K_2 \):

\[ T_j = \frac{100 \cdot 385}{385 + 50} = \frac{38500}{435} \]

### Step 9: Calculate the Junction Temperature
Now, calculating the value:

\[ T_j = \frac{38500}{435} \approx 88.51^\circ C \]

### Conclusion
Thus, the junction temperature \( T_j \) is approximately \( 88.51^\circ C \).