Question

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)

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