Final answer:
The rate of heat transfer through the wall per unit area, given the two layers and the temperature difference across them, can be calculated using the principle of thermal resistance in series. The final calculated value is approximately 51.4 W/m², which corresponds with option D.
Step-by-step explanation:
The question is asking to find the rate of heat transfer through a composite wall per unit area, given the thermal conductivities and thicknesses of two layers, A and B, as well as the temperature drop across the wall. According to Fourier's law of thermal conduction, the rate of heat transfer Q/t through a material is proportional to the temperature difference across it, the thermal conductivity of the material, and the area, and is inversely proportional to the thickness of the material. Using this principle, we can calculate the heat transfer rates through each layer separately and then combine them since they are in series. The formula for the rate of heat transfer through a single layer is Q/t = kA(T2 - T1)/L, where k is the thermal conductivity, A is the area, T2 - T1 is the temperature difference, and L is the thickness. The effective thermal resistance for layers in series is given by the sum of the individual resistances, that is R_total = (LA/KA) + (LB/KB). The overall rate of heat transfer per unit area (Q/t)/A is then ΔT/R_total, where ΔT is the temperature difference across the entire wall. Substituting the given thermal conductivities and layer thicknesses into the respective formulas, we first calculate the resistances: R_A = LA/KA = 0.08m/0.8W/m·K = 0.1 m·2/W and R_B = LB/KB = 0.05m/0.2W/m·K = 0.25 m·2/W. Adding these gives us the total resistance R_total = 0.1 m·2/W + 0.25 m·2/W = 0.35 m·2/W. The rate of heat transfer per unit area is then (18°C / 0.35 m·2/W), which equals about 51.4 W/m·2.