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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

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
Eunice Reply
what is a capacitor?
Raymond Reply
Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
Maria Reply
please solve
Sharon
8m/s²
Aishat
What is Thermodynamics
Muordit
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
Saheed Reply
50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
Ramon Reply
I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
Someone
about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
Someone
definitely of physics
Haryormhidey Reply
how many start and codon
Esrael Reply
what is field
Felix Reply
physics, biology and chemistry this is my Field
ALIYU
field is a region of space under the influence of some physical properties
Collete
what is ogarnic chemistry
WISDOM Reply
determine the slope giving that 3y+ 2x-14=0
WISDOM
Another formula for Acceleration
Belty Reply
a=v/t. a=f/m a
IHUMA
innocent
Adah
pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
Nassze Reply
how do lnternal energy measures
Esrael
Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
JALLAH Reply
No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Robert
like charges repel while unlike charges atttact
Raymond
What is specific heat capacity
Destiny Reply
Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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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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