heat= massair*specifichat*deltatemp
moles air=PV/RT then mass air= 27*molesair in grams.
Now, use the first equatio to find deltaTemp
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?
2 answers
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
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
1 answer