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 ° C respectively, then the temperature of the surface of contact in steady state is _____________ ° C.

Solution

Physics sir · 3 steps

In steady state the same heat flows through both plates, so set the two heat currents equal.

1. Same heat current. In steady state no heat piles up at the contact surface. So the heat per second through A equals the heat per second through B. Let the contact temperature be T.

   (K_A A (100 - T))/d = (K_B A (T - 0))/d

2. Cancel and solve. Area A and thickness d are the same for both plates, so they cancel. This leaves 84(100 - T) = 126 T, so 8400 = 210 T.
3. So. T = 8400/210 = 40°C, so the answer is 40.

The common slip: 60. Putting 126 with plate A and 84 with plate B gives 60. The bigger temperature drop belongs to the poorer conductor, which is A.

The idea to keep. For slabs in series, heat current is the same through each one. The better conductor needs a smaller temperature drop, so the junction sits closer to its far end.

Answer from the NTA final answer key. Solution by PrepWiser.