Discussion
A reversed carnot cycle refrigerator maintains a temperature of -5°C.The ambient air temperature is 35°C.The heat gained by the refrigerator at a continuous rate is 2.5 kJ/sec.What will be (COP) of refrigerator?
Answer: A
The COP is 6.7.
For a reversed Carnot refrigerator, with \(T_{L} = -5 + 273 = 268\) K and \(T_{H} = 35 + 273 = 308\) K:
\[ \begin{aligned} COP &= \frac{T_{L}}{T_{H} - T_{L}} \\ &= \frac{268}{40} = 6.7 \end{aligned} \]
The 2.5 kJ/s heat load is not needed for the COP; it would be used to find the power input, \(2.5/6.7 = 0.37\) kW. The nearby values 6.6, 6.5 and 6.4 come from rounding or from using 273.15 inconsistently.
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