Question

A slab consists of two identical plates of copper and brass. The free face of the brass is at \( 0^\circ C \) and that of copper at \( 100^\circ C \). If the thermal conductivities of brass and copper are in the ratio \( 1:4 \), then the temperature of the interface is:

• A. \( 20^\circ C \)
• B. \( 40^\circ C \)
• C. \( 60^\circ C \)
• D. \( 80^\circ C \)

Hint:

For steady-state heat conduction through composite slabs, use the equation \( K_1 (T_1 - T) = K_2 (T - T_2) \) to determine the interface temperature.

Answer 1

The Correct Option is D

Step 1: Understanding Heat Transfer Through Composite Slabs 
The heat transfer rate \( Q \) through a composite slab in steady-state condition is the same through both materials: \[ \frac{K_1 A (T_1 - T)}{d} = \frac{K_2 A (T - T_2)}{d}. \] where: - \( K_1, K_2 \) are the thermal conductivities of copper and brass, - \( T_1 = 100^\circ C \), \( T_2 = 0^\circ C \), - \( T \) is the temperature at the interface. 
Step 2: Applying Given Ratio 
Given that the ratio of thermal conductivities is: \[ K_{{brass}} : K_{{copper}} = 1:4. \] Let \( K_{{brass}} = K \) and \( K_{{copper}} = 4K \). Using the steady-state heat transfer equation: \[ \frac{4K (100 - T)}{d} = \frac{K (T - 0)}{d}. \] Step 3: Solving for Interface Temperature 
Canceling \( K \) and \( d \): \[ 4 (100 - T) = T. \] \[ 400 - 4T = T. \] \[ 400 = 5T. \] \[ T = 80^\circ C. \] Step 4: Conclusion 
Thus, the temperature at the interface is: \[ 80^\circ C. \]