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A refrigerator has a coefficient of performance equal to 5.00. The refrigerator takes in \(120 \mathrm{J}\) of energy from a cold reservoir in each cycle. Find (a) the work required in each cycle and (b) the energy expelled to the hot reservoir.

Short Answer

Expert verified
The work required in each cycle is 24 J and the energy expelled to the hot reservoir is 144 J.

Step by step solution

01

Find the work required in each cycle

First, rearrange the formula for the coefficient of performance to find the work done: \(W = Q_c/COP\). Substitute the values into the formula: \(W = 120 \, J/5.00 = 24 \, J\). So the work required in each cycle is 24 J.
02

Find the energy expelled to the hot reservoir

The energy expelled to the hot reservoir is the energy taken from the cold reservoir plus the work done on the system. Therefore, using the energy conservation law we have \(Q_h = Q_c + W\). Substituting in the given and calculated values we have: \(Q_h = 120 \, J + 24 \, J = 144 \, J\). So the energy expelled to the hot reservoir in each cycle is 144 J.

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Key Concepts

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

Understanding Thermodynamics in the Context of Refrigerators
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In the realm of refrigerators, thermodynamics plays a central role as it offers the principles governing the heat transfer which enables the cooling process. One of the fundamental principles is that heat flows spontaneously from a warmer body to a cooler one. However, in a refrigerator, this natural flow is reversed by the input of work, which is underpinned by the work-energy principle. This mechanical work is done by the compressor, effectively pumping heat from the colder interior of the fridge to the warmer outside environment. Understanding thermodynamics is critical to grasping how refrigerators can maintain a temperature lower than that of their surroundings.
Breaking Down the Refrigerator Cycle
A refrigerator cycle, or refrigeration cycle, is the process by which a refrigerator removes heat from an enclosed space and expels it to the surrounding environment. The cycle consists of several stages, involving a refrigerant circulating through the system: evaporation, compression, condensation, and expansion. During evaporation, the refrigerant absorbs heat from the fridge interior, turning from liquid to gas. It is then compressed, which increases its temperature. The hot gas then condenses, releasing the absorbed heat to the outside air, and turns back into a liquid. Upon expansion, the refrigerant cools down again, ready to absorb more heat from the refrigerator's interior. This process is made efficient through the use of the coefficient of performance (COP), which, in the given exercise, determines the amount of energy needed for each refrigeration cycle.
Energy Conservation Law in Refrigeration
The energy conservation law states that energy cannot be created or destroyed, only transformed from one form to another. When applied to refrigeration, this law explains how the energy within the system is handled. The work done by the refrigerator (to drive the refrigeration cycle) and the heat absorbed from the cold reservoir (interior of the fridge) is equal to the heat expelled to the hot reservoir (outside environment). Understanding this concept is key to solving problems related to the performance of refrigerators, such as calculating the work required for each cycle and the energy expelled to the hot reservoir, as exhibited in the exercise.
Applying the Work-Energy Principle to Refrigerators
The work-energy principle is pivotal in understanding how refrigerators function. It connects the work done on a system to the energy change within that system. For refrigerators, work is required to move heat from a lower to a higher temperature, which is counter to the natural direction of heat flow and necessitates energy input. The COP of a refrigerator reveals how effective the device is in transferring heat with each joule of work exerted. A higher COP indicates better efficiency, as less work is needed to transfer a specific amount of heat from the cold reservoir to the hot one. In the exercise provided, the COP helps us deduce the amount of work needed for the refrigeration cycle, thus applying the work-energy principle in a practical context.

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Most popular questions from this chapter

An electric power plant that would make use of the temperature gradient in the ocean has been proposed. The system is to operate between \(20.0^{\circ} \mathrm{C}\) (surface water temperature) and \(5.00^{\circ} \mathrm{C}\) (water temperature at a depth of about 1 \(\mathrm{km}\) ). (a) What is the maximum efficiency of such a system? (b) If the useful power output of the plant is \(75.0 \mathrm{MW},\) how much energy is taken in from the warm reservoir per hour? (c) In view of your answer to part (a), do you think such a system is worthwhile? Note that the "fuel" is free.

What is the maximum possible coefficient of performance of a heat pump that brings energy from outdoors at \(-3.00^{\circ} \mathrm{C}\) into a \(22.0^{\circ} \mathrm{C}\) house? Note that the work done to run the heat pump is also available to warm up the house.

Calculate the change in entropy of \(250 \mathrm{g}\) of water heated slowly from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} .\) (Suggestion: Note that \(d Q=m c d T)\)

A heat engine operates between two reservoirs at \(T_{2}=600 \mathrm{K}\) and \(T_{1}=350 \mathrm{K} .\) It takes in \(1000 \mathrm{J}\) of energy from the higher-temperature reservoir and performs 250 J of work. Find (a) the entropy change of the Universe \(\Delta S_{U}\) for this process and (b) the work \(W\) that could have been done by an ideal Carnot engine operating between these two reservoirs. (c) Show that the difference between the amounts of work done in parts (a) and (b) is \(T_{1} \Delta S_{U}\)

A refrigerator has a coefficient of performance of 3.00 The ice tray compartment is at \(-20.0^{\circ} \mathrm{C},\) and the room temperature is \(22.0^{\circ} \mathrm{C} .\) The refrigerator can convert \(30.0 \mathrm{g}\) of water at \(22.0^{\circ} \mathrm{C}\) to \(30.0 \mathrm{g}\) of ice at \(-20.0^{\circ} \mathrm{C}\) each minute. What input power is required? Give your answer in watts.

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