Question

.a room with floor measurement 7m by10m air of mass 250kg at a temperature of 34 degree celsius.The air is cooled until the temperature falls to 24 degrees celsius.(i)calculate the height of the room.(ii)quantity of energy extracted to cool the room.(iii)which is higher,the calculate value or actual energy needed to cool the room?give a reason for your answer.(shc of air=1010j/kg/k density of air 1.25kg/m cube)

Answer 1

Room floor area : 7 m * 10 m
mass of air :  250 kg 
Temperature T₀ = 34⁰ C
T₁ = 24° C

Air density 1.25 kg / m³   - let us take that this is the density at 34⁰ C
Volume of the room  = mass of air / density = 250 / 1.25 = 200 m³
Height of the room = volume / floor area = 200 / 70 = 2.855 m
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shc of air : specific heat capacity of air = 1010 J/ kg/ °K 
Heat extracted when air is cooled from 34° to 24°C : ΔQ = m s ΔT
           = 250 * 1010 * (34 - 24) = 2,525 KJ
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The actual energy needed (to be extracted) to cool the room is more than the calculated value.  One needs to cool the walls also, while cooling the air in the room.  The walls try to heat the room continuously.  The air outside the room also tries to heat the room.  So that additional heat is also to be extracted out.