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A room air conditioner has a coefficient of performance of 2.9 on a hot day and uses 850 \(\mathrm{W}\) of electrical power. (a) How many ioules of heat does the air conditioner remove from the room in one minute? (b) How many joules of heat does the air conditioner deliver to the hot outside air in one minute? (c) Explain why your answers to parts (a) and (b) are not the same.

Short Answer

Expert verified
(a) 147900 J, (b) 198900 J, (c) Energy use explains the difference.

Step by step solution

01

Identify Given Values

The coefficient of performance (COP) of the air conditioner is given as 2.9. The electrical power used is 850 W. We need to find the heat removed from the room ( Q_c ) and the heat delivered to the outside (Q_h).
02

Calculate Heat Removed from the Room

The formula relating the COP, heating power (Q_c), and electrical power (W) is given by:\[\text{COP} = \frac{\text{Heat removed from the room (Q_c)}}{\text{Electrical Power} (W)}.\]Using this, we can calculate Q_c in one minute. The electrical energy used in one minute:\[W \times t = 850 \times 60 = 51000 \text{ Joules}.\]So, \[Q_c = \text{COP} \times W \times t = 2.9 \times 850 \times 60 = 147900 \text{ Joules}.\]This is the energy the air conditioner removes from the air-conditioned room in one minute.
03

Calculate Heat Delivered to Outside

The heat delivered to the outside is the sum of the heat removed from the room and the electrical energy used:\[Q_h = Q_c + (W \times t) = 147900 + 51000 = 198900 \text{ Joules}.\]Thus, the air conditioner delivers 198900 Joules to the hot outside air in one minute.
04

Explain the Different Quantities of Heat

The difference between heat removed from the room and heat delivered outside is due to the added electrical energy. The air conditioner removes a certain amount of heat from the indoor air and, in doing so, consumes energy. That electrical energy gets converted to additional heat, which is released to the outside environment. Hence, the heat delivered to the outside is more than that removed from the room.

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Key Concepts

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

Thermodynamics
Thermodynamics is the science that deals with energy, its transformations, and its relation to matter. It's the branch of physics that investigates the relationships between heat, work, and energy transfer. In the context of the air conditioning system, thermodynamics helps us understand how the air conditioner functions as a heat pump.
The air conditioner uses the basic principles from thermodynamics, such as the laws of energy conservation and efficiency. The first law explains that energy cannot be created or destroyed, only transformed. This is crucial in understanding how the air conditioner moves heat from the inside to the outside.
In a thermodynamic cycle, the air conditioner's ability to transfer heat is influenced by its Coefficient of Performance (COP). The COP is a measurement of efficiency, defined as the ratio of heating or cooling provided to the electrical energy consumed. The given COP in this problem is used to determine how effectively the system removes heat from the room.
Heat Transfer
Heat transfer is the process by which heat moves from one place to another. In the context of air conditioning, this concept explains how the air conditioner extracts heat from the indoor environment and releases it outside.
There are three main mechanisms of heat transfer: conduction, convection, and radiation. In air conditioning systems, conduction and convection are primarily involved. The system uses a refrigerant to absorb heat from the room air (convection) and then transfers this heat to the environment (also via convection and conduction) through the heat exchanger.
In the exercise, the heat transfer involves two main quantities: the heat removed from the room ( Q_c ) and the heat delivered to the outside air ( Q_h ). These values illustrate how much thermal energy is being transferred through the system and how efficiently the energy input (electrical power) is being utilized.
Energy Conversion
Energy conversion in an air conditioning system is the transformation of electrical energy into heat energy, causing heat to move from cooler to warmer areas. This is contrary to the natural direction in which heat moves, and it is achieved through energy input from electricity.
The air conditioner relies on a mechanism called a heat pump which uses electrical energy to increase the pressure of a refrigerant. This process creates a cycle where heat is absorbed inside the home and released outside. In the step-by-step solution, calculating the electrical energy used (850 W) helps us determine how much extra thermal energy is produced in addition to what was taken from the room.
The difference in heat quantities observed in the initial exercise results from this conversion process. The air conditioner's design and operation increase thermal energy outside by the amount of electrical energy consumed, demonstrating how energy conversion affects overall system performance.
Air Conditioning System Analysis
Analyzing an air conditioning system involves understanding how these systems function and affect indoor environments. The air conditioner's main role is to maintain a desired room temperature by having an efficient cycle of heat removal and release.
The system analysis looks at how well the air conditioner performs under certain conditions. COP is a key performance indicator, as it gives insights into the ratio of cooling output relative to energy input. In the exercise, with a COP of 2.9, it shows how for every unit of electricity consumed, 2.9 units of heat are being removed from the room. This efficiency metric helps determine operating costs and energy consumption.
In terms of environmental impact, the analysis extends to how the refrigerant impacts heat transfer efficiency and how effectively heat is emitted to the outside atmosphere. Efficient air conditioning systems not only provide comfort but also do so in an energy-conscious manner, reducing overall environmental impact.

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Most popular questions from this chapter

You decide to take a nice hot bath but discover that your thoughtless roommate has used up most of the hot water. You fill the tub with 270 \(\mathrm{kg}\) of \(30.0^{\circ} \mathrm{C}\) water and attempt to warm it further by pouring in 5.00 \(\mathrm{kg}\) of boiling water from the stove. (a) Is this a reversible or an irreversible process? Use physical reasoning to explain. (b) Calculate the final temperature of the bath water. (c) Calculate the net change in entropy of the system (bath water \(+\) boiling water), assuming no heat exchange with the air or the tub itself.

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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?

A lonely party balloon with a volume of 2.40 \(\mathrm{L}\) and containing 0.100 \(\mathrm{mol}\) of air is left behind to drift in the temporarily uninhabited and depressurized International Space Station. Sunlight coming through a porthole heats and explodes the balloon, causing the air in it to undergo a free expansion into the empty station, whose total volume is 425 \(\mathrm{m}^{3} .\) Calculate the entropy change of the air during the expansion.

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)?

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