b) One wall of a house consists of plywood backed by insulation. The thickness of plywood is \( (0.019 \mathrm{~m}) \) and that of insulation is \( (0.076 \mathrm{~m}) \). The inside temperature is \( 25^{\circ} \mathrm{C} \) and outside is \( 4^{\circ} \mathrm{C} \). The thermal conductivities of the insulation and plywood are, respectively, 0.030 and \( 0.080 \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m} \cdot{ }^{\circ} \mathrm{C}\right) \), and the area of the wall is \( 35 \mathrm{~m}^{2} \). Calculate the amount of heat conducted through the wall in 1 hour.

To calculate the amount of heat conducted through the wall, we can use Fourier's law of heat conduction, which is given by the formula: \[ Q = \frac{k \cdot A \cdot \Delta T}{d} \] where: - \( Q \) is the heat transfer (in Joules), - \( k \) is the thermal conductivity (in \( \mathrm{J/(s \cdot m \cdot ^\circ C)} \)), - \( A \) is the area through which heat is being conducted (in \( \mathrm{m^2} \)), - \( \Delta T \) is the temperature difference across the material (in \( ^\circ C \)), - \( d \) is the thickness of the material (in meters). Since the wall consists of two materials (plywood and insulation), we need to calculate the heat transfer through each material separately and then sum them up. ### Step 1: Calculate the total thermal resistance The total thermal resistance \( R \) for the wall can be calculated as: \[ R = R_{\text{plywood}} + R_{\text{insulation}} \] where: \[ R_{\text{plywood}} = \frac{d_{\text{plywood}}}{k_{\text{plywood}} \cdot A} \] \[ R_{\text{insulation}} = \frac{d_{\text{insulation}}}{k_{\text{insulation}} \cdot A} \] ### Given values: - Thickness of plywood, \( d_{\text{plywood}} = 0.019 \, \text{m} \) - Thickness of insulation, \( d_{\text{insulation}} = 0.076 \, \text{m} \) - Thermal conductivity of plywood, \( k_{\text{plywood}} = 0.080 \, \text{J/(s \cdot m \cdot ^\circ C)} \) - Thermal conductivity of insulation, \( k_{\text{insulation}} = 0.030 \, \text{J/(s \cdot m \cdot ^\circ C)} \) - Area of the wall, \( A = 35 \, \text{m}^2 \) - Inside temperature, \( T_{\text{inside}} = 25^\circ C \) - Outside temperature, \( T_{\text{outside}} = 4^\circ C \) ### Step 2: Calculate the temperature difference \[ \Delta T = T_{\text{inside}} - T_{\text{outside}} = 25^\circ C - 4^\circ C = 21^\circ C \] ### Step 3: Calculate the thermal resistances 1. **Plywood:** \[ R_{\text{plywood}} = \frac{0.019 \, \text{m}}{0.080 \, \text{J/(s \cdot m \cdot ^\circ C)} \cdot 35 \, \text{m}^2} = \frac{0.019}{0.080 \cdot 35} = \frac{0.019}{2.8} \approx 0.00679 \, \text{°C/W} \] 2. **Insulation:** \[ R_{\text{insulation}} = \frac{0.076 \, \text{m}}{0.030 \, \text{J/(s \cdot m \cdot ^\circ C)} \cdot 35 \, \text{m}^2} = \frac{0.076}{0.030 \cdot 35} = \frac{0.076}{1.05} \approx 0.07238 \, \text{°C/W} \] ### Step 4: Total thermal resistance \[ R_{\text{total}} = R_{\text{plywood}} + R_{\text{insulation}} \approx 0.00679 + 0.07238 \approx 0.07917 \, \text{°C/W} \] ### Step 5: Calculate the heat transfer rate Using the total thermal resistance, we can calculate the heat transfer rate \( \dot{Q} \): \[ \dot{Q} = \frac{\Delta T}{R_{\text{total}}} = \frac{21 \, \text{°C}}{0.07917 \, \text{°C/W}} \approx 265.5 \, \text{W} \] ### Step 6: Calculate the total heat conducted in 1 hour To find the total heat conducted in 1 hour, we convert the power to energy: \[ Q = \dot{Q} \cdot t \] where \( t = 1 \, \text{hour} = 3600 \, \text{s} \): \[ Q = 265.5 \, \text{W} \cdot 3600 \, \text{s} \approx 956,000 \, \text{J} \] ### Final Answer The amount of heat conducted through the wall in 1 hour is approximately: \[ Q \approx 956,000 \, \text{J} \text{ or } 956 \, \text{kJ} \]