Question

An air conditioner operates on 800 W of power and has a performance coefficient of 2.80 with a room temperature of 21.0 ∘C and an outside temperature of 33.0 ∘C.

Calculate the rate of heat removal for this unit.

Calculate the rate at which heat is discharged to the outside air.

Calculate the total entropy change in the room if the air conditioner runs for 1 hour.

Calculate the total entropy change in the outside air for the same time period.

What is the net change in entropy for the system (room + outside air)?

Answer 1

Operating power = 800 W

Performance coefficient = 2.80

Room temperature = 21.0 oC

Outside temperature = 33.0 oC

(1). Rate of heat removal = Operating Power * Performance coefficient

= 800 * 2.80

= 2240 W

(2). Rate at which heat is discharged to the outside air:

=  Rate of heat removal + Operating Power

= 2240 + 800

= 3040 W

(3). Total entropy change in the room if the air conditioner runs for 1 hour:

Heat = 2240 W * 3600 s

= 8064000 J

Temperature = 21 oC = 294 K

Entropy = - 8064000 / 294

= - 2.74x10⁴ J/K

(4). Total entropy change in the outside air for the same time period:

Heat = 3040 W * 3600 s

= 10944000 J

Temperature = 33 oC = 306 K

Entropy = 10944000 / 306

=  3.57x10⁴ J/K

(5). Net change in entropy for the system:

= 3.57x10⁴ - 2.74x10⁴

= 0.83x10⁴ J/K