A hotel room is in thermal equilibrium with the rooms on either side and with the hallway on a third side. The room loses heat primarily through a 1.30 -cm- thick glass window that has a height of \(76.2 \mathrm{cm}\) and a width of $156 \mathrm{cm} .\( If the temperature inside the room is \)75^{\circ} \mathrm{F}$ and the temperature outside is \(32^{\circ} \mathrm{F}\), what is the approximate rate (in \(\mathrm{kJ} / \mathrm{s}\) ) at which heat must be added to the room to maintain a constant temperature of \(75^{\circ} \mathrm{F} ?\) Ignore the stagnant air layers on either side of the glass.

Inside temperature: \(T_{1} = \frac{5}{9}(70 - 32)\) \(T_{1} = \frac{5}{9}(38)\) \(T_{1} = 21.1 °C\) Outside temperature: \(T_{2} = \frac{5}{9}(30 - 32)\) \(T_{2} = \frac{5}{9}(-2)\) \(T_{2} = -1.1 °C\) #tag_title#Step 2: Convert window dimensions to meters #tag_content#Next, let's convert the window dimensions from inches to meters. The conversion factor is 1 inch = 0.0254 meters. Width: \(4 ft \times 12 in/ft \times 0.0254 m/in = 1.2192 m\) Height: \(6 ft \times 12 in/ft \times 0.0254 m/in = 1.8288 m\) Thickness: \(0.25 in \times 0.0254 m/in = 0.00635 m\) #tag_title#Step 3: Calculate the area of the window #tag_content#Now, let's calculate the area of the window: \(A = width \times height\) \(A = 1.2192 m \times 1.8288 m\) \(A = 2.2304 m^2\) #tag_title#Step 4: Calculate the rate of heat transfer #tag_content#Finally, let's apply the formula for the rate of heat transfer: \(q = k \frac{A(T_{1} - T_{2})}{d}\) Given thermal conductivity of glass, \(k = 0.96 \, W/mK\): \(q = 0.96 \frac{2.2304(21.1 - (-1.1))}{0.00635}\) \(q = 0.96 \frac{2.2304(22.2)}{0.00635}\) \(q = 7583.09 \, W\) #tag_title#Step 5: Convert the rate of heat transfer to kilojoules per second #tag_content#Finally, let's convert the rate of heat transfer from watts to kilojoules per second: \(q = 7583.09 \, W \times \frac{1 \, kJ}{1000 \, W}\) \(q = 7.583 \, kJ/s\) So the rate at which heat must be added to maintain a constant temperature in the room is approximately 7.583 kJ/s.