An air conditioner removes heat steadily from a house at a rate of 910 kJ/min while drawing electric power at a rate of 5.25 kW.
The correct answer and explanation is:
To find the coefficient of performance (COP) of the air conditioner, use the formula: COP=QLW\text{COP} = \frac{Q_L}{W}
Where:
- QLQ_L is the amount of heat removed per second (cooling effect), in kJ/s (equivalent to kW)
- WW is the electric power input, in kW
Step 1: Convert heat removal rate from kJ/min to kW
Given:
- QL=910 kJ/minQ_L = 910 \, \text{kJ/min}
- Convert to kJ/s (divide by 60):
QL=91060=15.17 kJ/s=15.17 kWQ_L = \frac{910}{60} = 15.17 \, \text{kJ/s} = 15.17 \, \text{kW}
Step 2: Use the formula for COP
Given:
- W=5.25 kWW = 5.25 \, \text{kW}
- QL=15.17 kWQ_L = 15.17 \, \text{kW}
COP=15.175.25≈2.89\text{COP} = \frac{15.17}{5.25} \approx 2.89
Final Answer: The coefficient of performance (COP) is approximately 2.89
Explanation:
An air conditioner works by transferring heat from the inside of a building to the outside. The coefficient of performance (COP) is a key measure of efficiency for cooling systems. It shows how much cooling is provided for each unit of electrical energy used. The higher the COP, the more efficient the system is.
In this case, the air conditioner removes 910 kilojoules of heat every minute. This was first converted to kilowatts, because power (work per unit time) is typically measured in kilowatts in engineering applications. Since 1 minute has 60 seconds, the rate of heat removal becomes 15.17 kW.
The electric power consumed by the air conditioner is 5.25 kW. Using the COP formula, dividing the cooling output (15.17 kW) by the power input (5.25 kW) gives a value of approximately 2.89. This means the air conditioner removes about 2.89 units of heat for every 1 unit of electricity it consumes.
A COP of 2.89 is typical for many modern air conditioning systems and indicates moderate energy efficiency. Higher COP values mean better performance and lower electricity costs for the same cooling effect.