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A window air conditioner uses 1125 W of electricity and has a coefficient of performance of 3.41. Assume that all of this 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 initially has the same temperature as the air outside (324 K), and the air conditioner will attempt to cool the room to 295 K. Assume the air in this room will be cooled at constant volume, with the specific heat of air at constant volume being 720 J/kg-K, and the density of air is 1.2 kg/m^3.

Required:

a. How much energy must be transferred out of this room to decrease the temperature of the air from 324 K to 295 K?

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

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

Answer :

Final answer:

The energy that needs to be transferred out of the room is approximately 3.79 MJ. The work done by the air conditioner during the process is 1.11 MJ. It will take approximately 16.45 minutes to cool the room using this air conditioner.

Explanation:

Let's start with your first question, how much energy needs to be transferred out of this room in order to decrease the temperature from 324 K to 295 K. Firstly, you need to find the mass of air in the room. The volume of the room is 7m * 7m * 3m = 147 m^3. Given that the density of the air is 1.2 kg/m^3, the mass of the air in the room is 147 m^3 * 1.2 kg/m^3 = 176.4 kg. Then, you apply the formula for heat transfer at constant volume, Q = mcΔT, where m is the mass, c is the specific heat at constant volume, and ΔT is the change in temperature. Q = 176.4 kg * 720 J/kg-K * (324 K - 295 K) = 3.79 million Joules (or 3.79 MJ).

For your second question, the work done by the air conditioner is the energy transfer divided by the coefficient of performance. W = Q / COP = 3.79 MJ / 3.41 = 1.11 MJ.

Lastly, the window air conditioner uses 1125 W, or 1.125 kW, of electricity. A kilowatt is equivalent to kilojoules per second, so it takes 1.11 MJ / 1.125 kJ/s = 987 seconds to cool the room. Converting to minutes, that is approximately 16.45 minutes.

Learn more about Cooling Process here:

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