power*timeinseconds=specifheatair*densityair*volumeair(Tf-Ti)
Tf=Ti+power*time/(specheatair*density*volume)
Tf=18.8+54*60*122/(840*6.4*14.5*3.5*density)
now change denstity to kg/m^3
density=1.25E-3g/cm^3*(100cm/m)^3*1kg/1000g= 1.25E-3*E6*E-3=1.25kg/m^3
Tf= about one C higher. Check my calcs
A class of 11 students taking an exam has a power output per student of 122 W. Assume that the initial temperature of the room is 18.8oC and that its dimensions are 6.40 m by 14.5 m by 3.50 m. What is the temperature (in oC, do not enter units) of the room at the end of 54.0 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 840 J/kg*oC, and its density is about 1.25E-3 g/cm3
** I keep getting different answers ranging from 30-40 and nothing is correct! HelP!
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