posted by linda .
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?
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