A refrigerator has a coefficient of performance of 2.25, runs on an input of 95 W of electrical power, and keeps its inside compartment at 5 degrees celsius. IF you put a dozen 1.0 L plastic bottles of water at 31 degrees celsius in the refrigerator, how long will it take for them to be cooled down to 5 degrees celsius? (ignore any heat that leaves the plastic)

To calculate the time it takes for the bottles of water to be cooled down, we need to use certain formulas and equations related to thermodynamics.

First, we need to calculate the amount of heat that needs to be transferred from the bottles of water to the refrigerator compartment.

The amount of heat transferred can be calculated using the equation:

Q = m * c * ΔT

Where:
Q is the amount of heat transferred
m is the mass of water
c is the specific heat capacity of water
ΔT is the change in temperature

In this case, we have 12 bottles of 1.0 L water, which is equal to 12 kg of water (assuming water density of 1 kg/L). The specific heat capacity of water is approximately 4.18 J/g°C.

Before calculating ΔT, we need to convert the temperatures from Celsius to Kelvin, as temperature differences should be in Kelvin.

Given that the initial temperature of the water is 31°C (which is 31 + 273.15 = 304.15 K) and the final temperature is 5°C (which is 5 + 273.15 = 278.15 K), we can calculate ΔT:

ΔT = T_final - T_initial
ΔT = 278.15 K - 304.15 K = -26 K

Now, we can calculate the amount of heat that needs to be transferred:

Q = m * c * ΔT
Q = 12 kg * 4.18 J/g°C * (-26 K) = -1275.36 kJ

Next, we need to calculate the work done by the refrigerator to extract this amount of heat. The work done can be calculated using the equation:

W = Q / COP

Where:
W is the work done
Q is the amount of heat transferred
COP is the coefficient of performance of the refrigerator

In this case, the coefficient of performance (COP) is given as 2.25.

W = -1275.36 kJ / 2.25
W = -566.93 kJ

The negative sign indicates that work is done on the system (refrigerator) to remove the heat.

Now, we have the amount of work done by the refrigerator. To find the time required, we can use the formula:

Power = Work / Time

Given that the electrical power input to the refrigerator is 95 W, we have:

Power = 95 W
Work = -566.93 kJ = -566.93 * 1000 J

Now, we can rearrange the equation to solve for time:

Time = Work / Power
Time = (-566.93 * 1000 J) / 95 W

Evaluating this expression, we can calculate the time required for the bottles of water to cool down to 5°C.