A heat engine extracts 55KJof heat from the hot reservoir each cycle and exhausts 40KJof heat. What are (a) the thermal efficiency and (b) the work done per cycle?

Short Answer

Expert verified

(a) The thermal efficiency is 27 %

(b) The work done per cycle is 15kJ.

Step by step solution

01

Step : 1 Introduction (part a)

(a) The ratio of work done per cycle to amount of heat retrieved from the warm reservoir Q1is the efficiency of the engine. The difference between the extracted heat and the work done per cycle equals the work done per cycle. Q1and exhausted heat Q2so we have

localid="1649241559768" η=WQ1

work done W per cycle is the difference of heat extracted and heat dumped into cold reservoir .

W=55kJ-40kJ.=15kJ.

Substitute 15 kJ for W and 55 kJ for Q1in the equation of efficiency above.

η=15kJ55kJ×100=27%

Q1
02

Step :2 Explanation (part b)

(b) The work done per cycle, as explained is difference of heat extracted and heat dumped into cold reservoir.

W=Q1Q2

localid="1649242118588" =55kJ-40kJ=15kJ

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Most popular questions from this chapter

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?

What are (a) WoutandQHand (b) the thermal efficiency for the heat engine shown in FIGUREEX21.14?

The ideal gas in a Carnot engine extracts 1000Jof heat energy during the isothermal expansion at 300°C. How much heat energy is exhausted during the isothermal compression at50°C?

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

Do the energy-transfer diagrams inFIGUREQ21.8represent possible heat engines? If not, what is wrong?

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