Question
Download Solution PDFA refrigerator with a COP of 2 removes heat from the refrigerated space at the rate of 100 kJ/min. The amount of heat rejected to the surroundings will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The COP of a refrigerator is defined as:
\( COP = \frac{Q_L}{W} \Rightarrow W = \frac{Q_L}{COP} \)
Total heat rejected: \( Q_H = Q_L + W \)
Given:
- \( COP = 2 \)
- \( Q_L = 100~\text{kJ/min} = 1.667~\text{kW} \)
Calculation:
\( W = \frac{1.667}{2} = 0.8335~\text{kW} \)
\( Q_H = 1.667 + 0.8335 = 2.5~\text{kW} \)
Last updated on Jul 15, 2025
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