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
Qc = m (3.35*10^5 + 15*4186) = heat removed from water in a day
W = 226 J/s * 24 * 3600 = 1.95*10^7 Joules per day of electricity or work in
COP = Qc/W
3.9 = m (397340) / 19500000
m = 191 Kg
To find the maximum amount of ice that the unit can produce in one day of continuous operation, we need to consider the energy transfer involved in the process.
First, let's calculate the energy required to freeze the water from temperature 15.0°C to 0.0°C. The formula for calculating energy is:
Q = mcΔT
Where:
Q = energy transfer (in Joules)
m = mass of water (in kg)
c = specific heat capacity of water (4186 J/(kg·C°))
ΔT = change in temperature (in °C)
In this case, the change in temperature is 15.0°C (from 15.0°C to 0.0°C). Assuming a mass of 1 kg of water, the energy required would be:
Q = (1 kg) * (4186 J/(kg·C°)) * (15.0°C)
Q = 62790 J
Next, let's calculate the energy required for the phase change from water to ice. The formula for energy in this case is:
Q = mL
Where:
Q = energy transfer (in Joules)
m = mass of water (in kg)
L = latent heat of fusion of water (3.35x10^5 J/kg)
Assuming we have 1 kg of water, the energy required for the phase change would be:
Q = (1 kg) * (3.35x10^5 J/kg)
Q = 3.35x10^5 J
Now, let's calculate the total energy required to freeze 1 kg of water:
Total energy = energy to change temperature + energy for phase change
Total energy = 62790 J + 3.35x10^5 J
Total energy = 4.02x10^5 J
Since the coefficient of performance (COP) of the ice maker is given as 3.90, we can calculate the input energy required for freezing 1 kg of water:
Input energy = Total energy / COP
Input energy = 4.02x10^5 J / 3.90
Input energy ≈ 1.03x10^5 J
The given wattage of the ice maker is 226 W, which represents the rate at which it does work. To find the time required to freeze 1 kg of water, we can use the formula:
Time = Energy / Power
Time = 1.03x10^5 J / 226 W
Time ≈ 456 seconds
Now, to find the maximum amount of ice that the unit can produce in one day of continuous operation, we need to calculate the number of 1 kg water batches that can be frozen in 24 hours:
Number of batches = 24 hours / (Time to freeze 1 kg of water)
Number of batches = 24 hours / (456 seconds)
Number of batches ≈ 198
Therefore, the maximum amount of ice that the unit can produce in one day of continuous operation is 198 kg.