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An air conditioner operates on 800 \(\mathrm{W}\) of power and has a performance coefficient of 2.80 with a room temperature of \(21.0^{\circ} \mathrm{C}\) and an outside temperature of \(35.0^{\circ} \mathrm{C}\) . (a) Calculate the rate of heat removal for this unit. (b) Calculate the rate at which heat is discharged to the outside air. (c) Calculate the total entropy change in the room if the air conditioner runs for 1 hour. Calculate the total entropy change in the outside air for the same time period. (d) What is the net change in entropy for the system (room + outside air)?

Short Answer

Expert verified
(a) 2240 W, (b) 3040 W, (c) Room: -27438.83 J/K, Outside: 35456.88 J/K, (d) Net: 8018.05 J/K.

Step by step solution

01

Understanding the Problem

The air conditioner operates with a known power input, and its performance coefficient is given as 2.80. The goal is to find the rate of heat removal from the room, the rate at which heat is discharged outside, and the changes in entropy of the room, the outside air, and the overall system.
02

Calculate Rate of Heat Removal

The coefficient of performance (COP) for an air conditioner is given by \(COP = \frac{Q_c}{W}\), where \(Q_c\) is the heat removed from the room, and \(W\) is the work or power input. We have \(COP = 2.80\) and \(W = 800 \, \mathrm{W}\). Therefore, \(Q_c = COP \times W = 2.80 \times 800 = 2240 \, \mathrm{W}\). Hence, the rate of heat removal is 2240 W.
03

Calculate Rate of Heat Discharge

The heat discharged to the outside, \(Q_h\), is given by \(Q_h = Q_c + W\), where \(Q_c = 2240 \, \mathrm{W}\) and \(W = 800 \, \mathrm{W}\). Thus, \(Q_h = 2240 + 800 = 3040 \, \mathrm{W}\). The rate at which heat is discharged to the outside is 3040 W.
04

Calculate Entropy Change in the Room

The change in entropy for the room, \(\Delta S_{room}\), when energy \(Q_c\) is removed at a constant temperature of \(T_{room} = 21.0^{\circ} C = 294.15 \, \mathrm{K}\), is \(\Delta S_{room} = -\frac{Q_c \cdot \Delta t}{T_{room}}\). For \(\Delta t\) of 1 hour (or 3600 seconds), \(\Delta S_{room} = -\frac{2240 \times 3600}{294.15} = -27438.83 \, \mathrm{J/K}\).
05

Calculate Entropy Change in the Outside Air

The change in entropy for the outside air, \(\Delta S_{outside}\), is \(\Delta S_{outside} = \frac{Q_h \cdot \Delta t}{T_{outside}}\), with \(T_{outside} = 35.0^{\circ} C = 308.15 \, \mathrm{K}\). Hence, \(\Delta S_{outside} = \frac{3040 \times 3600}{308.15} = 35456.88 \, \mathrm{J/K}\).
06

Calculate Net Entropy Change for the System

The net change in entropy of the system (room + outside air) is \(\Delta S_{system} = \Delta S_{room} + \Delta S_{outside}\). Therefore, \(\Delta S_{system} = -27438.83 + 35456.88 = 8018.05 \, \mathrm{J/K}\). This means the net change in entropy is positive, as expected in real processes.

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Key Concepts

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

Entropy Change
Entropy is a measure of the randomness or disorder of a system. In thermodynamics, it helps us understand how energy is distributed within a system. When discussing air conditioners or refrigerators, we often talk about how these devices transfer heat and cause changes in entropy. For the room in the exercise, heat is removed by the air conditioner, leading to a decrease in entropy. This is because energy is being extracted from a controlled space. The formula for entropy change when heat is removed is: \[ \Delta S = -\frac{Q_c \cdot \Delta t}{T} \] where \( Q_c \) is the heat removed, \( \Delta t \) is the time duration, and \( T \) is the temperature of the room. Conversely, when this energy is expelled to the outside, the outside environment experiences an increase in entropy, this time calculated as: \[ \Delta S = \frac{Q_h \cdot \Delta t}{T} \] This difference in entropy demonstrates why the net entropy change across a system is generally positive, confirming the natural trend towards disorder. Entropy changes are crucial because they are directly related to the efficiency and performance of thermodynamic cycles, such as those found in HVAC systems.
Coefficient of Performance
The coefficient of performance, or COP, is fundamental in assessing the effectiveness of devices like air conditioners or heat pumps. It represents the ratio of heat removed from the cold reservoir to the work input required to remove that heat. The COP is defined as: \[ \text{COP} = \frac{Q_c}{W} \] where \( Q_c \) is the heat removed from the cooled space and \( W \) is the work or energy input to the system. A higher COP indicates a more efficient system, meaning more heat is removed for each unit of work input. In the given example, with a COP of 2.80, it implies that for every watt of power consumed, the air conditioner removes 2.80 watts of heat from the room. Understanding COP helps in comparing different air conditioners and choosing the most efficient one, particularly important for energy conservation and cost reduction.
Heat Transfer
Heat transfer is a critical concept in thermodynamics, especially in the context of air conditioners which transfer heat from an indoor environment to an outdoor one. This process is typically described in terms of the rate at which heat is removed from the room and the rate at which it is discharged outside. For this exercise, the rate of heat removal by the air conditioner was calculated as: \[ Q_c = \text{COP} \times W \] and the rate of heat discharge as: \[ Q_h = Q_c + W \] These calculations are based on the energy conservation principle, where energy input plus heat removed equals the energy output. Heat transfer in this sense affects the climate control in homes and buildings and involves conduction, convection, and radiation processes over time. Efficient heat transfer systems ensure comfortable environments and better energy use.
Second Law of Thermodynamics
The second law of thermodynamics is one of the cornerstones of thermal science. It states that in any natural process, the total entropy of a closed system will generally increase over time. This is why the net entropy change in the example system, which includes both the room and the outside air, is positive. In practical scenarios, this law means that energy transformations are not 100% efficient due to the inevitable generation of entropy, often termed waste heat. This law is essential when designing thermodynamic cycles for devices that transfer energy, such as air conditioners and refrigerators. It highlights:
  • Inevitability of some energy loss when performing work
  • Tendency of systems towards equilibrium
  • Impossibility of creating a perpetual motion machine
By understanding the concepts laid out by this law, engineers can design more efficient systems that minimize energy losses and optimize thermal performance.

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Most popular questions from this chapter

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

An object of mass \(m_{1},\) specific heat \(c_{1},\) and temperature \(T_{1}\) is placed in contact with a second object of mass \(m_{2},\) specific heat \(c_{2},\) and temperature \(T_{2}>T_{1} .\) As a result, the temperature of the first object increases to \(T\) and the temperature of the second object decreases to \(T^{\prime}\) . (a) Show that the entropy increase of the system is $$\Delta S=m_{1} c_{1} \ln \frac{T}{T_{1}}+m_{2} c_{2} \ln \frac{T^{\prime}}{T_{2}}$$ and show that energy conservation requires that $$m_{1} c_{1}\left(T-T_{1}\right)=m_{2} c_{2}\left(T_{2}-T^{\prime}\right)$$ (b) Show that the entropy change \(\Delta S,\) considered as a function of \(T,\) is a maximum if \(T=T^{\prime},\) which is just the condition of thermodynamic equilibrium. (c) Discuss the result of part (b) in terms of the idea of entropy as a measure of disorder.

A typical coal-fired power plant generates 1000 \(\mathrm{MW}\) of usable power at an overall thermal efficiency of 40\(\% .\) (a) What is the rate of heat input to the plant? (b) The plant burns anthracite coal, which has a heat of combustion of \(2.65 \times 10^{7} \mathrm{J} / \mathrm{kg}\) . How much coal does the plant use per day, if it operates continuously? (c) At what rate is heat ejected into the cool reservoir, which is the nearby river? (d) The river's temperature is \(18.0^{\circ} \mathrm{C}\) before it reaches the power plant and \(18.5^{\circ} \mathrm{C}\) after it has received the plant's waste heat. Calculate the river's flow rate, in cubic meters per second. (e) By how much does the river's entropy increase each second?

A Carnot heat engine has a thermal efficiency of \(0.600,\) and the temperature of its hot reservoir is 800 \(\mathrm{K} .\) If 3000 \(\mathrm{J}\) of heat is rejected to the cold reservoir in one cycle, what is the work output of the engine during one cycle?

You design an engine that takes in \(1.50 \times 10^{4} \mathrm{J}\) of heat at 650 \(\mathrm{K}\) in each cycle and rejects heat at a temperature of 350 \(\mathrm{K}\). The engine completes 240 cycles in 1 minute. What is the theoretical maximum power output of your engine, in horsepower?

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