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A refrigerator has a coefficient of performance of 2.10. In each cycle it absorbs 3.10 \(\times\) 10\(^4\) 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
(a) 1.48 × 10^4 J, (b) 4.58 × 10^4 J.

Step by step solution

01

Understanding Coefficient of Performance

The coefficient of performance (COP) is a measure of a refrigerator's efficiency and is defined by the formula \( \text{COP} = \frac{Q_c}{W} \), where \( Q_c \) is the heat absorbed from the cold reservoir and \( W \) is the work (mechanical energy) required.
02

Given Values

We know from the problem statement that \( \text{COP} = 2.10 \) and \( Q_c = 3.10 \times 10^4 \text{ J} \). These are the key values needed to find the work \( W \).
03

Calculate Mechanical Energy Required

Rearrange the COP formula to solve for \( W \): \( W = \frac{Q_c}{\text{COP}} \). Substitute the given values: \( W = \frac{3.10 \times 10^4}{2.10} \). Calculate \( W \) to find the mechanical energy.
04

Calculation for Part a

\( W = \frac{3.10 \times 10^4}{2.10} \approx 1.48 \times 10^4 \text{ J} \). The mechanical energy required each cycle is approximately \( 1.48 \times 10^4 \text{ J} \).
05

Calculate Heat Discarded to High-Temperature Reservoir

Use the energy conservation principle: the heat discarded \( Q_h \) to the high-temperature reservoir is given by \( Q_h = Q_c + W \).
06

Calculation for Part b

Substitute the known values to find \( Q_h \): \( Q_h = 3.10 \times 10^4 + 1.48 \times 10^4 = 4.58 \times 10^4 \text{ J} \). The heat discarded to the high-temperature reservoir is \( 4.58 \times 10^4 \text{ J} \).

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Key Concepts

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

Understanding Thermodynamics
Thermodynamics is a foundational concept in physics that deals with the movement and transformation of energy. It is crucial for understanding the principles behind heat engines, refrigerators, and many other systems.

Thermodynamics is built on the concept of energy conservation. In essence, energy cannot be created or destroyed, only transformed from one form to another. There are four fundamental laws of thermodynamics, but for this exercise, the first and second laws are most relevant.
  • The First Law of Thermodynamics states that the total energy of an isolated system is constant, and energy can be transferred or converted but not destroyed.
  • The Second Law of Thermodynamics introduces the concept of entropy, stating that energy transformations are not 100% efficient, and some energy is always dissipated as heat.
In the context of a refrigerator, these principles help explain how the system moves heat from a cooler area to a warmer one, requiring work (mechanical energy) to do so. The coefficient of performance (COP) is a measure of this efficiency.
Explaining Heat Transfer in Refrigerators
Heat transfer is the process of thermal energy moving from one place to another. It's essential in the operation of refrigerators where heat is transferred away from the interior.

Refrigerators work on the principle of extracting heat from a low-temperature area (inside the fridge) and transferring it to a high-temperature area (outside the fridge). This heat transfer requires an input of mechanical energy.
  • A refrigerator absorbs heat from its contents, seen as the cold reservoir in thermodynamic terms.
  • This absorbed heat (\(Q_c\)) is then expelled to the surroundings, defined as the high-temperature reservoir.
  • The work (\(W\)) done by the refrigerator is derived from an external power source, usually electrical energy.
For efficient operation, understanding the relationship between these energy exchanges is essential. The heat rejected to the high-temperature reservoir is always larger than the heat absorbed from the cold one, due to added mechanical energy.
Efficiency Calculation and Coefficient of Performance
Efficiency in thermodynamic cycles, like those in refrigerators, is assessed using different metrics. For a refrigerator, the Coefficient of Performance (COP) is key.

The COP of a refrigerator is a ratio that compares the heat removed from the cold reservoir to the work input required:\[\text{COP} = \frac{Q_c}{W}\]
  • Higher COP values signify greater efficiency, meaning the refrigerator requires less work to remove a specific amount of heat.
In our exercise, the provided COP and the heat absorbed from the cold reservoir allow us to calculate the work required (\(W\)) using the formula rearranged as:\[W = \frac{Q_c}{\text{COP}}\]
  • With a COP of 2.10 and \(Q_c = 3.10 \times 10^4 \text{ J}\), we find the work required is approximately \(1.48 \times 10^4 \text{ J}\).
  • The efficiency encapsulated in COP highlights how much mechanical energy is utilized effectively to move heat.
Ultimately, efficient energy transfer saves energy and reduces costs, which is why calculating efficiency is so crucial in thermodynamics.

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Most popular questions from this chapter

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation, the heat rejected by the engine melts 0.0400 kg of ice. During this time, how much work \(W\) is performed by the engine?

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\(^\circ\)C and 6\(^\circ\)C, respectively. (a) What is the maximum theoretical efficiency of this power plant? (b) If the power plant is to produce 210 \(kW\) of power, at what rate must heat be 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\(^\circ\)C. What must be the flow rate of cold water through the system? Give your answer in kg/h and in \(L\)/h.

A 4.50-kg block of ice at 0.00\(^\circ\)C falls into the ocean and melts. The average temperature of the ocean is 3.50\(^\circ\)C, including all the deep water. By how much does the change of this ice to water at 3.50\(^\circ\)C alter the entropy of the world? Does the entropy increase or decrease? (\(Hint\): Do you think that the ocean temperature will change appreciably as the ice melts?)

A gasoline engine takes in 1.61 \(\times\) 10\(^4\) J of heat and delivers 3700 J of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of 4.60 \(\times\) 10\(^4\) J/g. (a) What is the thermal efficiency? (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?

A cylinder contains oxygen at a pressure of 2.00 atm. The volume is 4.00 \(L\), and the temperature is 300 \(K\). Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes: (i) Heated at constant pressure from the initial state (state 1) to state 2, which has \(T\) = 450 K. (ii) Cooled at constant volume to 250 \(K\) (state 3). (iii) Compressed at constant temperature to a volume of 4.00 \(L\) (state 4). (iv) Heated at constant volume to 300 \(K\), which takes the system back to state 1. (a) Show these four processes in a \(pV\)-diagram, giving the numerical values of \(p\) and \(V\) in each of the four states. (b) Calculate \(Q\) and \(W\) for each of the four processes. (c) Calculate the net work done by the oxygen in the complete cycle. (d) What is the efficiency of this device as a heat engine? How does this compare to the efficiency of a Carnot cycle engine operating between the same minimum and maximum temperatures of 250 \(K\) and 450 \(K\)?

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