Question

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)

Answer 1

COP = 0.090

Step-by-step 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