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A window air conditioner uses 1010 W of electricity and has a coefficient of performance of 3.88. 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 (324 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 324 K to 295 K?
2) How much work will be done by the air conditioner during the process of cooling the room?J3) How much time in minutes will it take for the room to cool using this air conditioner?minutes4)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 324 K to 295 K?minutes

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fortuneonyemuwa

Answer:

Explanation:

1.) [tex]Q=MC_v\delta T\\=\rho \times v \times C_v \times \delta T\\=1.2 \times 7 \times 7 \times 3 \times 720 \times (324-295)\\=3683232J[/tex]

2.) COP

[tex]= \frac{Q_{abs}}{W}=3.88\\\\work=949286.6J[/tex]

3.) Rate of heat absorption

   COP[tex] \times Power\\[/tex]

[tex]=3.88\times 1010watt=3918.8watt[/tex]

Therefore, cooling in minutes=[tex]\frac{\frac{3683232}{3918.8}}{60}=15.66mins[/tex]

COP cannot cycle= [tex]\frac{295}{324-295}\\\\=10.17=\frac{Q_{abs}}{power}[/tex]

Rate of heat absorption [tex]=(1010\times 10.17)watt=10271.7min[/tex]

cooling time in minutes with cannot cycle[tex]=\frac{\frac{3683232}{10271.7}}{60}=5.98mins[/tex]