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An ideal refrigerator does \(150 \mathrm{~J}\) of work to remove \(560 \mathrm{~J}\) as heat from its cold compartment. (a) What is the refrigerator's coefficient of performance? (b) How much heat per cycle is exhausted to the kitchen?

Short Answer

Expert verified
(a) The COP is 3.73. (b) The heat exhausted is 710 J per cycle.

Step by step solution

01

Understanding the Problem

We need to find the coefficient of performance (COP) of the refrigerator and the heat exhausted to the kitchen. This involves using the formulas related to refrigerators, including work done and heat transfers.
02

Refrigerator's Coefficient of Performance (COP)

The coefficient of performance for a refrigerator is given by the formula: \[ \text{COP} = \frac{Q_c}{W} \] where \( Q_c = 560 \mathrm{~J} \) (heat removed from cold compartment) and \( W = 150 \mathrm{~J} \) (work done). Substitute the values into the formula: \[ \text{COP} = \frac{560}{150} = 3.73 \]
03

Calculate Exhausted Heat

The heat exhausted to the kitchen (\( Q_h \)) can be found by using the energy conservation equation for refrigerators: \[ Q_h = Q_c + W \] Now substitute \( Q_c = 560 \mathrm{~J} \) and \( W = 150 \mathrm{~J} \) into the equation: \[ Q_h = 560 + 150 = 710 \mathrm{~J} \]

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Key Concepts

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

Coefficient of Performance
The Coefficient of Performance (COP) is a measure of the efficiency of a refrigerator or a heat pump. It tells us how well a refrigerator performs by comparing the amount of heat it moves with the amount of work it requires to do so. You can think of COP as a 'bang for your buck' measure in terms of energy.

The formula to calculate the coefficient of performance for a refrigerator is:
\[\text{COP} = \frac{Q_c}{W}\]where
  • \( Q_c \) is the heat removed from the refrigerator's cold compartment
  • \( W \) is the work input required
For instance, in our problem, the refrigerator removes \( 560 \mathrm{~J} \) of heat with an energy input of \( 150 \mathrm{~J} \) of work. By substituting these values into the COP formula, we can calculate:
\[\text{COP} = \frac{560}{150} = 3.73\]This means the refrigerator is quite efficient as it manages to remove 3.73 times the energy it consumes as work.
Refrigerator Cycle
The refrigerator cycle is the process through which a refrigerator removes heat from its interior to maintain a cool environment. This cycle relies on the principles of thermodynamics and energy transfer.

In this cycle, refrigerant absorbs heat from the internal compartment, thereby cooling it. This absorbed heat, along with the heat generated by work done by the motor, is then expelled to the outside environment, often into the kitchen.

Steps in the Refrigerator Cycle

  • **Evaporation:** The refrigerant absorbs heat as it evaporates inside the refrigerator, cooling the interior.
  • **Compression:** The gaseous refrigerant is compressed, which increases its temperature.
  • **Condensation:** The heat is released to the surroundings as the refrigerant condenses back to a liquid state.
  • **Expansion:** The refrigerant is expanded, cooling it further, and the cycle repeats.
Energy Conservation
Energy conservation is a fundamental principle in physics, stating that energy cannot be created or destroyed; it can only change forms. This principle is crucial in understanding the operation of a refrigerator.

For any refrigerator, the energy balance can be represented as:
\[ Q_h = Q_c + W \]where
  • \( Q_h \) is the heat exhausted to the environment
  • \( Q_c \) is the heat extracted from the refrigerator's cold compartment
  • \( W \) is the work input
Using energy conservation, the equation ensures that the energy entering the refrigerator is equal to the energy leaving it, after accounting for the work done. In the given exercise, the exhausted heat \( (Q_h) \) will be the sum of the heat removed \( (Q_c) \) and the work done \( (W) \). Calculating, we find:
\[Q_h = 560 + 150 = 710 \mathrm{~J} \]This confirms that the cycle maintains the energy equilibrium as expected.

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Most popular questions from this chapter

A Carnot engine absorbs \(52 \mathrm{~kJ}\) as heat and exhausts \(36 \mathrm{~kJ}\) as heat in each cycle. Calculate (a) the engine's efficiency and (b) the work done per cycle in kilojoules.

A Carnot engine whose high-temperature reservoir is at \(400 \mathrm{~K}\) has an efficiency of \(30.0 \% .\) By how much should the temperature of the low- temperature reservoir be changed to increase the efficiency to \(40.0 \% ?\)

A mixture of \(1773 \mathrm{~g}\) of water and \(227 \mathrm{~g}\) of ice is in an initial equilibrium state at \(0.000^{\circ} \mathrm{C}\). The mixture is then, in a reversible process, brought to a second equilibrium state where the water-ice ratio, by mass, is 1.00: 1.00 at \(0.000^{\circ} \mathrm{C}\). (a) Calculate the entropy change of the system during this process. (The heat of fusion for water is \(333 \mathrm{~kJ} / \mathrm{kg}\).) (b) The system is then returned to the initial equilibrium state in an irreversible process (say, by using a Bunsen burner). Calculate the entropy change of the system during this process. (c) Are your answers consistent with the second law of thermodynamics?

A Carnot engine has an efficiency of \(22.0 \% .\) It operates between constant- temperature reservoirs differing in temperature by \(75.0 \mathrm{C}^{\circ} .\) What is the temperature of the (a) lower-temperature and (b) higher- temperature reservoir?

An insulated Thermos contains \(130 \mathrm{~g}\) of water at \(80.0^{\circ} \mathrm{C}\). You put in a \(12.0 \mathrm{~g}\) ice cube at \(0^{\circ} \mathrm{C}\) to form a system of ice + original water. (a) What is the equilibrium temperature of the system? What are the entropy changes of the water that was originally the ice cube (b) as it melts and (c) as it warms to the equilibrium temperature? (d) What is the entropy change of the original water as it cools to the equilibrium temperature? (e) What is the net entropy change of the ice + original water system as it reaches the equilibrium temperature?

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