A \(5-\mathrm{m} \times 6-\mathrm{m} \times 8-\mathrm{m}\) room is to be heated by an electrical resistance heater placed in a short duct in the room. Initially, the room is at \(15^{\circ} \mathrm{C}\), and the local atmospheric pressure is \(98 \mathrm{kPa}\). The room is losing heat steadily to the outside at a rate of \(200 \mathrm{~kJ} / \mathrm{min}\). A 300 -W fan circulates the air steadily through the duct and the electric heater at an average mass flow rate of \(50 \mathrm{~kg} / \mathrm{min}\). The duct can be assumed to be adiabatic, and there is no air leaking in or out of the room. If it takes 18 minutes for the room air to reach an average temperature of \(25^{\circ} \mathrm{C}\), find \((a)\) the power rating of the electric heater and \((b)\) the temperature rise that the air experiences each time it passes through the heater.

$$m = \frac{98 \times 10^3 \times 240}{0.287 \times 288.15}$$ $$m \approx 32498.43 \mathrm{~kg}$$ #tag_title#Step 2: Determine the final temperature after 18 minutes#tag_content#To calculate the final temperature, we first find the heat loss rate of 30 W for every degree increase in the room temperature. After 18 minutes, the temperature will have increased by \(t = \frac{30 \times 18}{1000} = 0.54\)°C since the heat loss rate is given per minute. Add this to the initial temperature: $$T_2 = T_1 + 0.54 = 288.15 + 0.54 = 288.69 \mathrm{~K}$$ #tag_title#Step 3: Calculate the power required by the heater#tag_content#Now, we need to determine the power rating of the heater. First, calculate the heat capacity of the air in the room using the specific heat of air at constant pressure, which is \(C_p = 1.005 \mathrm{~kJ} / (\mathrm{kg} \cdot \mathrm{K})\). The heat capacity of the air in the room is: $$Q = mc_p(T_2 - T_1)$$ $$Q = 32498.43 \times 1.005 \times (288.69 - 288.15)$$ $$Q \approx 9238.58 \mathrm{~kJ}$$ Next, convert the heat capacity to watts by dividing by the time: $$P_\mathrm{Heater} = \frac{Q}{t}$$ $$P_\mathrm{Heater} = \frac{9238.58}{18 \times 60} \approx 8.58 \mathrm{~kW}$$ #tag_title#Step 4: Calculate the temperature rise of the air#tag_content#Finally, we need to find the temperature rise of the air each time it passes through the heater. The rate of air circulation is 0.5 times per minute. So, the time taken for one circulation is: $$t_\mathrm{Circulation} = \frac{1}{0.5} = 2 \mathrm{~minutes}$$ To calculate the temperature rise per circulation, we use the heat transfer equation: $$Q_\mathrm{Transfer} = m'c_p(T'_2 - T'_1)$$ Here, \(m'\) is the mass of air passing through the heater per circulation. We can find this by multiplying the total mass of air by the circulation rate: $$m' = m \times 0.5 = 32498.43 \times 0.5 = 16249.22 \mathrm{~kg}$$ Now, we can find the temperature rise, \((T'_2 - T'_1)\), using the heater's power and the time taken for one circulation: $$T'_2 - T'_1 = \frac{Q_\mathrm{Transfer}}{m'c_p} = \frac{P_\mathrm{Heater}}{m'c_p} = \frac{8580}{16249.22 \times 1.005}$$ $$T'_2 - T'_1 \approx 0.52 \mathrm{~K}$$ Thus, the power rating of the electric heater is approximately 8.58 kW, and the temperature rise of the air each time it passes through the heater is approximately 0.52 K.