A window air conditioner uses 1242 W of electricity and has a coefficient of per

Question

A window air conditioner uses 1242 W of electricity and has a coefficient of performance of 3.74. For simplicity, let's assume that all of that electrical energy is transformed into work done by the air conditioner to cool a room with dimensions of 7 m x 7 m x 3 m. The air inside this room will initially have the same temperature as the air outside (320 K) and the air conditioner will attempt to cool the room to 295 K. For the purposes of estimation, assume that the air in this room will be cooled at constant volume, that the specific heat of the air at constant volume is 720 J/kg-K, and that the density of the air is 1.2 kg/m3.

1)How much energy must be transferred out of this room in order to decrease the temperature of the air from 320 K to 295 K?

2)How much work will be done by the air conditioner during the process of cooling the room?

3)How much time in minutes will it take for the room to cool using this air conditioner?

4)How much time in minutes would the cooling process take if instead we used an air conditioner that uses a Carnot cycle that operates between 320 K and 295 K?

Explanation / Answer

Volume of room = 7*7*3=147m3;

Mass of air = Density*volume=1.2*147=176.4kg

Energy transferred out = Mass of air*specific heat capacity * change in temperature=176.4*720*(320-295)=3175200J

COP=Heat out/Work done;

Work done = Heat out/COP=3175200/3.74=848983.9572J

Rate of heat out = COP *1242 = 3.74 * 1242 =4645.08J/s;

Time = Heat transferred out/rate =3175200/4645.08= 683.56 s =11.39 mins;

COP of carnot cycle =Tc/(Th-Tc)=11.8;

therefore rate = 11.8*1242=14655.6;

Therefore Time = 3175200/14655.6=216.654s=3.61mins

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