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.)

Answer 1

The coefficient of performance of the refrigerator is 2.25. The heat extracted from the cold reservoir is 303.75 J/s. The time it takes to cool the water to 5°C is 12.48 seconds.

Step-by-step explanation:

The coefficient of performance (COP) of a refrigerator is given by the equation COP = Qc/W, where Qc is the amount of heat extracted from the cold reservoir and W is the work done by the refrigerator. In this case, the COP is 2.25. Since the work done by the refrigerator is 135 W, we can use the equation COP = Qc/W to find Qc. Rearranging the equation, we have Qc = COP * W = 2.25 * 135 = 303.75 J/s.

The rate of heat transfer is given by the equation Qc = m * c * (Tc - Th), where m is the mass of the water, c is the specific heat capacity of water, Tc is the initial temperature of the water, and Th is the temperature of the refrigerator. In this case, there are a dozen 1.0-L (1000 mL) bottles of water, so the mass is 12 * 1000 g = 12000 g = 12 kg. The specific heat capacity of water is approximately 4.18 J/g°C. The initial temperature of the water is 31°C, and the temperature of the refrigerator is 5°C. Plugging these values into the equation, we get Qc = 12 * 4.18 * (31 - 5) = 1685.92 J.

Finally, to find the time it takes for the water to be cooled down to 5°C, we can use the equation Qc = P * t, where P is the power input to the refrigerator and t is the time. Rearranging the equation, we have t = Qc / P = 1685.92 / 135 = 12.48 seconds.

Answer 2

To calculate the time it takes for 12 bottles of water to cool down in the refrigerator, we need to calculate the amount of heat energy that needs to be removed. By using the formula Q=m*C*ΔT, where Q is the heat energy, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature, we determine that each bottle requires 108,836 J of heat energy to be removed. Multiplying this by the number of bottles gives a total of 1,306,032 J. Dividing this by the power consumed by the refrigerator, which is 135 W, we find that it takes approximately 9679.5 seconds for the bottles to cool down.

Step-by-step explanation:

To determine how long it will take for the bottles of water to cool down in the refrigerator, we need to calculate the amount of heat energy that needs to be removed from the bottles. The formula to calculate the amount of heat energy is Q=m*C*ΔT, where Q is the heat energy, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Given that each bottle has a mass of 1.0 kg (since 1 liter of water has a mass of 1 kg) and the initial temperature of the bottles is 31°C, the change in temperature is ΔT = (31°C - 5°C) = 26°C. We also know that the specific heat capacity of water is approximately 4,186 J/(kg·°C).

Therefore, the amount of heat energy that needs to be removed from each bottle is Q = (1 kg) * (4,186 J/(kg·°C)) * (26°C) = 108,836 J.

Now, let's calculate the total amount of heat energy that needs to be removed from all 12 bottles. Since each bottle needs to have 108,836 J of heat energy removed, the total amount of heat energy is 12 * 108,836 J = 1,306,032 J.

Lastly, we can find the time it takes to remove this amount of heat energy by dividing it by the power consumed by the refrigerator. The power is given as 135 W. Therefore, the time it takes to cool down the bottles is 1,306,032 J / 135 W = 9679.5 seconds.