Question
A certain refrigerator requires 35 J of work to remove 190 J of heat from its...
A certain refrigerator requires 35 J of work to remove 190 J of heat from its interior.
(a) What is its coefficient of performance?
(b) How much heat is ejected to the surroundings at 16 °C?
J
(c) If the refrigerator cycle is reversible, what is the
temperature inside the refrigerator?
°C
Homework Answers
Answer #1
Heat removed from the interior by the refrigerator = QC = 190 J
Work required by the refrigerator = W = 35 J
COP = 5.428
Heat ejected to the surroundings = QH
QH = QC + W
QH = 190 + 35
QH = 225 J
Temperature of the surroundings = TH = 16oC = 289 K
Temperature inside the refrigerator = TC
The refrigerator cycle is reversible that is it runs on carnot cycle.
1568.692 - 5.428TC = TC
6.428TC = 1568.692
TC = 244.04 K
TC = -28.96oC
a) Coefficient of performance = 5.428
b) Heat ejected to the surroundings = 225 J
c) Temperature inside the refrigerator = -28.96 oC
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