A well sealed room contains 60 kg of air at 200 kPa and 25 deg C. Now solar ener

Question

A well sealed room contains 60 kg of air at 200 kPa and 25 deg 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, what will the air temperature be in 30 minutes?

Similarly, A room contains 75 kg of air at 100 kPa and 15 deg C. The room has a 250 W refrigerator, a 120 W television, a 1.8 kW electric resistance heater, and a 50 W fan. During a cold day, it is observed that the appliances all run continuously but the air temperature remains constant. What is the rate of heat loss from the room?

Explanation / Answer

Q=0.8kJ/s*30*60s assume the fan only do work and not lost work Q=1440kJ heat capatity of air 1.012kJ/(kg*K) T=25C+1440kJ/(60kg*1.01kJ*kg^-1*C^-1) =48.7deg C refrigerator,television,heater all increase the heat of the room. Unless your refrigerator's heat dispatch is outside the room like a AC. Q=250W+120W+1800W=2170W=2170J/s since no heat change, Qin =Q out = 2170J/s

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