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)
Question
Answer:
t = 4299.7 s = 1.19 h
Explanation:
First we find the cooling rate of the refrigerator:
COP = Cooling Rate/Electric Input
Cooling Rate = (COP)(Electric Input)
Cooling Rate = (2.25)(135 W)
Cooling Rate = 303.75 W
Now, we find the energy required to be removed to cool the water in bottles:
E = m C ΔT
where,
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 J
Now, the time required to cool the water is given as:
Cooling Rate = E/t
t = E/Cooling Rate
t = 1306032 J/303.75 W
t = 4299.7 s = 1.19 h