Question

Two plates A and B have thermal conductivities 84 \(Wm^{-1}K^{-1}\) and 126 \(Wm^{-1}K^{-1}\) respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of A and B are kept at 100 °C and 0 ºC respectively, then the temperature of the surface of contact in steady state is ________°C.

Answer 1

Correct Answer: 40

Let the temperature of the contact surface be \( T \). The heat current through plate A is equal to the heat current through plate B in the steady state: \[ K_A A \frac{(T_A - T)}{L} = K_B A \frac{(T - T_B)}{L}, \] where \( K_A \) and \( K_B \) are the thermal conductivities, \( T_A = 100^\circ \text{C} \), \( T_B = 0^\circ \text{C} \), and \( L \) is the thickness. Substitute the given values: \[ 84 \cdot (100 - T) = 126 \cdot (T - 0). \] Simplify: \[ 8400 - 84T = 126T. \] Rearrange: \[ 8400 = 210T. \] Solve for \( T \): \[ T = \frac{8400}{210} = 40^\circ \text{C}. \] Thus, the temperature of the contact surface is \( \boxed{40^\circ \text{C}} \).