4.9

Question

Estimate the maximum possible COP of a household air conditioner. Use any reasonable values for the reservoir temperatures.

Step-by-Step Solution

Verified

Answer

COP=21

1Step 1: Given

Assume temp inside room T_c=25^o C = 298 K

Assume temp outside room T_h=40^o C = 313K

Find COP.

2Step 2: Explanation

Coefficient of performance (COP) $= \frac{T_{c}}{T_{h} - T_{c}}$
Substitute the values of temperature
$COP = \frac{T_{c}}{T_{h} - T_{c}} = \frac{313 K}{(313 K) - (298 K)} = 20.866$ = 21 ( rounded)

4.8

4.10

Other exercises in this chapter

4.7

Why must you put an air conditioner in the window of a building, rather than in the middle of a room?

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4.8

Can you cool off your kitchen by leaving the refrigerator door open? Explain.

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4.10

Suppose that heat leaks into your kitchen refrigerator at an average rate of 300 watts. Assuming ideal operation, how much power must it draw from the wall?

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4.11

What is the maximum possible COP for a cyclic refrigerator operating between a high-temperature reservoir at 1K and a low-temperature reservoir at 0.01 K ?

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