An old, rusted air conditioner in a 2000 ft 2 house containing R-12 has lost all
ID: 903986 • Letter: A
Question
An old, rusted air conditioner in a 2000 ft2 house containing R-12 has lost all of its refrigerant to the atmosphere. Given that cooling is measured in terms of (short) tons of refrigerant (where one (short) ton is the amount of heat that would melt 1 (short) ton of ice in 24 h):
what is the rate of cooling in terms of kJ/h, and
how many ozone molecules were potentially destroyed by the release of the refrigerant?
Given:
A general rule for sizing air conditioners: (House area (ft2) x 25)/12000 - 0.5 = tons of refrigerant cooling
A general rule for cooling loads is 3 lbs of refrigerant required for 1 ton of cooling.
Heat of fusion = 80 calories/g
Explanation / Answer
Tons of Refrigerant cooliing =2000*25/12000-0.5 =3.66 tons of Refrigerant
Ton is the amount of heat that melts 200 lb of ice in 24 hr
Heat of fusion = 80calories/g =80*10-3Kcal/g=80*10-3*1800btu/lb =144 btu/lb
for melting 2000 lb of ice in 24 hr = 144*2000/24 = btu/hr=12000 btu/hr =12000*1.055 Kj/hr=12660.31Kj/hr
Refrigerant required= 3.66 tons of Refrigerant = 3.66 tons=3.66*2000 lbs = 7320 lbs
moles of Refrigerant= 7320/121=60.49 lbmoles/hr
moles of Ozone destroyed (1:1 ratio)= 60.49 moles/ hr of O3= 60.49*6.023*1023 molecules=364.4*1023 molecules