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.

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.

cdquestions Admin · Accepted Answer

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}} $.

Correct Answer: 40

Solution and Explanation

Top Questions on Heat Transfer

Questions Asked in JEE Main exam