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A refrigerator is such that it extracts 20kJ of heat at a lower temperature and give out 50kJ of heat to the surrounding, what is the coefficient of performance (K)
Correct answer is '0.67'. Can you explain this answer?
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Verified Answer
A refrigerator is such that it extracts 20kJ of heat at a lower temper...
W = QK – QL
= 20/30
K = 0.67
The correct answer is: 0.67
Most Upvoted Answer
A refrigerator is such that it extracts 20kJ of heat at a lower temper...
Explanation:
The coefficient of performance (K) of a refrigerator is a measure of its efficiency and is defined as the ratio of the heat extracted from the refrigerator to the work done on the refrigerator.
K = Qc / W
where K is the coefficient of performance, Qc is the heat extracted from the refrigerator, and W is the work done on the refrigerator.
In this case, the refrigerator extracts 20 kJ of heat at a lower temperature and gives out 50 kJ of heat to the surroundings.
Heat extracted from the refrigerator (Qc) = 20 kJ
Heat given out to the surroundings (Qh) = 50 kJ
We can calculate the work done on the refrigerator using the first law of thermodynamics, which states that the work done on a system is equal to the heat added to the system minus the heat extracted from the system.
W = Qh - Qc
Substituting the values, we get:
W = 50 kJ - 20 kJ = 30 kJ
Now, we can calculate the coefficient of performance (K) using the formula:
K = Qc / W
Substituting the values, we get:
K = 20 kJ / 30 kJ = 0.67
Therefore, the coefficient of performance (K) of the refrigerator is 0.67.
The coefficient of performance (K) of a refrigerator is a measure of its efficiency and is defined as the ratio of the heat extracted from the refrigerator to the work done on the refrigerator.
K = Qc / W
where K is the coefficient of performance, Qc is the heat extracted from the refrigerator, and W is the work done on the refrigerator.
In this case, the refrigerator extracts 20 kJ of heat at a lower temperature and gives out 50 kJ of heat to the surroundings.
Heat extracted from the refrigerator (Qc) = 20 kJ
Heat given out to the surroundings (Qh) = 50 kJ
We can calculate the work done on the refrigerator using the first law of thermodynamics, which states that the work done on a system is equal to the heat added to the system minus the heat extracted from the system.
W = Qh - Qc
Substituting the values, we get:
W = 50 kJ - 20 kJ = 30 kJ
Now, we can calculate the coefficient of performance (K) using the formula:
K = Qc / W
Substituting the values, we get:
K = 20 kJ / 30 kJ = 0.67
Therefore, the coefficient of performance (K) of the refrigerator is 0.67.
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A refrigerator is such that it extracts 20kJ of heat at a lower temperature and give out 50kJ of heat to the surrounding, what is the coefficient of performance (K)Correct answer is '0.67'. Can you explain this answer?
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