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In one cycle, a freezer uses \(785 \mathrm{~J}\) of electrical energy in order to remove \(1750 \mathrm{~J}\) of heat from its freezer compartment at \(10^{\circ} \mathrm{F}\). (a) What is the coefficient of performance of this freezer? (b) How much heat does it expel into the room during this cycle?

Short Answer

Expert verified
(a) The COP is 2.23. (b) The heat expelled into the room is 2535 J.

Step by step solution

01

Understanding the Coefficient of Performance (COP)

The coefficient of performance for a refrigeration cycle is calculated using the formula: \( \text{COP} = \frac{Q_L}{W} \), where \( Q_L \) is the heat removed from the low temperature reservoir (freezer compartment), and \( W \) is the work done (electrical energy used).
02

Calculate the Coefficient of Performance (COP)

Given \( Q_L = 1750 \mathrm{~J} \) and \( W = 785 \mathrm{~J} \), substitute these values into the formula: \( \text{COP} = \frac{1750}{785} \approx 2.23 \).
03

Understanding Heat Expelled into the Room

The heat expelled into the room, \( Q_H \), can be calculated using the relationship \( Q_H = Q_L + W \). This accounts for the energy used to remove the heat and the work done by the system.
04

Calculate the Heat Expelled into the Room

Substitute the given values into the equation: \( Q_H = 1750 \mathrm{~J} + 785 \mathrm{~J} = 2535 \mathrm{~J} \). This is the total heat expelled into the room during one cycle.

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Key Concepts

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

Refrigeration Cycle
The Refrigeration Cycle is the process used by refrigerators, air conditioners, and freezers to transfer heat from a cooler area (such as inside a freezer) to a warmer one (like a room). This is contrary to natural heat flow, which goes from warm to cool. In the cycle, mechanical work is applied to move heat in the desired direction.
In simple terms, a refrigeration cycle uses mechanical processes to absorb heat from inside the freezer compartment and release it into the surrounding environment. This creates a cooling effect. The process involves compression, condensation, expansion, and evaporation. These steps allow the appliance to maintain a cold interior temperature.
  • **Compression:** Increases the pressure and temperature of the refrigerant gas.
  • **Condensation:** The heated gas transfers heat to the air outside, becoming a liquid.
  • **Expansion:** The liquid refrigerant travels through an expansion valve and its pressure decreases.
  • **Evaporation:** The low-pressure liquid absorbs heat from the freezer, becoming a gas.
This cycle repeats continuously to keep the freezer at the desired low temperature.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat, work, energy, and how these quantities interact in systems. It provides the principles underlying the concepts of temperature and energy transfer. The laws of thermodynamics describe the movement of energy in a system:
  • **First Law of Thermodynamics:** Energy cannot be created or destroyed, only transformed. This means the total energy of an isolated system is conserved.
  • **Second Law of Thermodynamics:** Heat naturally moves from a hot object to a cold one, not vice versa. Refrigerators, therefore, require work to reverse this natural flow.
In the context of refrigeration, these laws explain why electrical energy is needed to transfer heat from inside the freezer to the outside. The First Law ensures that the energy used (electrical work) plus the heat removed from inside must equal the total heat expelled to the surroundings. The Second Law explains why work must be done to force heat to move in the unnatural direction.
Heat Transfer
Heat transfer plays a crucial role in the refrigeration process. It involves the movement of thermal energy from one place to another. This movement occurs through three main mechanisms: conduction, convection, and radiation.
  • **Conduction:** The transfer of heat through materials. In a fridge, this can happen when thermal energy moves through the freezer walls.
  • **Convection:** Involves the motion of fluid or air to transfer heat. The refrigerator's airflow helps equalize temperatures.
  • **Radiation:** Transfer of energy through electromagnetic waves. It's less significant in typical household refrigeration systems but can still occur.
In a refrigeration cycle, the primary concern is the efficient absorption and expulsion of heat. The evaporator coils inside absorb heat from the freezer's interior. This process brings down the temperature, allowing food to stay frozen. When the refrigerant exits the freezer section, it moves to the condenser coils, where it releases absorbed heat to the exterior. The efficiency of these processes significantly determines how well a refrigeration cycle functions.
Energy Efficiency
Energy efficiency is about using less energy to perform the same task. In terms of refrigeration, this means achieving lower temperatures with minimal energy use. A key measure of a refrigeration system’s efficiency is the Coefficient of Performance (COP).The COP measures how effectively a refrigeration system uses electrical energy to transfer heat. It's defined by the formula: \[ \text{COP} = \frac{Q_L}{W} \]where \(Q_L\) is the heat removed from the freezer and \(W\) is the work or electrical energy used.A higher COP indicates a more efficient system. For example, if a refrigerator has a COP of 2.23, as calculated in the original problem, it means that for every unit of energy consumed (785 J), it removes 2.23 units of heat (1750 J).In addition to the COP, other factors affect energy efficiency:
  • **Insulation:** Better insulation reduces unwanted heat transfer, needing less energy to maintain low temperatures.
  • **Compressor Efficiency:** A more efficient compressor reduces energy consumption.
  • **Seal Quality:** Ensures that the cool air stays inside and warm air remains outside.
By understanding and improving these aspects, a refrigeration system can operate more efficiently, conserving energy and reducing operational costs.

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Most popular questions from this chapter

A diesel engine performs \(2200 \mathrm{~J}\) of mechanical work and discards \(4300 \mathrm{~J}\) of heat each cycle. (a) How much heat must be supplied to the engine in each cycle? (b) What is the thermal efficiency of the engine?

A person having skin of surface area \(1.85 \mathrm{~m}^{2}\) and temperature \(30.0^{\circ} \mathrm{C}\) is resting in an insulated room where the ambient air temperature is \(20.0^{\circ} \mathrm{C}\). In this state, a person gets rid of excess heat by radiation. By how much does the person change the entropy of the air in this room each second? (Recall that the room radiates back into the person and that the emissivity of the skin is \(1.00 .)\)

A refrigerator has a coefficient of performance of \(K=2.0 .\) Each cycle, it absorbs \(3.40 \times 10^{4} \mathrm{~J}\) of heat from the cold reservoir. The refrigerator is driven by a Carnot engine that has an efficiency of \(e=0.5 .\) (a) How much mechanical energy is required each cycle to operate the refrigerator? (b) During each cycle, how much heat flows into the Carnot engine?

A Carnot engine is operated between two heat reservoirs at temperatures of \(520 \mathrm{~K}\) and \(300 \mathrm{~K}\). (a) If the engine receives \(6.45 \mathrm{~kJ}\) of heat energy from the reservoir at \(520 \mathrm{~K}\) in each cycle, how many joules per cycle does it reject to the reservoir at \(300 \mathrm{~K} ?\) (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?

A gasoline engine takes in \(1.61 \times 10^{4} \mathrm{~J}\) of heat and delivers \(3700 \mathrm{~J}\) of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of \(4.60 \times 10^{4} \mathrm{~J} / \mathrm{g}\). (a) What is the thermal efficiency of the engine? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?

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