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A refrigerator has a coefficient of performance of \(2.10 .\) In each cycle it absorbs \(3.10 \times 10^{4} \mathrm{~J}\) of heat from the cold reservoir. (a) How much mechanical energy is required each cycle to operate the refrigerator? (b) During each cycle, how much heat is discarded to the high-temperature reservoir?

Short Answer

Expert verified
The mechanical energy required to operate the refrigerator each cycle is approximately \(1.48 \times 10^{4} \mathrm{~J}\), and the heat discarded to the higher temperature reservoir is approximately \(4.58 \times 10^{4} \mathrm{~J}\).

Step by step solution

01

Calculate the Mechanical Work

The mechanical energy or work (W) required to operate the refrigerator can be calculated using the relation between the coefficient of performance (K) and the heat absorbed from the cold reservoir (Q_c). This can be rearranged from the formula \( K = Q_c/W \) to solve for W: \( W = Q_c / K \). Then, substitute the given values into the equation: \( W = 3.10 \times 10^{4} \mathrm{~J} / 2.10 \).
02

Calculate Heat Discarded

The heat discarded to the high-temperature reservoir (Q_h) can be calculated using the conservation of energy of the system. This means that the energy put into the system (W) plus the energy removed from the cold reservoir (Q_c), should equal the heat given off to the warmer reservoir (Q_h). This is mathematically represented by the equation: \( Q_h = Q_c + W \). Substituting our known values: \( Q_h = 3.10 \times 10^{4} \mathrm{~J} + W \).

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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 crucial concept in thermodynamics, especially when it comes to heating and cooling systems like refrigerators. It measures the efficiency of a refrigerator, telling us how effectively it uses energy to transfer heat. For refrigerators, the COP is defined as the ratio of the heat absorbed from the cold reservoir (\(Q_c\)) to the work done (\(W\)):
  • Formula: \[K = \frac{Q_c}{W}\]
  • The higher the COP, the more efficient the device.
In the given exercise, the COP is 2.10, indicating that for every unit of work input, the device is able to absorb 2.10 units of heat from the cold reservoir.
This characteristic is key to understanding how fridges conserve energy while maintaining their cooling effect. When using the formula to find the work required, you rearrange it to solve for \(W\) which gives: \[W = \frac{Q_c}{K}\]. Knowing the COP allows us to efficiently determine the energy requirements for different cycles, underscoring its importance in the analysis of cooling systems.
Conservation of Energy
The principle of Conservation of Energy is at the heart of understanding how heat is transferred and managed in thermodynamic systems like refrigerators. This principle states that energy cannot be created or destroyed, only transformed from one form to another.
In the context of the refrigerator exercise, this means that all energy put into the system manifests either as useful energy transferred between reservoirs or wasted energy:
  • The work input (\(W\)) is added to the absorbed heat from the cold reservoir (\(Q_c\)) to determine the heat discharged to the environment (\(Q_h\)).
  • The formula representing this balance is: \[Q_h = Q_c + W\].
This equation poignantly illustrates the conservation of energy in a thermodynamic cycle. If a refrigerator absorbs a certain amount of heat from the cold reservoir and does a specific amount of work, it must release a total quantity of heat to the warm environment that is equal to the sum of these energies. Without this energy balance, the system wouldn't function correctly and efficiently.
Heat Transfer
Heat Transfer is the process of energy moving from one body or substance to another due to a temperature difference. In refrigeration, this concept is fundamental. Within a refrigerator, heat transfer occurs from a colder area to a hotter area, which is essentially the reverse of the natural flow of heat.
  • This is achieved by the absorption of heat (\(Q_c\)) from the interior, cooling it down.
  • Subsequently, this absorbed heat is expelled to the environment (\(Q_h\)), which is at a higher temperature than the interior.
The devices utilizing this principle, like refrigerators, require a work input to achieve the reversal of natural heat flow. Here's how it happens: the refrigerant inside the fridge absorbs heat as it evaporates, effectively lowering the internal temperature of the fridge.
Later, as the refrigerant is compressed and heats up, it releases this heat to the environment outside the refrigerator. Understanding this process highlights the significance of both the work and COP in maintaining efficient operation. The ability to control and manage heat transfer via these principles ensures that refrigeration systems maintain desired temperatures while optimizing energy use.

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Most popular questions from this chapter

A Carnot engine performs \(2.5 \times 10^{4} \mathrm{~J}\) of work in each cycle and has an cfficicncy of \(66 \%\). (a) How much heat docs the cngine extract from its heat source in each cycle? (b) If the engine exhausts heat at room temperature \(\left(20.0^{\circ} \mathrm{C}\right),\) what is the temperature of its heat source?

A Carnot engine whose high-temperature reservoir is at \(620 \mathrm{~K}\) takes in \(550 \mathrm{~J}\) of heat at this temperature in each cycle and gives up \(335 \mathrm{~J}\) to the low-temperature rescrvoir. (a) How much mechanical work does the cngine perform during cach cycle? What is (b) the temperature of the low-temperature reservoir; (c) the thermal cfficicncy of the cycle?

Compare the entropy change of the warmer water to that of the colder water during one cycle of the heat engine, assuming an ideal Carnot cycle. (a) The entropy does not change during one cycle in either case. (b) The cntropy of both increases, but the entropy of the colder water increases by more because its initial temperature is lower. (c) The cntropy of the warmer water decreases by more than the cntropy of the colder water increases, because some of the heat removed from the warmer water goes to the work done by the engine. (d) The entropy of the warmer water decreases by the same amount that the entropy of the colder water increases.

Carnot refrigcrator \(A\) has a \(16 \%\) highcr cocfficicnt of performance than Carnot refrigerator \(B\). The temperature difference between the hot and cold reservoirs is \(30 \%\) greater for \(B\) than for \(A\). If the coldreservoir temperature for refrigerator \(B\) is \(180 \mathrm{~K},\) what is the cold- reservoir tempcrature for refrigerator \(A ?\)

In a Carnot engine the hot reservoir is \(72.0 \mathrm{C}^{\circ}\) warmer than the cold reservoir. The engine's efficiency is \(12.5 \% .\) What are the Kelvin temperatures of the two reservoirs?

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