Can you help me please?

Question

A freezer maintains an interior temperature inside of −18.0°C and has a coefficient of performance of 3.00. The freezer sits in a room with a temperature of 24.0°C. The freezer is able to completely convert 22.0 g of liquid water at 24.0°C to ice at −18.0°C in one minute. What input power (in watts) does the freezer require? (The specific heat of liquid water is 4.186 J/(g · °C), the specific heat of ice is 2.090 J/(g · °C), and the latent heat of fusion of water is 334 J/g.)

What If? In reality, only part of the power consumption of a freezer is used to make ice. The remainder is used to maintain the temperature of the rest of the freezer. Suppose, however, that 100% of a freezer's typical power consumption of 160 W is available to make ice. The freezer has the same coefficient of performance as given above. How many grams per minute of water at 24.0°C could this freezer convert to ice at −18.0°C?

Answer 1

The coefficient of performance (COP) of a refrigerator is given by the equation:

COP = QL/W

where QL is the amount of heat removed from the cold reservoir (the freezer in this case) and W is the work done by the refrigerator.

The heat removed from the cold reservoir is the heat required to freeze the water. We can calculate the heat required as follows:

Q = (mass of water) x (heat of fusion) + (mass of ice) x (specific heat of ice) x (final temperature of ice)

Here, mass of ice is the same as mass of water since all the water will be frozen into ice.

Q = (22 g) x (334 J/g) + (22 g) x (2.090 J/(g · °C)) x (-18.0°C)

Q = -7357.48 J (note that the negative sign indicates heat removal)

Now, we can calculate the COP:

COP = QL/W = Q/(W + Q)

Since the problem mentions that the COP is 3.00, we can use this to solve for W:

3.00 = -7357.48 J / (W - 7357.48 J)

W = 2452.49 J

Now, we can calculate the input power required by the freezer:

P = W / t

where t is the time taken to freeze the water, which is given as 1 minute.

P = 2452.49 J / 60 s

P = 40.87 W

Therefore, the input power required by the freezer is 40.87 W.

For the second part of the question, if 100% of the power consumption of the freezer is available to make ice, then the input power available for freezing water is 160 W.

Using the same method as before, we can calculate the heat required to freeze the water as:

Q = (mass of water) x (heat of fusion) + (mass of ice) x (specific heat of ice) x (final temperature of ice)

Q = m x (334 J/g) + m x (2.090 J/(g · °C)) x (-18.0°C)

Q = -37.26 m J (note that the negative sign indicates heat removal)

Setting COP = 3.00 as before, we have:

3.00 = -37.26 m J / (160 W - 37.26 m J)

Solving for m, we get:

m = 0.15 g/min

Therefore, if 100% of the power consumption of the freezer is available for making ice, the freezer can convert 0.15 g/min of water at 24.0°C to ice at −18.0°C.