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DiKsa [7]
3 years ago
11

In one cycle, a freezer uses 800 J of electrical energy in order to remove 1735 J of heat from its freezer compartment at 10.0°F

. Part A)
What is the coefficient of performance of this freezer?
Part B)
How much heat does it expel into the room during this cycle?
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
8 0

Answer:

Explanation:

A)

W = work done by the freezer = 800 J

Q = heat removed from the freezer = 1735 J

Q' = Heat expelled into the room

Coefficient of performance is given as

\beta = \frac{Q}{W}

inserting the values

\beta = \frac{1735}{800}

\beta = 2.2

B)

Heat expelled is given as

Q' = W + Q

Q' = 800 + 1735

Q' = 2535 J

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