In Problems 65through 68you are given the equation(s) used to solve a problem. For each of these, you are to

a. Write a realistic problem for which this is the correct equation(s).

b. Finish the solution of the problem.

0.20=1-QC/QH

Wout=QH-QC=20J

Short Answer

Expert verified

a. The heat will be wanted to identify.

b. The heat value is100J.

Step by step solution

01

Step 1: 

Camot's engine is efficiency is,

The heat engine work done is,

W = Q11 - QC ................................................(2)

Here, Qc is the heat removed from the engine and Q11is the heat input to the engine.

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

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?

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?

The power output of a car engine running at 2400rpmis 500kW. How much (a) work is done and (b) heat is exhausted per cycle if the engine's thermal efficiency is 20%? Give your answers inKJ.

A Carnot heat engine with thermal efficiency 13is run backward as a Carnot refrigerator. What is the refrigerator’s coefficient of performance?

(T2+273)K=200kPa500kPa×1×(400+273)K

In Problems 67 through 70 you are given the equation(s) used to solve a problem. For each of these, you are to

a. Write a realistic problem for which this is the correct equation(s).

b. Draw a pVdiagram.

c. Finish the solution of the problem.

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