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The motor in a refrigerator has a power of 293 W. If the freezing compartment is at 270 K and the outside air is at 294 K, and assuming the efficiency of a Carnot refrigerator, what is the maximum amount of energy that can be extracted as heat from the freezing compartment in 19.0 min?

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Final answer:

The maximum amount of energy that can be extracted as heat from the freezing compartment of a refrigerator, with a power of 293 W and operating for 19 minutes under Carnot efficiency, is 4,088,971 J.

Step-by-step explanation:

Given that the power of the motor is 293 W and the operation time is 19.0 minutes, we first need to calculate the work done by the refrigerator in that time:

Work done (W) = Power (P) × Time (t)

W = 293 W × (19 min × 60 seconds/min) = 333,740 J

Now, we'll use the Carnot refrigerator efficiency formula to find the maximum amount of energy that can be extracted as heat (Qc):

ε = 1 - (Tc/Th) where ε is the efficiency, Tc is the temperature of the cold reservoir (freezing compartment), and Th is the temperature of the hot reservoir (outside air).

ε = 1 - (270 K / 294 K) = 0.0816

The efficiency of the Carnot cycle is also the work divided by the heat extracted from the cold reservoir:

ε = W / Qc ⇒ Qc = W / ε

Qc = 333,740 J / 0.0816 = 4,088,971 J

The maximum amount of energy that can be extracted as heat from the freezing compartment in 19.0 minutes is 4,088,971 J.

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