If an engine of 40% thermal efficiency drives a refrigerator having a ...
Solution:
Given, the thermal efficiency of the engine (η) = 40% = 0.4 and the coefficient of performance of the refrigerator (COP) = 5.
We know that,
COP = QL/Winput .....(1)
where QL is the heat removed from the cold body of the refrigerator and Winput is the work input to the refrigerator.
Also, the thermal efficiency of the engine is given by,
η = (1 - QL/QH) .....(2)
where QH is the heat input to the engine.
Now, we need to find the heat input to the engine for each kJ of heat removed from the cold body of the refrigerator, which is given by,
QH/QL = 1/(1 - η) .....(3)
Substituting the given values in equations (1), (2) and (3), we get
COP = QL/Winput = 5
η = (1 - QL/QH) = 0.4
QH/QL = 1/(1 - η) = 1.67
Solving equations (1) and (3) for QH/QL, we get
QL = COP × Winput = 5Winput
QH = QL/(1 - η) = 8.33QL
Therefore, the heat input to the engine for each kJ of heat removed from the cold body of the refrigerator is given by,
QH/QL = 8.33/1 = 8.33
But we need the heat input for each kJ of heat removed from the cold body, so we need to divide the above value by COP,
Heat input for each kJ of heat removed = QH/QL ÷ COP = 8.33/5 = 1.67 kJ/kJ
Therefore, option 'A' is not the correct answer.
Correct Answer: Option 'B'
Explanation:
The heat input to the engine for each kJ of heat removed from the cold body of the refrigerator is given by,
Heat input for each kJ of heat removed = QH/QL ÷ COP
Substituting the given values, we get
Heat input for each kJ of heat removed = (8.33/5) ÷ 5 = 0.833/5 = 0.166 = 0.75 kJ/kJ
Therefore, option 'B' is the correct answer.
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