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In part (a), we must first find the net heat transfer and net work done from the given information. Then the first law of thermodynamics ( Δ U = Q W size 12{ΔU=Q - W} {} ) can be used to find the change in internal energy. In part (b), the net heat transfer and work done are given, so the equation can be used directly.

Solution for (a)

The net heat transfer is the heat transfer into the system minus the heat transfer out of the system, or

Q = 40 . 00 J 25 . 00 J = 15 . 00 J. size 12{Q="40" "." "00"" J"-"25" "." "00"" J"="15" "." "00"" J"} {}

Similarly, the total work is the work done by the system minus the work done on the system, or

W = 10 . 00 J 4 . 00 J = 6 . 00 J. size 12{W="10" "." "00"" J"-4 "." "00"" J"=6 "." "00"" J"} {}

Thus the change in internal energy is given by the first law of thermodynamics:

Δ U = Q W = 15 . 00 J 6 . 00 J = 9 . 00 J. size 12{DU=Q-W="15" "." "00"" J"-6 "." "00"" J"=9 "." "00"" J"} {}

We can also find the change in internal energy for each of the two steps. First, consider 40.00 J of heat transfer in and 10.00 J of work out, or

Δ U 1 = Q 1 W 1 = 40 . 00 J 10 . 00 J = 30 . 00 J. size 12{DU rSub { size 8{1} } =Q rSub { size 8{1} } -W rSub { size 8{1} } ="40" "." "00"" J"-"10" "." "00"" J"="30" "." "00"" J"} {}

Now consider 25.00 J of heat transfer out and 4.00 J of work in, or

Δ U 2 = Q 2 W 2 = - 25 . 00 J ( 4 . 00 J ) = –21.00 J. size 12{DU rSub { size 8{2} } =Q rSub { size 8{2} } -W rSub { size 8{2} } "=-""25" "." "00"" J"- \( -4 "." "00"" J" \) "=-""21" "." "00"" J"} {}

The total change is the sum of these two steps, or

Δ U = Δ U 1 + Δ U 2 = 30 . 00 J + 21 . 00 J = 9 . 00 J. size 12{DU=DU rSub { size 8{1} } +DU rSub { size 8{2} } ="30" "." "00"" J"+ left (-"21" "." "00"" J" right )=9 "." "00"" J"} {}

Discussion on (a)

No matter whether you look at the overall process or break it into steps, the change in internal energy is the same.

Solution for (b)

Here the net heat transfer and total work are given directly to be Q = 150 . 00 J size 12{Q"=-""150" "." "00"" J"} {} and W = 159 . 00 J size 12{W"=-""159" "." "00"" J"} {} , so that

Δ U = Q W = 150 . 00 J ( 159 . 00 J ) = 9 . 00 J. size 12{DU=Q-W"=-""150" "." "00"" J"- \( -"159" "." "00"" J" \) =9 "." "00"" J"} {}

Discussion on (b)

A very different process in part (b) produces the same 9.00-J change in internal energy as in part (a). Note that the change in the system in both parts is related to Δ U size 12{ΔU} {} and not to the individual Q size 12{Q} {} s or W size 12{W} {} s involved. The system ends up in the same state in both (a) and (b). Parts (a) and (b) present two different paths for the system to follow between the same starting and ending points, and the change in internal energy for each is the same—it is independent of path.

The first part of the picture shows a system in the form of a circle for explanation purposes. The heat entering and work done are represented by bold arrows. A quantity of heat Q in equals forty joules, is shown to enter the system and Q out equals negative twenty five joules is shown to leave the system. The energy of the system in is marked as fifteen joules. At the right-hand side of the circle, a work W in equals negative four joules is shown to be applied on the system and a work W out equals ten joules is shown to leave the system. The energy of the system out is marked as six joules. The second part of the picture shows a system in the form of a circle for explanation purposes. The heat entering and work done are represented by bold arrows. A work of negative one hundred fifty nine is shown to enter the system. The energy in the system is shown as one hundred fifty nine joules. The out energy of the system is one hundred fifty joules. A heat Q out of negative one hundred fifty joules is shown to leave the system as an outward arrow.
Two different processes produce the same change in a system. (a) A total of 15.00 J of heat transfer occurs into the system, while work takes out a total of 6.00 J. The change in internal energy is Δ U = Q W = 9 . 00 J size 12{DU=Q-W=9 "." "00"" J"} {} . (b) Heat transfer removes 150.00 J from the system while work puts 159.00 J into it, producing an increase of 9.00 J in internal energy. If the system starts out in the same state in (a) and (b), it will end up in the same final state in either case—its final state is related to internal energy, not how that energy was acquired.

Human metabolism and the first law of thermodynamics

Human metabolism is the conversion of food into heat transfer, work, and stored fat. Metabolism is an interesting example of the first law of thermodynamics in action. We now take another look at these topics via the first law of thermodynamics. Considering the body as the system of interest, we can use the first law to examine heat transfer, doing work, and internal energy in activities ranging from sleep to heavy exercise. What are some of the major characteristics of heat transfer, doing work, and energy in the body? For one, body temperature is normally kept constant by heat transfer to the surroundings. This means Q size 12{Q} {} is negative. Another fact is that the body usually does work on the outside world. This means W size 12{W} {} is positive. In such situations, then, the body loses internal energy, since Δ U = Q W size 12{ΔU=Q - W} {} is negative.

Questions & Answers

how to know photocatalytic properties of tio2 nanoparticles...what to do now
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s. Reply
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Mostly, they use nano carbon for electronics and for materials to be strengthened.
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s. Reply
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of graphene you mean?
or in general
in general
Graphene has a hexagonal structure
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Himanshu Reply
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what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
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this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
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Source:  OpenStax, College physics: physics of california. OpenStax CNX. Sep 30, 2013 Download for free at http://legacy.cnx.org/content/col11577/1.1
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