A brick wall \( (k=0.9 \mathrm{~W} / \mathrm{m} . \mathrm{K}) \) of thickness 0.18 m separates the warm air in a room from the cold ambient air. On a particular winter day, the outside air temperature is \( -5^{\circ} \mathrm{C} \) and the room needs to be maintained at \( 27^{\circ} \mathrm{C} \). The heat transfer coefficient associated with outside air is \( 20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K} \). Neglecting the convective resistance of the air inside the room, the heat loss, in \( \mathrm{W} / \mathrm{m}^{2} \), is . . (round off to the nearest integer).

To determine the heat loss through the brick wall, we first calculate the total thermal resistance. The formula for total thermal resistance \( R_{\text{total}} \) is the sum of the convective resistance at the outside surface \( R_{\text{conv, outside}} \) and the conductive resistance of the wall \( R_{\text{conductive}} \): 1. **Calculating Convective Resistance (Outside)**: \[ R_{\text{conv, outside}} = \frac{1}{h_{outside}} = \frac{1}{20 \, \mathrm{W/m^2 \cdot K}} = 0.05 \, \mathrm{m^2 \cdot K/W} \] 2. **Calculating Conductive Resistance (Wall)**: \[ R_{\text{conductive}} = \frac{d}{k} = \frac{0.18 \, \mathrm{m}}{0.9 \, \mathrm{W/m \cdot K}} = 0.2 \, \mathrm{m^2 \cdot K/W} \] 3. **Total Resistance**: \[ R_{\text{total}} = R_{\text{conv, outside}} + R_{\text{conductive}} = 0.05 \, \mathrm{m^2 \cdot K/W} + 0.2 \, \mathrm{m^2 \cdot K/W} = 0.25 \, \mathrm{m^2 \cdot K/W} \] 4. **Temperature Difference**: \[ \Delta T = T_{inside} - T_{outside} = 27^{\circ}C - (-5^{\circ}C) = 32 \, \mathrm{K} \] 5. **Calculating Heat Loss (Q)**: Using Fourier’s law of heat conduction and Newton's law of cooling, \[ Q = \frac{\Delta T}{R_{\text{total}}} = \frac{32 \, \mathrm{K}}{0.25 \, \mathrm{m^2 \cdot K/W}} = 128 \, \mathrm{W/m^2} \] Thus, the heat loss through the brick wall rounding off to the nearest integer is **128 W/m²**.