A closed auditorium of volume 27100 m^3 is filled with 2210 people at the beginning of a show, and the air in the space is at a temperature of 292 K and a pressure of 1.013·105 Pa. If there were no ventilation, by how much would the temperature (in K) of the air rise during the 1.5-h show if each person metabolizes at a rate of 73.1 W?
To determine the temperature rise of the air in the closed auditorium during the show, we need to calculate the heat energy generated by the metabolism of each person and then use the equation for heat transfer.
Step 1: Find the total heat energy generated by metabolism.
The heat energy generated by each person's metabolism can be calculated using the formula:
Q = Pt
where Q is the heat energy, P is the power (metabolic rate), and t is the time.
Given:
P = 73.1 W (metabolic rate per person)
t = 1.5 hours = 1.5 × 3600 seconds = 5400 seconds (conversion to seconds)
Using the formula, we can calculate Q:
Q = 73.1 W × 5400 s = 394,740 J
Step 2: Calculate the change in temperature.
The change in temperature can be determined using the equation for heat transfer:
ΔQ = mcΔT
where ΔQ is the heat energy transferred, m is the mass of the air, c is the specific heat capacity of air, and ΔT is the change in temperature.
Given:
ΔQ = 394,740 J (from Step 1)
m = Volume × Density
= 27,100 m^3 × ρ (density of air)
The density of air can be obtained using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
From the ideal gas law, we can solve for the density of air (ρ):
ρ = n/V = P/RT
Given:
P = 1.013 × 10^5 Pa
V = 27,100 m^3
R = 8.314 J/(mol·K)
Using these values, we can calculate the density of air (ρ):
ρ = (1.013 × 10^5 Pa) / (8.314 J/(mol·K) × 292 K)
ρ ≈ 1.242 kg/m^3
Now we can substitute the density value into the mass equation:
m = 27,100 m^3 × 1.242 kg/m^3
m ≈ 33,621 kg
The specific heat capacity of air, c, is approximately 1005 J/(kg·K).
Using these values, we can calculate the change in temperature (ΔT):
ΔT = ΔQ / (mc)
= 394,740 J / (33,621 kg × 1005 J/(kg·K))
≈ 0.012 K
Therefore, the temperature of the air in the closed auditorium would rise by approximately 0.012 K during the 1.5-hour show if there were no ventilation.