Discussion
The COP of a heat pump operating between 6°C and 37°C will be:
Answer: B
The COP is 10.
For a reversible heat pump, with temperatures in kelvin, \(T_{H} = 37 + 273 = 310\) K and \(T_{L} = 6 + 273 = 279\) K:
\[ \begin{aligned} COP_{HP} &= \frac{T_{H}}{T_{H} - T_{L}} \\ &= \frac{310}{31} = 10 \end{aligned} \]
Using the refrigerator formula \(T_{L}/(T_{H} - T_{L}) = 279/31 = 9\) gives one less, because \(COP_{HP} = COP_{R} + 1\). Using degrees Celsius instead of kelvin gives nonsense values. 8, 12 and 14 do not come from the correct temperatures.
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