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 1.013·10^5 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?
heat= massair*specifichat*deltatemp
moles air=PV/RT then mass air= 27*molesair in grams.
Now, use the first equatio to find deltaTemp
Assume that the persons' bodies remain at the same temperature and that the heat they generate goes into the air. That is not a bad assumption.
Q = heat generated
= 2210*73.1 J/s * 5400 s
= 8.72*10^8 J
The specific heat at constant volume of the air is
Cv = (5/2)R
= 4.97 calories/mole K
= 20.77 J/mole K
Since there is no ventilation, it is assumed that the air cannot get out as the temperature rises. If the air can escape, then use
Cp = (7/2)R
Use n = PV/RT for the number of moles
n = 27.1*10^3*1.013·10^5/(8.317*292) = 1.13*10^6 moles
delta T = Q/(n*Cv) = ? degrees K or C
To find the temperature rise during the show, we need to calculate the amount of heat generated by the metabolizing of each person and then use the heat capacity formula to determine the temperature increase.
Step 1: Calculate the heat generated by each person:
The heat generated by each person can be calculated using the formula: Q = P × t, where Q is the heat generated, P is the power, and t is the time. In this case, P = 73.1 W and t = 1.5 hours = 1.5 × 3600 seconds = 5400 seconds.
So, Q = 73.1 W × 5400 sec = 394740 J
Step 2: Calculate the total heat generated by all the people:
To find the total heat generated, we multiply the individual heat by the number of people: Q_total = Q × number of people.
Q_total = 394740 J × 2210 people = 872040740 J.
Step 3: Calculate the temperature increase using the heat capacity formula:
The heat capacity of the air in the auditorium can be approximated using the formula: Q = mcΔT, where Q is the heat added, m is the mass, c is the specific heat capacity, and ΔT is the temperature change.
We can assume the mass of air as the volume multiplied by the density of air. The density of air at standard conditions (1 atm and 273 K) is approximately 1.225 kg/m^3.
m = volume × density = 27100 m^3 × 1.225 kg/m^3 = 33227.5 kg
The specific heat capacity of air at constant pressure is approximately 1005 J/(kg·K).
Now, let's solve for ΔT:
Q_total = mcΔT
872040740 J = 33227.5 kg × 1005 J/(kg·K) × ΔT
ΔT = 872040740 J / (33227.5 kg × 1005 J/(kg·K))
ΔT = 26.12 K
Therefore, the temperature of the air in the auditorium is estimated to rise by approximately 26.12 K during the 1.5-hour show if there is no ventilation.