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

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speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.