A window air conditioner is running to reduce the room temperature to the temperature setting. The operation time and the total heat released to the ambient need to be determined.

Assumption:

      Assume energy of the system remains constant.

(1) Heat transferred to the air conditioner Q_in

In order to cool the room to the setting temperature, heat needs to be removed from the room, which is transferred to the air conditioner, and it is the energy entering the system.

      Q_in = (50,000 J/^oF) (90^oF - 75^oF)
            = 750,000 J

(2) Total work input for the air conditioner W_in

According to the definition of COP, the amount of work input needed for the air conditioner is given by

      COP = Q_in / W_in

      W_in = Q_in / COP = (750,000 J) / 2.5
             = 300,000 J

The power input for the air conditioner is 900 W. So the operation time is:

      t = W_in / = (300,000 J) / (900 J/s)
        = 333.3 s = 5.555 min

This is a short time, and thus his room must be very small (maybe 10'x10'x7'). But then most student rooms are small.

(3) The total heat released to the ambient Q_out

The energy equation can be simplified according to the assumptions.

• Since energy of the system remains constant, ΔE_system is 0.
• No work is done by the air conditioner to its ambient , hence W_out = 0.
• The system is a closed system, hence no mass crossed its boundaries. That is,
  
  E_mass,in - E_mass,out = 0.

      Q_out = Q_in + W_in = 750,000 + 300,000
             = 1,050,000 J = 1,050 kJ