An air conditioner removes 5.0×105J/minof heat from a house and exhausts 8.0×105J/minto the hot outdoors.

a. How much power does the air conditioner’s compressor require?

b. What is the air conditioner’s coefficient of performance?

Short Answer

Expert verified

(A) Air conditioner's require 5kW,

(B) Air conditioner coefficient of performance1.7

Step by step solution

01

Step :1 Introduction  (part a)

(a) For a time of t=60swe have Qc=5105JandQh=8105J. The corresponding powers are therefore

Pc=Qct=510560=8333WPh=Qht=810560=13333W

Pcis the power at which heat is removed and Ph is the power at which heat is thrown into the environment . It will be sum of heat extracted and electrical power used up.

The power used up by the system can be calculated by

localid="1648567318756">P=PhPc=133338333=5000W

02

Step :2 Explanation (part b)

(b) The coefficient performance can be calculated as

KPowerextracedpowerconsumed

K=QcW=PctPt=PcP

Numerically we will have

K=833350001.7

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 Vaia!

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

Do the energy-transfer diagrams inFIGUREQ21.9represent possible refrigerators? If not, what is wrong?

How much heat is exhausted to the cold reservoir by the heat engine shown in FIGUREEX21.16?


FigureP21.46shows a Carnot heat engine driving a Carnot refrigerator.

a. Determine Q2,Q3and Q4.

b. Is Q3greater than, less than, or equal toQ1?

c. Do these two devices, when operated together in this way, violate the second law?


What are (a) the heat extracted from the cold reservoir and (b) the coefficient of performance for the refrigerator shown in Figure Ex-21.20?

A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine running backward is called a heat pump if its purpose is to exhaust warm air into the hot reservoir. Heat pumps are widely used for home heating. You can think of a heat pump as a refrigerator that is cooling the already cold outdoors and, with its exhaust heat QH, warming the indoors. Perhaps this seems a little silly, but consider the following. Electricity can be directly used to heat a home by passing an electric current through a heating coil. This is a direct, 100%conversion of work to heat. That is, 15kWof electric power (generated by doing work at the rate of 15kJ/sat the power plant) produces heat energy inside the home at a rate of 15kJ/s. Suppose that the neighbor’s home has a heat pump with a coefficient of performance of 5.0, a realistic value. Note that “what you get” with a heat pump is heat delivered, QH, so a heat pump’s coefficient of performance is defined asK=QH/Win.

a. How much electric power inkWdoes the heat pump use to deliver 15kJ/sof heat energy to the house?

b. An average price for electricity is about 40MJper dollar. A furnace or heat pump will run typically 250 hours per month during the winter. What does one month’s heating cost in the home with a 15kW electric heater and in the home of the neighbor who uses a heat pump?

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.

Sign-up for free