A 4-m-high and 6-m-wide wall consists of a long \(18-\mathrm{cm} \times\) 30 -cm cross section of horizontal bricks \((k=0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) separated by \(3-\mathrm{cm}\)-thick plaster layers \((k=0.22 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). There are also 2 -cm-thick plaster layers on each side of the wall, and a 2-cmthick rigid foam \((k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) on the inner side of the wall. The indoor and the outdoor temperatures are \(22^{\circ} \mathrm{C}\) and \(-4^{\circ} \mathrm{C}\), and the convection heat transfer coefficients on the inner and the outer sides are \(h_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{2}=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Assuming one-dimensional heat transfer and disregarding radiation, determine the rate of heat transfer through the wall.

The temperature difference between the indoor and outdoor is given by \(\Delta T = T_{indoors} - T_{outdoors} = 22^{\circ} \mathrm{C} - (-4^{\circ} \mathrm{C}) = 26 \mathrm{K}\).

We need to calculate the thermal resistance of each layer. The formula for thermal resistance is \(R = \frac{L}{kA}\), where \(L\) is the thickness of each layer, \(k\) is the thermal conductivity, and \(A\) is the area. The area of the wall, \(A\) is 4 m height * 6 m wide = 24 \(m^2\). Thermal resistance of bricks layer: \(R_{brick} = \frac{0.18\mathrm{m}}{0.72 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \times 24\mathrm{m^2}} = 0.1042 \mathrm{K} \cdot \mathrm{m}^2/\mathrm{W}\) Thermal resistance of plaster layers (both sides combined): \(R_{plaster} = \frac{0.05\mathrm{m}}{0.22 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \times 24\mathrm{m^2}} = 0.0955 \mathrm{K} \cdot \mathrm{m}^2/\mathrm{W}\) Thermal resistance of foam layer: \(R_{foam} = \frac{0.02\mathrm{m}}{0.026 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \times 24\mathrm{m^2}} = 0.0321 \mathrm{K} \cdot \mathrm{m}^2/\mathrm{W}\) Thermal resistance of convection on the inner side: \(R_{h1} = \frac{1}{10 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}} \times 24\mathrm{m^2}} = 0.0042 \mathrm{K} \cdot \mathrm{m}^2/\mathrm{W}\) Thermal resistance of convection on outer side: \(R_{h2} = \frac{1}{20 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}} \times 24\mathrm{m^2}} = 0.0021 \mathrm{K} \cdot \mathrm{m}^2/\mathrm{W}\)

To get the total thermal resistance, we sum the thermal resistance of all layers: \(R_{total} = R_{brick} + R_{plaster} + R_{foam} + R_{h1} + R_{h2} = 0.1042 + 0.0955 + 0.0321 + 0.0042 + 0.0021 = 0.2381 \mathrm{K} \cdot \mathrm{m}^2/\mathrm{W}\).

Now, we can calculate the rate of heat transfer by using the formula \(Q = \frac{\Delta T}{R_{total}} = \frac{26 \mathrm{K}}{0.2381 \mathrm{K} \cdot \mathrm{m}^2/\mathrm{W}} = 109.18 \mathrm{W}\). Therefore, the rate of heat transfer through the wall is approximately \(109.18 \mathrm{W}\).