/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 An air conditioner keeps the ins... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An air conditioner keeps the inside of a house at a temperature of \(19.0^{\circ} \mathrm{C}\) when the outdoor temperature is \(33.0^{\circ} \mathrm{C} .\) Heat, leaking into the house at the rate of 10500 joules per second, is removed by the air \(r\) conditioner. Assuming that the air conditioner is a Carnot air conditioner, what is the work per second that must be done by the electrical energy in order to keep the inside temperature constant?

Short Answer

Expert verified
The work needed per second is approximately 503.11 J/s.

Step by step solution

01

Understand the Problem

We need to find the work done per second (also known as power) by a Carnot air conditioner to maintain a constant inside temperature, given heat flows into the house at a rate of 10500 joules per second. The temperatures of the interior and exterior are required to compute the Carnot efficiency.
02

Convert Temperatures to Kelvin

The Carnot cycle calculations require temperatures in Kelvin. Convert the inside and outside temperatures: \[ T_{inside} = 19.0^{\circ}C + 273.15 = 292.15 K \]\[ T_{outside} = 33.0^{\circ}C + 273.15 = 306.15 K \]
03

Calculate the Carnot Efficiency

For a Carnot refrigerator (which is how a Carnot air conditioner can be viewed), its coefficient of performance (COP) is given by \[ \text{COP} = \frac{T_{inside}}{T_{outside} - T_{inside}} \]Substitute the values to find the COP: \[ \text{COP} = \frac{292.15}{306.15 - 292.15} = \frac{292.15}{14} \approx 20.87 \]
04

Relate COP to Power Needed

The COP tells us how many units of energy are moved per unit of work done by the air conditioner. The formula is \[ \text{COP} = \frac{Q_{in}}{W} \]where \(Q_{in}\) is the heat removed from the house and \(W\) is the work done. Rearranging gives:\[ W = \frac{Q_{in}}{\text{COP}} \]Substitute \(Q_{in} = 10500 \text{ J/s}\) and \(\text{COP} \approx 20.87\):\[ W = \frac{10500}{20.87} \approx 503.11 \text{ J/s} \]
05

Conclude with the Final Answer

The power required to keep the inside temperature constant, given the heat leakage rate and Carnot COP, is approximately \(503.11 \text{ J/s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Air Conditioning Efficiency
Air conditioning efficiency is a measure of how effectively an air conditioning system cools a space, relative to the energy it consumes. In the context of the Carnot cycle, this efficiency is dictated by the system's ability to perform work while moving heat from a lower temperature (inside the house) to a higher temperature (outside). The concept of efficiency here is represented by the Coefficient of Performance (COP) of a Carnot refrigerator, which we utilize to assess how much cooling effect (or heat absorption) is achieved per unit of work. This is crucial for determining how much electricity is needed by an air conditioner, which ultimately affects energy costs and environmental impact. A higher COP indicates a more efficient air conditioner, thus reducing costs for maintaining a comfortable indoor climate.
Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. When considering an air conditioning unit, thermodynamics explains how energy is transferred out of the house as heat is extracted to lower the interior temperature. The laws of thermodynamics govern the operation of air conditioners, notably the first and second laws:
  • The first law of thermodynamics, also known as the law of energy conservation, implies that the total energy absorbed by the system (from outside sources) is equivalent to the sum of the energy used to run the unit and the energy expelled as waste (heat).
  • The second law emphasizes that heat cannot spontaneously flow from a colder body to a warmer one without work being done, explaining why air conditioners require an energy input to move heat from a cool interior to a warmer exterior.
This framework is essential for understanding the limitations and potentials of air conditioning technology.
Carnot Refrigerator
A Carnot refrigerator or air conditioner is a theoretical model that represents the most efficient possible refrigeration cycle operating between two temperatures. It is based on the Carnot cycle, named after Sadi Carnot, who formulated laws that set the upper boundary for the efficiency of heat engines. The key feature of a Carnot refrigerator is its Coefficient of Performance (COP), which we calculate as the ratio of the heat removed from the cooler reservoir (inside temperature) to the work required to transfer this heat. Thus, for the air conditioner described in the exercise, the Carnot refrigerator serves as an ideal standard against which real-world refrigeration cycles can be measured. Though real appliances are not capable of achieving Carnot's theoretical efficiency due to unavoidable inefficiencies, understanding the Carnot cycle helps thermodynamic engineers design more effective cooling systems.
Temperature Conversion in Kelvin
Temperature conversion to Kelvin is fundamental when performing calculations in thermodynamics, because Kelvin is the SI unit for temperature and it avoids negative temperatures, simplifying mathematical operations. The Kelvin scale starts at absolute zero, the point at which all molecular motion ceases, thus allowing for consistent temperature measurements in energy calculations. For this exercise, converting Celsius to Kelvin involves adding 273.15 to the Celsius value:
  • For an indoor temperature of 19°C, the Kelvin equivalent is \[ T_{inside} = 19 + 273.15 = 292.15 \, K \]
  • Similarly, for the outdoor temperature of 33°C, the conversion is \[ T_{outside} = 33 + 273.15 = 306.15 \, K \]
These Kelvin values are essential for accurate computation in the Carnot cycle, enabling the precise calculation of the air conditioner's performance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The temperatures indoors and outdoors are 299 and \(312 \mathrm{K}\), respectively. A Carnot air conditioner deposits \(6.12 \times 10^{5} \mathrm{J}\) of heat outdoors. How much heat is removed from the house?

An irreversible engine operates between temperatures of 852 and 314 K. It absorbs 1285 J of heat from the hot reservoir and does 264 J of work. (a) What is the change \(\Delta S_{\text {univere } ~}\) in the entropy of the universe associated with the operation of this engine? (b) If the engine were reversible, what would be the magnitude \(|W|\) of the work it would have done, assuming that it operated between the same temperatures and absorbed the same heat as the irreversible engine? (c) Using the results of parts (a) and (b), find the difference between the work produced by the reversible and irreversible engines.

When a .22-caliber rifle is fired, the expanding gas from the burning gunpowder creates a pressure behind the bullet. This pressure cause the force that pushes the bullet through the barrel. The barrel has length of \(0.61 \mathrm{m}\) and an opening whose radius is \(2.8 \times 10^{-3} \mathrm{m} .\) A bulle (mass \(=2.6 \times 10^{-3} \mathrm{kg}\) ) has a speed of \(370 \mathrm{m} / \mathrm{s}\) after passing through th barrel. Ignore friction and determine the average pressure of the expanc ing gas.

A piece of aluminum has a volume of \(1.4 \times 10^{-3} \mathrm{m}^{3} .\) The coeffi-cient of volume expansion for aluminum is \(\beta=69 \times 10^{-6}\left(\mathrm{C}^{\circ}\right)^{-1} .\) The temperature of this object is raised from 20 to \(320^{\circ} \mathrm{C}\). How much work is done by the expanding aluminum if the air pressure is \(1.01 \times 10^{5} \mathrm{Pa} ?\)

One-half mole of a monatomic ideal gas absorbs \(1200 \mathrm{J}\) of heat while 2500 J of work is done by the gas. (a) What is the temperature change of the gas? (b) Is the change an increase or a decrease?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.