To solve the problem of how much heat the air conditioner removes from the room and how much it delivers to the outside air, we can use the concept of the coefficient of performance (COP). The COP is a measure of the efficiency of the air conditioner, indicating how much heat is removed from the space for each unit of electrical energy consumed.
Understanding the Coefficient of Performance
The coefficient of performance (COP) is defined as the ratio of the heat removed from the room (Q_c) to the work input (W) done by the air conditioner. Mathematically, this is expressed as:
COP = Q_c / W
From the problem, we know that the COP is 2.9 and the power consumption (work input) is 850 W. To find the heat removed from the room, we can rearrange the formula:
Q_c = COP × W
Calculating Heat Removed from the Room
First, we need to calculate the heat removed in joules. Since power is measured in watts (joules per second), we can find the total energy removed in one minute (60 seconds) by multiplying the power by the time:
W = 850 W
Time = 60 seconds
Now, we can calculate the total work done in one minute:
Total Work = W × Time = 850 W × 60 s = 51,000 J
Next, we can find the heat removed from the room:
Q_c = COP × W = 2.9 × 51,000 J = 147,900 J
Heat Delivered to the Outside Air
To find out how much heat is delivered to the outside air (Q_h), we can use the relationship between the heat removed and the work input:
Q_h = Q_c + W
Substituting the values we have:
Q_h = 147,900 J + 51,000 J = 198,900 J
Summary of Results
- Heat removed from the room in one minute: 147,900 J
- Heat delivered to the outside air in one minute: 198,900 J
This analysis shows how the efficiency of the air conditioner, as indicated by its COP, directly influences the amount of heat it can remove from a space while consuming a certain amount of electrical power. Understanding these relationships is crucial for evaluating the performance of HVAC systems in real-world applications.
