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0 = Q W , size 12{0=Q - W} {}

so that

W = Q . size 12{W=Q} {}

Thus the net work done by the system equals the net heat transfer into the system, or

W = Q h Q c (cyclical process), size 12{W=Q rSub { size 8{h} } - Q rSub { size 8{c} } } {}

just as shown schematically in [link] (b). The problem is that in all processes, there is some heat transfer Q c size 12{Q rSub { size 8{c} } } {} to the environment—and usually a very significant amount at that.

In the conversion of energy to work, we are always faced with the problem of getting less out than we put in. We define conversion efficiency Eff size 12{ ital "Eff"} {} to be the ratio of useful work output to the energy input (or, in other words, the ratio of what we get to what we spend). In that spirit, we define the efficiency of a heat engine to be its net work output W size 12{W} {} divided by heat transfer to the engine Q h size 12{Q rSub { size 8{h} } } {} ; that is,

Eff = W Q h . size 12{ ital "Eff"= { {W} over {Q rSub { size 8{h} } } } } {}

Since W = Q h Q c size 12{W=Q rSub { size 8{h} } -Q rSub { size 8{c} } } {} in a cyclical process, we can also express this as

Eff = Q h Q c Q h = 1 Q c Q h (cyclical process), size 12{ ital "Eff"= { {Q rSub { size 8{h} } - Q rSub { size 8{c} } } over {Q rSub { size 8{h} } } } =1 - { {Q rSub { size 8{c} } } over {Q rSub { size 8{h} } } } } {}

making it clear that an efficiency of 1, or 100%, is possible only if there is no heat transfer to the environment ( Q c = 0 size 12{Q rSub { size 8{c} } =0} {} ). Note that all Q size 12{Q} {} s are positive. The direction of heat transfer is indicated by a plus or minus sign. For example, Q c size 12{Q rSub { size 8{c} } } {} is out of the system and so is preceded by a minus sign.

Daily work done by a coal-fired power station, its efficiency and carbon dioxide emissions

A coal-fired power station is a huge heat engine. It uses heat transfer from burning coal to do work to turn turbines, which are used to generate electricity. In a single day, a large coal power station has 2 . 50 × 10 14 J size 12{2 "." "50" times "10" rSup { size 8{"14"} } J} {} of heat transfer from coal and 1 . 48 × 10 14 J size 12{1 "." "48" times "10" rSup { size 8{"14"} } J} {} of heat transfer into the environment. (a) What is the work done by the power station? (b) What is the efficiency of the power station? (c) In the combustion process, the following chemical reaction occurs: C + O 2 CO 2 size 12{C+O rSub { size 8{2} } rightarrow "CO" rSub { size 8{2} } } {} . This implies that every 12 kg of coal puts 12 kg + 16 kg + 16 kg = 44 kg of carbon dioxide into the atmosphere. Assuming that 1 kg of coal can provide 2 . 5 × 10 6 J size 12{2 "." 5 times "10" rSup { size 8{6} } J} {} of heat transfer upon combustion, how much CO 2 size 12{"CO" rSub { size 8{2} } } {} is emitted per day by this power plant?

Strategy for (a)

We can use W = Q h Q c size 12{W=Q rSub { size 8{h} } - Q rSub { size 8{c} } } {} to find the work output W size 12{W} {} , assuming a cyclical process is used in the power station. In this process, water is boiled under pressure to form high-temperature steam, which is used to run steam turbine-generators, and then condensed back to water to start the cycle again.

Solution for (a)

Work output is given by:

W = Q h Q c . size 12{W=Q rSub { size 8{h} } - Q rSub { size 8{c} } } {}

Substituting the given values:

W = 2 . 50 × 10 14 J 1 . 48 × 10 14 J = 1 . 02 × 10 14 J . alignl { stack { size 12{W=2 "." "50"´"10" rSup { size 8{"14"} } " J" +- 1 "." "48"´"10" rSup { size 8{"14"} } " J"} {} #=1 "." "02"´"10" rSup { size 8{"14"} } " J" "." {} } } {}

Strategy for (b)

The efficiency can be calculated with Eff = W Q h size 12{ ital "Eff"= { {W} over {Q rSub { size 8{h} } } } } {} since Q h size 12{Q rSub { size 8{h} } } {} is given and work W size 12{W} {} was found in the first part of this example.

Solution for (b)

Efficiency is given by: Eff = W Q h size 12{ ital "Eff"= { {W} over {Q rSub { size 8{h} } } } } {} . The work W was just found to be 1.02 × 10 14 J , and Q h size 12{Q rSub { size 8{h} } } {} is given, so the efficiency is

Eff = 1 . 02 × 10 14 J 2 . 50 × 10 14 J = 0 . 408 , or  40 . 8% alignl { stack { size 12{ ital "Eff"= { {1 "." "02" times "10" rSup { size 8{"14"} } J} over {2 "." "50" times "10" rSup { size 8{"14"} } J} } } {} #=0 "." "408"", or ""40" "." 8% {} } } {}

Strategy for (c)

The daily consumption of coal is calculated using the information that each day there is 2 . 50 × 10 14 J size 12{2 "." "50"´"10" rSup { size 8{"14"} } " J"} {} of heat transfer from coal. In the combustion process, we have C + O 2 CO 2 size 12{C+O rSub { size 8{2} } rightarrow "CO" rSub { size 8{2} } } {} . So every 12 kg of coal puts 12 kg + 16 kg + 16 kg = 44 kg of CO 2 size 12{"CO" rSub { size 8{2} } } {} into the atmosphere.

Solution for (c)

The daily coal consumption is

2 . 50 × 10 14 J 2 . 50 × 10 6 J/kg = 1 . 0 × 10 8 kg. size 12{ { {2 "." "50"´"10" rSup { size 8{"14"} } " J"} over {2 "." "50"´"10" rSup { size 8{6} } " J/kg"} } =1 "." 0´"10" rSup { size 8{7} } " J/kg"} {}

Assuming that the coal is pure and that all the coal goes toward producing carbon dioxide, the carbon dioxide produced per day is

Practice Key Terms 4

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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