EFFECT}{Work\ Input} = \frac{T_L}{T_H\;-\;T_L}\)where TH = ABSOLUTE higher temperature, TL = Absolute lower temperatureCalculation:Given:TH = 27°C = 300 K, TL = -23°C = 250 K, Refrigeration Effect = 2.5 kJ/s.\(COP = \frac{250}{300\;-\;250}\)\(COP = \frac{250}{50} = 5\)Now, \(COP=\frac{Refrigeration\ Effect}{Work\ Input}\)\(5=\frac{2.5}{Work\ Input}\)\(Work\ Input=\frac{2.5}{5}\)Work Input = 0.5 kWHence the least power necessary to PUMP this heat out CONTINUOUSLY is 0.5 kW.