The wattage of a commercial ice maker is 226 W and is the rate at which it does work. The ice maker operates just like a refrigerator and has a coefficient of performance of 3.90. The water going into the unit has a temperature of 15.0°C, and the ice maker produces ice cubes at 0.0°C. Ignoring the work needed to keep stored ice from melting, find the maximum amount (in kg) of ice that the unit can produce in one day of continuous operation. Water has a specific heat capacity 4186 J/(kg·C°) and a latent heat of fusion of 3.35x105 J/kg

Question

The wattage of a commercial ice maker is 226 W and is the rate at which it does work. The ice maker operates just like a refrigerator and has a coefficient of performance of 3.90. The water going into the unit has a temperature of 15.0°C, and the ice maker produces ice cubes at 0.0°C. Ignoring the work needed to keep stored ice from melting, find the maximum amount (in kg) of ice that the unit can produce in one day of continuous operation. Water has a specific heat capacity 4186 J/(kg·C°) and a latent heat of fusion of 3.35x105 J/kg

Answer 1

To find the maximum amount of ice that the unit can produce in one day of continuous operation, we need to use the concept of heat transfer.

First, let's find the heat absorbed by the ice maker from the water to produce ice. The heat transfer formula is:

Heat absorbed = Mass of water * Specific heat capacity of water * (Final temperature - Initial temperature)

In this case, the initial temperature of the water is 15.0°C, and the final temperature is 0.0°C. The specific heat capacity of water is 4186 J/(kg·C°).

Heat absorbed = Mass of water * 4186 J/(kg·C°) * (0.0°C - 15.0°C)

Now, we need to convert the heat absorbed to the heat released by the ice maker, which is the work done by the ice maker.

Given that the ice maker has a power consumption of 226 W, we can use the formula for power:

Power = Work done / Time

Rearranging the formula, we get:

Work done = Power * Time

Since we are interested in one day of continuous operation, the time is 24 hours or 86400 seconds.

Work done = 226 W * 86400 s

The coefficient of performance (COP) of the ice maker is given as 3.90, which is the ratio of heat absorbed to work done:

COP = Heat absorbed / Work done

Rearranging the formula, we get:

Heat absorbed = COP * Work done

Now, we can equate the two formulas for heat absorbed and solve for the mass of water:

Mass of water * 4186 J/(kg·C°) * (0.0°C - 15.0°C) = COP * Work done

Mass of water = (COP * Work done) / (4186 J/(kg·C°) * (0.0°C - 15.0°C))

Now, the final step is to convert the mass of water to the mass of ice. Water has a latent heat of fusion of 3.35x10^5 J/kg, which is the amount of energy needed to convert water from a liquid to a solid (ice) at its melting point.

Mass of ice = Mass of water * (1 latent heat of fusion / specific heat capacity of water)

Mass of ice = Mass of water * (3.35x10^5 J/kg / 4186 J/(kg·C°))

By substituting the values and solving the equation, we can find the maximum amount of ice that the unit can produce in one day of continuous operation.