A thermoelectric refrigerator is powered by a 16-V power supply that draws 2.9 A of current. If the refrigerator cools down 3.1 kg of water from 25 °C to 11 °C in 12 hours, what is the average COP of the refrigerator? Take the specific heat of water as 4180 J/kg. K. (3 decimal digits)

COP = 0.090

Explanation:

The general formula for COP is:

COP = Desired Output/Required Input

Here,

Desired Output = Heat removed from water while cooling

Desired Output = (Specific Heat of Water) (Mass of Water) (Change in Temperature) / Time

Desired Output = [ (4180 J/kg. k) (3.1 kg) (25 - 11) k]/[ (12 hr) (3600 sec/hr) ]

Desired Output = 4.199 W

And the required input can be given as electrical power:

Required Input = Electrical Power = (Current) (Voltage)

Required Input = (2.9 A) (16 V) = 46.4 W

Therefore:

COP = 4.199 W/46.4 W

COP = 0.090