There are 200 people in a \(2000-\mathrm{m}^{2}\) room, lighted with \(30 \mathrm{~W} / \mathrm{m}^{2}\). Estimate the maximum temperature increase, in \({ }^{\circ} \mathrm{C}\), if the ventilation system fails for \(20 \mathrm{~min}\). Each person generates \(400 \mathrm{~kJ} / \mathrm{h}\). The room is \(3 \mathrm{~m}\) high. a) \(5.6\) c) \(8.6\) b) \(6.8\) d) \(13.4\)

The lighting power density is given as \(30~W/m^2\), and the room's area is \(2000~m^2\). So, the total lighting power can be calculated as: \(\text{Total Lighting Power (P)} = \text{Lighting Power Density} \times \text{Room Area}\) \(P = 30 \frac{\text{W}}{\text{m}^2} \times 2000~\text{m}^2 = 60,000~\text{W}\)

Every one of the 200 people generates \(400~kJ/h\). To get the total energy generation from people per second (watts) inside the room, we need to multiply it by the number of people and convert it to watts: \(\text{Total People Power (Pp)} = \text{Power per Person} \times \text{Number of People} \times \frac{1~\text{hour}}{3600~\text{s}}\) \(Pp = 400000 \frac{\text{J}}{\text{hour}} \times 200 \times \frac{1~\text{hour}}{3600~\text{s}} = 44,444.44~\text{W}\)

We can add the energy generation from lighting and people to get the total power generation: \(\text{Total Power Generation} = \text{Total Lighting Power} + \text{Total People Power}\) \(\text{Total Power Generation} = 60,000~\text{W} + 44,444.44~\text{W} = 104,444.44~\text{W}\)

Now, we need to determine how much energy will be generated within 20 minutes without ventilation: \(\text{Energy Generated} = \text{Total Power Generation} \times \text{Time}\) \(\text{Energy Generated} = 104,444.44~\text{W} \times 20 \times 60 \text{s} = 125,333,333.2~\text{J}\)

Given the room's length, width, and height, we can calculate the room's volume: \(\text{Volume} = \text{Area} \times \text{Height}\) \(\text{Volume} = 2000~\text{m}^2 \times 3~\text{m} = 6000~\text{m}^3\)

To find the temperature increase, we will use the equation: \(\Delta T = \frac{\text{Energy Generated}}{\text{Volume} \times \text{Specific Heat Capacity} \times \text{Air Density}}\) The specific heat capacity of air is approximately \(1002~\frac{\text{J}}{\text{kg} \cdot {}^{\circ}C}\) and air density is approximately \(1.18~\frac{\text{kg}}{\text{m}^3}\). Plugging in the values: \(\Delta T = \frac{125,333,333.2~\text{J}}{6000~\text{m}^3 \times 1002~\frac{\text{J}}{\text{kg} \cdot {}^{\circ}C} \times 1.18~\frac{\text{kg}}{\text{m}^3}}\) \(\Delta T \approx 6.8^{\circ}\text{C}\) Therefore, the maximum temperature increase if the ventilation system fails for 20 minutes is approximately \(6.8^{\circ}\text{C}\) (option b).