A 12-m-long and 5-m-high wall is constructed of two layers of \(1-\mathrm{cm}\)-thick sheetrock \((k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) spaced \(16 \mathrm{~cm}\) by wood studs \((k=0.11 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose cross section is \(16 \mathrm{~cm} \times 5 \mathrm{~cm}\). The studs are placed vertically \(60 \mathrm{~cm}\) apart, and the space between them is filled with fiberglass insulation \((k=0.034 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The house is maintained at \(20^{\circ} \mathrm{C}\) and the ambient temperature outside is \(-9^{\circ} \mathrm{C}\). Taking the heat transfer coefficients at the inner and outer surfaces of the house to be \(8.3\) and \(34 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine \((a)\) the thermal resistance of the wall considering a representative section of it and (b) the rate of heat transfer through the wall.

Here's what we know: - Sheetrock thickness: 1cm (0.01 m); k=0.17 W/m.K - Wood studs: 16 cm x 5 cm cross-section; k=0.11 W/m.K; placed 60 cm (0.6 m) apart - Fiberglass insulation: k=0.034 W/m.K - Wall dimensions: 12 m x 5 m - Indoor temperature: 20°C - Outdoor temperature: -9°C - Inner surface heat transfer coefficient: 8.3 W/m².K - Outer surface heat transfer coefficient: 34 W/m².K

We will separate the calculations for the sheetrock, wood studs, and fiberglass insulation: (1) Sheetrock resistance: The thermal resistance is given by: $R_S = \frac{t_s}{k_s A_s} \\ R_s = \frac{0.01\,\text{m}}{0.17\,\text{W/m.K} \cdot * (12\,\text{m}\cdot 5\,\text{m})}$ (2) Wood studs resistance: The distance between studs is 0.6 m. If we consider a representative section of the wall between two studs 1 m high, the wood stud cross-sectional area is 5 cm x 1 m = 0.05 m². Then, the thermal resistance is: $R_{WS} = \frac{t_{WS}}{k_{WS} A_{WS}} \\ R_{WS} = \frac{0.16\,\text{m}}{0.11\,\text{W/m.K} \cdot * 0.05\,\text{m}^{2}}$ (3) Fiberglass insulation resistance: The geometric resistance per unit length of the fiberglass insulation (since it's the same height as the wall) will be: $R_{FI} = \frac{t_{FI}}{k_{FI} A_{FI}} \\ R_{FI} = \frac{0.16\,\text{m}}{0.034\,\text{W/m.K} \cdot * (0.6\,\text{m} - 0.05\,\text{m})}$ (4) Inner and outer surface resistance: The inner surface resistance is given by: $R_{in} = \frac{1}{h_{in}A_{in}} \\ R_{in} = \frac{1}{8.3\,\frac{\text{W}}{\text{m}^{2}\cdot \text{K}} \cdot (12\,\text{m}\cdot 5\,\text{m})}$ The outer surface resistance is given by: $R_{out} = \frac{1}{h_{out}A_{out}} \\ R_{out} = \frac{1}{34\,\frac{\text{W}}{\text{m}^{2}\cdot \text{K}} \cdot(12\,\text{m}\cdot 5\,\text{m})}$

Considering that every 0.6 m, there is a wood stud, and the space between studs is filled with insulation, we can calculate the overall thermal resistance of a representative section (1 m high): \(R = R_{in} + \frac{0.05\,\text{m}}{0.6\,\text{m}} R_{WS} + \frac{0.55\,\text{m}}{0.6\,\text{m}} R_{FI} + R_S + R_{out}\)

The rate of heat transfer through the wall can be found using the equation: \(q = \frac{\Delta T}{R}\), where \(\Delta T = T_{in} - T_{out} = 20°C - (-9°C)\) So, \(q = \frac{(20°C - (-9°C))}{R}\)