A freezer has a coefficient of performance of 6.30. It is advertised as using 486 kWh/yr. Note: One kilowatt-hour (kWh) is an amount of energy equal to running a 1-kW appliance for one hour.

a) On average, how much energy does it use in a single day?
I got 4.79 x 10^6 J

(b) On average, how much energy does it remove from the refrigerator in a single day?
I got 30177000 J

Now I'm stuck on this part:
What maximum mass of water at 17.1°C could the freezer freeze in a single day? (The latent heat of fusion of water is 3.33 x 10^5 J/kg, and its specific heat is 4186 J/kg · °C.) answer in kg

Compute the amount of heat that must be removed to freeze one kg of water that is initially 17.1 C. Assume the ice temperature stays at 0 C, once it is formed (although this is not necessarily the case)

That would be 3.35*10^5 J + 17.1*4186 = 4.046*10^5 J/kg

Finally, divide your computed heat removal by that heat per mass number.

Thank you!

To find the maximum mass of water the freezer could freeze in a single day, we need to determine the amount of energy transferred during the freezing process.

Step 1: Calculate the energy removed from the refrigerator in a single day in joules.
From part (b), we found that the energy removed is 30,177,000 J per day.

Step 2: Determine the mass of water that could be frozen with this amount of energy.
The latent heat of fusion (L) is the energy required to change a substance from a solid to a liquid, or vice versa. It is given as 3.33 x 10^5 J/kg.

The energy required to freeze a given mass (m) of water can be calculated using the formula:
Energy = mass x latent heat of fusion

Given that the energy removed from the refrigerator is 30,177,000 J and the latent heat of fusion is 3.33 x 10^5 J/kg, we can solve for the mass (m):
30,177,000 J = m x 3.33 x 10^5 J/kg

To solve for the mass, divide both sides of the equation by the latent heat of fusion:
m = 30,177,000 J / (3.33 x 10^5 J/kg)

m = 90.79 kg

Therefore, the maximum mass of water that the freezer could freeze in a single day is 90.79 kg.

To find the maximum mass of water the freezer can freeze in a single day, we need to calculate the total heat energy that the freezer can remove from the water.

The formula for calculating the heat energy required to freeze a substance is:

Q = m * L

Where:
Q is the heat energy (in joules)
m is the mass of the substance (in kg)
L is the latent heat of fusion (in J/kg)

In this case, we want to find the maximum mass of water that can be frozen in a day by the freezer. We'll assume that the temperature of the water starts at the freezing point (0°C) and needs to be lowered to its freezing point.

First, we need to calculate the amount of energy required to lower the temperature of the water to 0°C. The formula for calculating this is:

Q1 = m * c * ΔT

Where:
Q1 is the heat energy (in joules)
m is the mass of the water (in kg)
c is the specific heat capacity of water (in J/kg · °C)
ΔT is the change in temperature (in °C)

In this case, the initial temperature of the water is 17.1°C, and we need to lower it to 0°C. Using the specific heat capacity of water, which is 4186 J/kg · °C, we can calculate Q1:

Q1 = m * 4186 * (0 - 17.1)

Next, we need to calculate the heat energy required to freeze the water at 0°C, which is given by:

Q2 = m * L

Where:
Q2 is the heat energy (in joules)
m is the mass of the water (in kg)
L is the latent heat of fusion of water (in J/kg)

Using the values provided, we can calculate Q2:

Q2 = m * 3.33 x 10^5

Now, the total heat energy removed by the freezer in a day is the sum of Q1 and Q2:

Q_total = Q1 + Q2

Since the freezer has a coefficient of performance (COP) of 6.30, it removes Q_total amount of heat energy from the refrigerator. The COP is defined as the ratio of useful energy output to the energy input. In this case, the useful energy output is the heat energy removed from the refrigerator, and the energy input is the electrical energy consumed by the freezer.

Given that the freezer uses 486 kWh/yr, we can convert this to joules:

Energy_input = 486 kWh * (3.6 x 10^6 J/kWh)

Now, using the formula for the coefficient of performance:

COP = Q_total / Energy_input

We can solve this equation for Q_total:

Q_total = COP * Energy_input

Now that we have Q_total, we can substitute it back in the equation Q_total = Q1 + Q2 and solve for the mass of water (m):

m = Q_total / (Q_total / Q2 + Q2 / Q1)

Substituting the values we have:

m = Q_total / (Q_total / (m * 3.33 x 10^5) + (m * 3.33 x 10^5) / (m * 4186 * (0 - 17.1)))

Now, we can simplify and solve for m.