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50 students are sitting in the room of 5*10*3m^3. The air inside the room is at 27C and 1atm pressure. If each student loses 100 Watt per second assuming the walls, ceiling, floor and all materials present inside the room are perfectly insulated and neglecting the loss air to the outside as the temperature is raised, how much rise in temperature will be noticed in 10minutes?

Sunil Kumar FP
6 years ago
we have power=work done/time taken
work done is the energy released by the student
total energy released by the student in 1second=50*100 J
energy released by the student in 10 minutes=5000*10*60 J=3*10^6J
Since heat released by the students will lead to increase in internal energy,therefore
dU=change in internal energy=nCvdT
dT is the change in temperature
number of mole of air=PV/RT =1*5*10^6 /(.0821*300) =2*10^5mole
change in internal energy=heat released
3*10^6=2*10^5*(T-300)*718
T=300+.02 K
=300.02 K