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Strategy

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 did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
what is the solution
bill
how did you arrive at this answer?
bill
-24m+3+3mÁ^2
Susan
i really want to learn
Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
Hw did u arrive to this answer.
Aphelele
hi
Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
Hi
Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
Marry Reply
how far
Abubakar
cool u
Enock
state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
Abegail Reply
hello
BenJay
hi
Method
I am eliacin, I need your help in maths
Rood
how can I help
Sir
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Amoon
however, may I ask you some questions about Algarba?
Amoon
hi
Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
difference between rational and irrational numbers
Arundhati Reply
When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
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Solomon Reply
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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