A refrigerator has a coefficient of performance of 2.25, runs on an input of 135 W of electrical power, and keeps its inside compartment at 5°C. If you put a dozen 1.0 L plastic bottles of water at 31°C into this refrigerator, how long will it take for them to be cooled down to 5°C? (Ignore any heat that leaves the plastic.). 1 calorie= 4.186J/(g⋅K)

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Answer:t = 4299.7 s = 1.19 h

Explanation:First we find the cooling rate of the refrigerator:

COP = Cooling Rate/Electric InputCooling Rate = (COP)(Electric Input)Cooling Rate = (2.25)(135 W)Cooling Rate = 303.75 WNow, we find the energy required to be removed to cool the water in bottles:

E = m C ΔTwhere,

E = Energy required to be removed = ?

m = mass of water = Vρ = (12 x 1 L)(1 kg/L) = 12 kg

C = Specific heat of water = 4186 J/kg °C

ΔT = Change in Temperature = 31°C – 5°C = 26°C

Therefore,

E = (12 kg)(4186 J/kg °C)(26 °C)E = 1306032 JNow, the time required to cool the water is given as:

Cooling Rate = E/tt = E/Cooling Ratet = 1306032 J/303.75 Wt = 4299.7 s = 1.19 h