A refrigerator consumes 14 kW of electricity and has a COP o

A refrigerator consumes 1.4 kW of electricity and has a COP of 1.5. How long will it take to cool 50 kg of bananas from 24 to 13°C? Assume the specific heat of the bananas is 4 kJ/kgK. Is this a best-case or worst-case approximation? Why?

Solution

COP K = 1.5

Mass of banana m = 50 kg

temprature decrease dT = 24 - 13 =11 ^o C

the specific heat of the bananas C= 4 kJ/kgK

                                                 = 4000 J / kg K

Heat removed Q \' = mC dT

                         = 50 x 4000x11

                         = 2.2 x10 ⁶ J

We know K = Q/W

From this Q = KW

                  = 1.5 x 1.4 kilo watt

                  = 1.5 x1.4 x10 ³ watt

                  = 2.1 x10 ³ watt

Required time t = Q \' / Q

                        = (2.2x10 ⁶)/(2.1x10 ³)

                        = 1047.6 s

                        = 1047.6/60 min

                         = 17.46 min

time taken to cool the bananas is good

So, it is best case.