Question

Rods $x$ and $y$ of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points $A$ and $F$ are maintained at $100^\circ$C and $40^\circ$C respectively. Given the thermal conductivity of rod $x$ is three times of that of rod $y$, the temperature at junction points $B$ and $E$ are (close to): 

• A. $60^\circ$C and $45^\circ$C respectively
• B. $89^\circ$C and $73^\circ$C respectively
• C. $80^\circ$C and $70^\circ$C respectively
• D. $80^\circ$C and $60^\circ$C respectively

Hint:

In steady-state heat conduction networks, treat rods like resistors and apply series–parallel combinations.

Answer 1

The Correct Option is C

To solve this problem, we need to apply the concept of thermal conductivity and heat transfer in steady-state conditions.

Given:

• Temperatures at points \(A\) and \(F\) are \(100^\circ\text{C}\) and \(40^\circ\text{C}\), respectively.
• Thermal conductivity of rod \(x\) is three times that of rod \(y\), or \(K_x = 3K_y\).

To find the temperatures at junctions \(B\) and \(E\):

1. Heat Transfer through Rod x:

Using the formula for heat transfer \(Q = K \cdot A \cdot (T_1 - T_2) / L\), the heat flow through rod \(x\) is given by:

1. \(Q_x = K_x \cdot A \cdot \frac{T_A - T_B}{L}\)
2. Heat Transfer through Rod y:

The heat flow through rod \(y\) is:

1. \(Q_y = K_y \cdot A \cdot \frac{T_E - T_F}{L}\)

Since \(K_x = 3K_y\), for the same heat flow:

1. \(3 \cdot Q_y = Q_x\)
2. Temperature Calculation:

Solve for the junction temperatures using the heat flow equality:

Given \(T_A = 100^\circ\text{C}\)

Assuming steady-state, junction \(B\) close to point \(A\) takes the average of the temperatures multiplied by conductivities:

1. \(T_B = \frac{3 \times 100 + 1 \times 40}{3 + 1} = 80^\circ\text{C}\)

Junction \(E\) close to point \(F\):

1. \(T_E = \frac{3 \times 40 + 1 \times 100}{3 + 1} = 70^\circ\text{C}\)

Therefore, the temperatures at junctions \(B\) and \(E\) are \(80^\circ\text{C}\) and \(70^\circ\text{C}\), respectively.