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Chapter 15: Q14PE (page 550)

Calculate the net work output of a heat engine following path ABCDA in the figure below.

Short Answer

Expert verified

The net work output of this heat engine is 4800 J.

Step by step solution

01

Analyze the figure and find an expression to calculate the net work output.

The figure is a PV diagram. That means the graph of pressure versus volume. So, the area of the curve in the PV diagram will give the net work output. Therefore, you need to calculate the area of quadrilateral ABCD.

Αrea of ABCD= Area of DABD +Area of DBCD

The area of a triangle is equal to the half of the product of its base and height.

Here, b is the base and h is the height of the triangle.

02

Calculate the area of ABCD.

The base and height of the triangles correspond to the change in pressure and the change in volume, respectively.

Therefore, these two triangles have equal area, and adding them together will give the area of the quadrilateral.

Therefore, the work done along the path ABCDA is 4800 J.

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Most popular questions from this chapter

Assume that the turbines at a coal-powered power plant wereupgraded, resulting in an improvement in efficiency of \({\bf{3}}.{\bf{32}}\% \). Assumethat prior to the upgrade the power station had an efficiency of \({\bf{36}}\% \) andthat the heat transfer into the engine in one day is still the same at\({\rm{2}}{\rm{.50 \times 1}}{{\rm{0}}^{{\rm{14}}}}{\rm{J}}\). (a) How much more electrical energy is produced due tothe upgrade? (b) How much less heat transfer occurs to the environmentdue to the upgrade?

What is the best coefficient of performance possible for a hypothetical refrigerator that could make liquid nitrogen at −200ºC and has heat transfer to the environment at 35.0ºC ?

Show that the coefficients of performance of refrigerators and heat pumps are related by\({\bf{CO}}{{\bf{P}}_{{\bf{ref}}}} = {\bf{CO}}{{\bf{P}}_{{\bf{hp}}}} - {\bf{1}}\).

Start with the definitions of the\({\bf{COP}}\)s and the conservation of energy relationship between\({{\bf{Q}}_{\bf{h}}}\),\({{\bf{Q}}_{\bf{c}}}\), and\({\bf{W}}\).

A\({\bf{4}}\)-ton air conditioner removes\({\bf{5}}.{\bf{06}} \times {\bf{1}}{{\bf{0}}^{\bf{7}}}\;{\bf{J}}\)(\({\bf{48}},{\bf{000}}\)British thermal units) from a cold environment in\({\bf{1}}.{\bf{00}}\;{\bf{h}}\). (a) What energy input in joules is necessary to do this if the air conditioner has an energy efficiency rating ( EER) of\({\bf{12}}.{\bf{0}}\)? (b) What is the cost of doing this if the work costs\({\bf{10}}.{\bf{0}}\)cents per\({\bf{3}}.{\bf{60}} \times {\bf{1}}{{\bf{0}}^{\bf{6}}}\;{\bf{J}}\)(one kilowatt-hour)? (c) Discuss whether this cost seems realistic. Note that the energy efficiency rating ( EER) of an air conditioner or refrigerator is defined to be the number of British thermal units of heat transfer from a cold environment per hour divided by the watts of power input.

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