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A refrigerating machine having a coefficient of performance equal to 2 is used to remove heat at the rate of 1200 kJ/min. What is the power required for this machine?
  • a)
    80 kW
  • b)
    60 kW
  • c)
    20 kW
  • d)
    10 kW
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A refrigerating machine having a coefficient of performance equal to 2...
Given data:
Coefficient of performance (COP) = 2
Heat removed rate = 1200 kJ/min

Calculating power required:
- The coefficient of performance (COP) of a refrigerating machine is given by COP = QL / W, where QL is the heat removed rate and W is the power required.
- Given COP = 2 and QL = 1200 kJ/min, we can rearrange the formula to find W: W = QL / COP
- W = 1200 kJ/min / 2 = 600 kJ/min
- To convert kJ/min to kW, we divide by 60 (1 min = 1/60 hour): W = 600 kJ/min / 60 = 10 kW
Therefore, the power required for this refrigerating machine is 10 kW, which corresponds to option 'D'.
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