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

Short Answer

Expert verified

a.The work done by the gas will be numerically equal to the area under the graph is10J,Qhis124J.

b.The efficiency for a heat engine is8%.

Step by step solution

01

Calculation work done (part a)

a.

The work done by the gas will be quantitatively equal to the area beneath the graph.

Work is,

Wout=12·200·103·100·10-6=10J

The heat injected into the system will be converted to mechanical energy which is equal to 10 J and heat rejected out which is equal to 114 J.

We can write it in the following way.

QH=Wout+Qc

QH=10+114=124J

02

Calculation for efficiency of heat engine(part b)

b.

The efficiency can be calculated as,

η=WoutQH

η=10124=8%

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Most popular questions from this chapter

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FIGURE P21.60is the pVdiagram of Example 21.2, but now the device is operated in reverse.

a. During which processes is heat transferred into the gas?

b. Is thisQH, heat extracted from a hot reservoir, or QC, heat extracted from a cold reservoir? Explain.

c. Determine the values ofQHandQC.

Hint: The calculations have been done in Example 21.2and do not need to be repeated. Instead, you need to determine which processes now contribute to QHand which to QC.

d. Is the area inside the curve Winor Wout? What is its value?

e. The device is now being operated in a ccw cycle. Is it a refrigerator? Explain.

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