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: \[ 80^\circ C. \]

Answer 2

To find the temperature of the interface between the brass and copper plates, we need to consider the heat conduction through the slab. Since the system is at steady state, the heat flow through both brass and copper must be equal. The formula for heat conduction is given by Fourier's law:

\(Q = \frac{K \cdot A \cdot \Delta T}{d}\), where:

• \(Q\) is the rate of heat transfer,
• \(K\) is the thermal conductivity,
• \(A\) is the cross-sectional area,
• \(\Delta T\) is the temperature difference across the material,
• \(d\) is the thickness of the material.

Given that the thermal conductivities of brass and copper are in the ratio \(1:4\), let the conductivity of brass be \(K\) and that of copper be \(4K\).

The temperatures of the free faces are: brass at \(0^\circ C\) and copper at \(100^\circ C\). Let the temperature at the interface be \(T\).

For brass plate from \(0^\circ C\) to \(T^\circ C\):

\(Q_b = \frac{K \cdot A \cdot T}{d}\)

For copper plate from \(T^\circ C\) to \(100^\circ C\):

\(Q_c = \frac{4K \cdot A \cdot (100 - T)}{d}\)

In steady state, \(Q_b = Q_c\):

\(\frac{K \cdot A \cdot T}{d} = \frac{4K \cdot A \cdot (100 - T)}{d}\)

Canceling the common terms \(K, A, d\), we get:

\(T = 4(100 - T)\)

\(T = 400 - 4T\)

\(5T = 400\)

\(T = 80^\circ C\)

Thus, the temperature of the interface is \(80^\circ C\).