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A certain brand of freezer is advertised to use 730 \(\mathrm{kW} \cdot \mathrm{h}\) of energy per year. (a) Assuming the freezer operates for 5 hours each day, how much power does it require while operating? (b) If the freezer keeps its interior at a temperature of \(-5.0^{\circ} \mathrm{C}\) in a \(20.0^{\circ} \mathrm{C}\) room, what is its theoretical maximum performance coefficient? (c) What is the theoretical maximum amount of ice this freezer could make in an hour, starting with water at \(20.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
a) 0.4 kW; b) 10.726; c) 37 kg ice/hour.

Step by step solution

01

Calculate Daily Energy Consumption

The freezer uses 730 kWh per year. Given there are 365 days in a year, the daily energy consumption is calculated as \( \frac{730 \text{ kWh}}{365 \text{ days}} = 2 \text{ kWh/day} \).
02

Determine Power Requirement

The freezer operates for 5 hours each day. Therefore, the power requirement while operating is \( \frac{2 \text{ kWh}}{5 \text{ hours}} = 0.4 \text{ kW} \).
03

Calculate Temperature Difference for Coefficient of Performance

The interior of the freezer is maintained at \(-5.0^{\circ} \mathrm{C}\) and the room temperature is \(20.0^{\circ} \mathrm{C}\). Hence, the temperature difference is \(20.0 - (-5.0) = 25.0^{\circ} \mathrm{C}\).
04

Convert Temperatures to Kelvin

Convert the temperatures from Celsius to Kelvin: \(-5.0^{\circ} \mathrm{C} = 268.15 \mathrm{K}\) and \(20.0^{\circ} \mathrm{C} = 293.15 \mathrm{K}\).
05

Calculate Coefficient of Performance

The theoretical maximum performance coefficient (CoP) for a Carnot refrigerator is given by \( \mathrm{CoP} = \frac{T_c}{T_h - T_c} \), where \( T_c = 268.15 \mathrm{K} \) and \( T_h = 293.15 \mathrm{K} \). So, \( \mathrm{CoP} = \frac{268.15}{293.15 - 268.15} = \frac{268.15}{25.0} = 10.726 \).
06

Convert Ice Cooling and Freezing to Energy Needed

To find the energy to cool and freeze water, use the specific heat of water (4.18 J/g°C) and latent heat of fusion (334 J/g). For 1 kg (1000 g) of water to cool from 20°C to 0°C: \( Q = m \, c \, \Delta T = 1000 \, \times \, 4.18 \, \times \, 20 = 83600 \, \mathrm{J} \). For freezing: \( Q = m \, \times \, 334 \, = 334000 \, \mathrm{J} \). Total energy: \( 83600 \,+ \, 334000 = 417600 \, \mathrm{J} \).
07

Calculate Maximum Ice Production

Power used in an hour is \(0.4 \times 3600 \, \mathrm{J/s} = 1440000 \, \mathrm{J} \). Use CoP to find heat removed: \( Q_{removed} = CoP \times P \space t = 10.726 \times 1440000 = 15445440 \, \mathrm{J} \). Amount of ice: \( \frac{15445440}{417600} = 37 \, \mathrm{kg} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carnot Refrigerator
A Carnot refrigerator is a theoretical device that operates on the Carnot cycle, serving as the most efficient refrigeration cycle possible according to the second law of thermodynamics. While no real refrigerator can achieve this perfect efficiency, it acts as a useful benchmark.

The Carnot cycle is made up of two isothermal processes and two adiabatic processes. The cycle comprises the expansion and compression of the refrigerant in a manner that eliminates any net work or heat flow over one complete cycle. In a refrigeration context, these processes work to remove heat from a cold reservoir and expel it to a hot reservoir.

A Carnot refrigerator, therefore, represents an ideal scenario: it removes the maximum amount of heat with the minimum amount of work required, maximizing the efficiency of the refrigeration process. The performance of any real refrigerator can be compared to the Carnot refrigerator through its Coefficient of Performance (COP).
Coefficient of Performance
The coefficient of performance (COP) of a refrigerator is a measure of its efficiency. It tells us how much heat is removed from the refrigerated space compared to the work input required to remove it. In a sense, it shows how effectively a refrigerator uses energy.

The formula for the COP of an ideal Carnot refrigerator is given by:\[ \text{COP} = \frac{T_c}{T_h - T_c} \]where \( T_c \) is the absolute temperature of the cold reservoir and \( T_h \) is the absolute temperature of the hot reservoir, both in Kelvin.

For a higher COP, the temperature difference between \( T_h \) and \( T_c \) should be minimized. Thus, refrigerators work more efficiently when the temperature of the refrigerated space is not drastically lower than the ambient temperature.

In practical terms, a higher COP means less energy is required to achieve the cooling effect, which translates to lower operating costs and less environmental impact. Understanding COP offers insights into optimizing refrigeration systems for better performance and energy conservation.
Latent Heat of Fusion
Latent heat of fusion is the amount of energy required to change a substance from the solid phase to the liquid phase without changing its temperature. For water, this energy is significant because it marks the phase change from solid ice to liquid water.

To calculate the energy necessary to freeze water and turn it into ice, we must account for the specific heat of water, which is the energy required to change the temperature of water before freezing, and the latent heat of fusion. The specific heat of water is 4.18 J/g°C, and the latent heat of fusion for ice is 334 J/g.
To convert 1 kg of water at \(20^{\circ}\, \mathrm{C}\) to ice at \(0^{\circ}\, \mathrm{C}\), we first cool the water from \(20^{\circ}\, \mathrm{C}\) to \(0^{\circ}\, \mathrm{C}\):\[ Q_1 = m \times c \times \Delta T = 1000 \times 4.18 \times 20 = 83600 \, \mathrm{J} \]

Then, the energy required to change the phase from water to ice is:\[ Q_2 = m \times \text{latent heat of fusion} = 1000 \times 334 = 334000 \, \mathrm{J} \]

Thus, the total energy needed is \(417600\, \mathrm{J}\), illustrating the substantial amount of energy required for phase change compared to simply cooling the water down.

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Most popular questions from this chapter

As a budding mechanical engineer, you are called upon to design a Carnot engine that has 2.00 mol of a monatomic ideal gas as its working substance and operates from a hightemperature reservoir at \(500^{\circ} \mathrm{C}\) . The engine is to lift a 15.0 -kg weight 2.00 \(\mathrm{m}\) per cycle, using 500 \(\mathrm{J}\) of heat input. The gas in the engine chamber can have a minimum volume of 5.00 \(\mathrm{L}\) during the cycle. (a) Draw a \(p V\) -diagram for this cycle. Show in your diagram where heat enters and leaves the gas. (b) What must be the temperature of the cold reservoir? (c) What is the thermal efficiency of the engine? (d) How much heat energy does this engine waste per cycle? (e) What is the maximum pressure that the gas chamber will have to withstand?

Entropy Change from Digesting Fat. Digesting fat produces 9.3 food calories per gram of fat, and typically 80\(\%\) of this energy goes to heat when metabolized. (One food calorie is 1000 calories and therefore equals 4186 \(\mathrm{J} . )\) The body then moves all this heat to the surface by a combination of thermal conductivity and motion of the blood. The internal temperature of the body (where digestion occurs) is normally \(37^{\circ} \mathrm{C},\) and the surface is usually about \(30^{\circ} \mathrm{C} .\) By how much do the digestion and metabolism of a \(2.50-\mathrm{g}\) pat of butter change your body's entropy? Does it increase or decrease?

An experimental power plant at the Natural Energy Laboratory of Hawaii generates electricity from the temperature gradient of the ocean. The surface and deep-water temperatures are \(27^{\circ} \mathrm{C}\) and \(6^{\circ} \mathrm{C},\) respectively. (a) What is the maximum theoretical efficiency of this power plant? (b) If the power plant is to produce 210 \(\mathrm{kW}\) of power, at what rate must he extracted from the warm water? At what rate must heat be absorbed by the cold water? Assume the maximum theoretical efficiency. (c) The cold water that enters the plant leaves it at a temperature of \(10^{\circ} \mathrm{C}\) . What must be the flow rate of cold water through the system? Give your answer in \(\mathrm{kg} / \mathrm{h}\) and in \(\mathrm{L} / \mathrm{h}\) .

You decide to take a nice hot bath but discover that your thoughtless roommate has used up most of the hot water. You fill the tub with 270 \(\mathrm{kg}\) of \(30.0^{\circ} \mathrm{C}\) water and attempt to warm it further by pouring in 5.00 \(\mathrm{kg}\) of boiling water from the stove. (a) Is this a reversible or an irreversible process? Use physical reasoning to explain. (b) Calculate the final temperature of the bath water. (c) Calculate the net change in entropy of the system (bath water \(+\) boiling water), assuming no heat exchange with the air or the tub itself.

A \(10.0\) -L gas tank containing 3.20 moles of ideal He gas at \(20.0^{\circ} \mathrm{C}\) is placed inside a completely evacuated, insulated bell jar of volume 35.0 L. A small hole in the tank allows the He to leak out into the jar until the gas reaches a final equilibrium state with no more leakage. (a) What is the change in entropy of this system due to the leaking of the gas? (b) Is the process reversible or irreversible? How do you know?

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