Question

A class of 10 students taking an exam has a power output per student of 114 W. Assume that the initial temperature of the room is 18.6oC and that its dimensions are 5.90 m by 15.0 m by 3.10 m. What is the temperature (in oC, do not enter units) of the room at the end of 50.0 min if all the heat remains in the air in the room and none is added by an outside source? The specific heat of air is 833 J/kg*oC, and its density is about 1.29E-3 g/cm3.

Answer 1

given,
density of air = (1.29*10^-3) g/cm^3
= 1.29 kg/m^3

let, volume of room be v
v = (length)*(breath)*(height)
= (5.90*15.0*3.10) m^3
= 274.35 m^3

so,
mass of air = v*(density of air)
= 274.35*1.29
= 353.9 kg

time = 50 min
= 3000 sec

energy per student in 50 min = (power per student)*time
= 114*3000
= 342000 J

total energy = (energy per student)*(total number of student)
= 342000*10
= 3420000 J

use,
total energy = (mass of air)*(specific heat of air)*(T2 - T1)
3420000 = 353.9*833*(T2 - 18.6)
3420000 = 294798.7*(T2 - 18.6)
(T2 - 18.6) = 11.6
T2 = (11.6 + 18.6) oC
= 30.2 oC

Answer: 30.2 oC