Question

When a man returns to his well-sealed house on a summer day, he finds that the house is at 35 ⁰c. he turns on the air conditioner, which cools the entire house to 20 ⁰c in 30 min. if the cop of the air-conditioning system is 2.8, determine the power drawn by the air conditioner. assume the entire mass within the house is equivalent to 800 kg of air for which cv 0.72 kj/kg·⁰c and cp 1.0 kj/kg·⁰c

Answer 1

The power drawn by the air conditioner which cools the house from 35°C to 20°C in 30 minutes with a COP of 2.8 is 2.380 kW.

Step-by-step explanation:

To determine the power drawn by the air conditioner, we need to calculate the total amount of heat removed (Q) from the house and then use the coefficient of performance (COP) of the air conditioner.

Coefficient of performance (COP) is defined as the amount of heat removed from the cold reservoir (Qc) divided by the work (W) done by the system: COP = Qc/W.

Firstly, we calculate the heat removed using the formula Q = m * cp * ΔT, where:

• m = mass of air = 800 kg
• cp = specific heat capacity at constant pressure = 1.0 kJ/kg·°C
• ΔT = change in temperature = 35°C - 20°C = 15°C

Q = 800 kg * 1.0 kJ/kg·°C * 15°C = 12000 kJ

Next, we convert the total heat removed (Q) from kJ to kWh since power is typically measured in kW:

1 kWh = 3600 kJ, so Q in kWh = 12000 kJ / 3600 kJ/kWh = 3.333 kWh (because 1 kWh = 3600 kJ)

Using the COP, we can now calculate the work done (W) by the air conditioner in kWh: W = Q/COP

W = 3.333 kWh / 2.8 = 1.190 kWh

Since the air conditioner does this work in 30 minutes (0.5 hours), we can find the power (P) in kW as follows: P = W/time (in hours), so:

P = 1.190 kWh / 0.5 h = 2.380 kW

Thus, the power drawn by the air conditioner is 2.380 kW.