A wall has two layers of materials A and B; each made of a different material. Both the layers have the same thickness. The thermal conductivity of materialA is twice that of B. Under the equilibrium, the temperature difference across the wall is 36°C. The temperature difference across the layer A is __________ °C.

ANS:B - 12

To solve this problem, we can use the concept of thermal resistance. The thermal resistance (R) of a material is calculated as the thickness (d) divided by the product of the material's thermal conductivity (k) and its area (A): R=k⋅Ad​ For each layer of the wall, the thermal resistance can be calculated as: RA​=kA​⋅Ad​ =RB​=kB​⋅Ad​ Given that the thermal conductivity of material A (kA​) is twice that of material B (kB​), and both layers have the same thickness and area, we can simplify the equation to: =12RA​=21​RB​ Now, let's denote the temperature difference across layer A as ΔTA​ and across layer B as ΔTB​. According to the equation for thermal resistance in series: Δtotal=Δ+ΔΔTtotal​=ΔTA​+ΔTB​ Given that the total temperature difference across the wall (ΔtotalΔTtotal​) is 36°C and =12RA​=21​RB​, we can set up the equation: 36°=Δ+2Δ36°C=ΔTA​+2ΔTA​ 36°=3Δ36°C=3ΔTA​ Solving for ΔTA​: Δ=36°3=12°ΔTA​=336°C​=12°C So, the temperature difference across layer A is Δ=12°ΔTA​=12°C​.