A heat engine with a thermal efficiency of 40%does 100Jof work per cycle. How much heat is (a) extracted from the hot reservoir and (b) exhausted to the cold reservoir per cycle?

Short Answer

Expert verified

(a) The heat extracted from the hot reservoir is 250J.

(b) The heat exhausted to the cold reservoir is 150J.

Step by step solution

01

Step :1  Hot reservoir (part a)

(a) The work done per cycle equals Wand the received heat i.e. the heat extracted from the hot reservoir Q1,efficiency of the engine can be expressed as follows.

η=WQ1

This yields

Q1=Wη

Substitute 100 W for W and .4 for ηin the equation above(η=40%=0.4)

Q1=100W0.4=250W.

02

Step :2 Difference between heat extracted and work done   (part b)

(b) The difference in heat extracted from the hot reservoir and heat exhausted to the cold reservoir.Q2per cycle is equal to the total work done per cycle. Therefore

W=Q1Q2

localid="1648496332888">Q2=Q1W=250W-100W=150J

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

A gas following the pVtrajectory of FIGURE EX21.12does 60Jof work per cycle. What is pmax?



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?

A 32%-efficient electric power plant produces 900MWof electric power and discharges waste heat into 20°C ocean water. Suppose the waste heat could be used to heat homes during the winter instead of being discharged into the ocean. A typical American house requires an average of 20kW for heating. How many homes could be heated with the waste heat of this one power plant?

The electric output of a power plant is 750MW. Cooling water flows through the power plant at the rate 1.0*108L/h. The cooling water enters the plant at 16°Cand exits at27°C. What is the power plant’s thermal efficiency?

A heat engine satisfies Wout=Qout Why is there no ΔEth term in this relationship?

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