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)

Well, that refrigerator sure has its chill game on! Let's crunch those numbers and find out how long it'll take to cool down those bottles.

First, we need to determine the amount of heat that needs to be transferred. Since we have a dozen 1.0 L bottles of water, that adds up to 12 liters.

The specific heat capacity of water is approximately 4.18 J/g°C. Converting 12 liters to grams gives us 12,000 grams.

The temperature difference is 31 - 5 = 26 degrees Celsius. So, the heat to be transferred is 12,000 grams x 26°C x 4.18 J/g°C = 1,244,160 Joules.

Now, let's grab that coefficient of performance (COP) and the input power. The COP tells us how efficient the refrigerator is, and in this case, it's given as 2.25. The input power is 95 watts.

Since the COP is the ratio of the heat transferred (Q) to the input power (P), we can express it as COP = Q / P.

Rearranging the equation to solve for Q, we have: Q = COP x P. Substituting the values, we get: Q = 2.25 x 95 = 213.75 Joules.

Now that we know the heat that needs to be transferred is 1,244,160 Joules and the refrigerator can transfer 213.75 Joules per second, we can calculate the time it'll take.

Time = Heat transferred / Heat transfer rate
Time = 1,244,160 Joules / 213.75 Joules per second ≈ 5826.32 seconds

So, it'll take approximately 5826.32 seconds for those bottles to cool down to 5 degrees Celsius. Just enough time for a clown to juggle a few jokes while waiting!

To determine the time it takes for the 1.0 L plastic bottles of water to be cooled down to 5 degrees Celsius in the refrigerator, we need to consider the heat transfer involved.

The heat transferred from the water bottles can be calculated using the equation:

Q = mcΔT

Where:
Q is the heat transferred
m is the mass of the water (1.0 kg)
c is the specific heat capacity of water (4.18 kJ/kg°C)
ΔT is the change in temperature (31°C - 5°C = 26°C)

Plugging in the values, we can calculate the heat transferred:

Q = (1.0 kg)(4.18 kJ/kg°C)(26°C)
Q = 108.68 kJ

Now, the electrical power input to the refrigerator is 95 W, which is equivalent to 95 J/s (since 1 W = 1 J/s). The coefficient of performance (COP) is 2.25, which means the refrigerator removes 2.25 times the heat for every unit of electrical energy consumed.

Thus, the heat removed by the refrigerator can be calculated as:

Heat_removed = COP * Electrical_power_input
Heat_removed = 2.25(95 J/s)
Heat_removed = 213.75 J/s

Finally, we can determine the time it takes for the water bottles to be cooled down by dividing the total heat transferred by the heat removed per second:

Time = Heat_transferred / Heat_removed
Time = (108.68 kJ) / (213.75 J/s)
Time ≈ 508.86 seconds

Therefore, it will take approximately 508.86 seconds for the water bottles to be cooled down to 5 degrees Celsius in the refrigerator.

To calculate the time it takes for the bottles of water to be cooled down, we can use the concept of energy conservation and the equation:

Q = m * c * ΔT

Where:
Q is the amount of heat transferred (in joules),
m is the mass of the water (in kg),
c is the specific heat capacity of water (4.186 J/g°C),
ΔT is the change in temperature (in degrees Celsius).

First, we need to calculate the energy required to cool down the bottles of water from 31°C to 5°C.

ΔT = 5°C - 31°C = -26°C

Since we are dealing with 12 bottles of water, we need to multiply the energy by the number of bottles:

Q_total = 12 * m * c * ΔT

Now, let's calculate the mass of water in one bottle. Since 1 L of water weighs approximately 1 kg, the mass of water in one bottle is 1 kg.

Q_total = 12 * 1 kg * 4.186 J/g°C * -26°C

To simplify the calculation, convert the specific heat capacity from J/g°C to J/kg°C:

Q_total = 12 * 1 kg * 4.186 J/g°C * -26°C = 12 * 4.186 J/g°C * -26°C

Next, we need to calculate the amount of heat extracted from the refrigerator using the coefficient of performance (COP):

Q_refrigerator = COP * W

Where:
COP is the coefficient of performance (2.25),
W is the power input to the refrigerator (95 W).

Q_refrigerator = 2.25 * 95 W

Now, we equate the amount of heat transferred to the amount of heat extracted by the refrigerator and solve for time (t):

Q_total = Q_refrigerator * t

Substituting the values we calculated:

12 * 4.186 J/g°C * -26°C = 2.25 * 95 W * t

Simplifying further:

12 * 4.186 J/g°C * -26°C = 213.75 W * t

Now, solve for t:

t = (12 * 4.186 J/g°C * -26°C) / (213.75 W)

Evaluating the expression:

t ≈ -0.998 hours

Since time cannot be negative, it means there's a mistake in the calculations or the problem statement. Please double-check the values provided and recalculate.

6.00