Question

a well-sealed room contains 60 kg of air at 25°c. now solar energy enters the room at an average rate of 0.8 kj/s while a 120-w fan is turned on to circulate the air in the room. if heat transfer through the walls is negligible the air temperature in the room in 30 min will be (cv = 0.718 kj/kg-°c and cp = 1.005 kj/kg-°c)

Answer 1

Temperature of the room after 30 min is 53°C

First we will calculate the total amount of heat entered in the room due to solar heat as well as due to fan

Q=(0.8*1000+120)*30*60

=1656000 J

=1656 kJ

heat is added in the room at constant pressure, therefore the temperature of the room after 30 min is calculated by the formula

Q=mC_{p}($T_{2}-T_{1})$

Plugging the value in the above equation

1656=60*1.005*(T_{2)-25)

T=52.46≈53°C