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Now let us take a look at the change in entropy of a Carnot engine and its heat reservoirs for one full cycle. The hot reservoir has a loss of entropy Δ S h = Q h / T h size 12{ΔS rSub { size 8{h} } = - Q rSub { size 8{h} } /T rSub { size 8{h} } } {} , because heat transfer occurs out of it (remember that when heat transfers out, then Q size 12{Q} {} has a negative sign). The cold reservoir has a gain of entropy Δ S c = Q c / T c size 12{ΔS rSub { size 8{c} } =Q rSub { size 8{c} } /T rSub { size 8{c} } } {} , because heat transfer occurs into it. (We assume the reservoirs are sufficiently large that their temperatures are constant.) So the total change in entropy is

Δ S tot = Δ S h + Δ S c . size 12{DS rSub { size 8{"tot"} } =DS rSub { size 8{h} } +DS rSub { size 8{c} } "." } {}

Thus, since we know that Q h / T h = Q c / T c size 12{Q rSub { size 8{h} } /T rSub { size 8{h} } =Q rSub { size 8{c} } /T rSub { size 8{c} } } {} for a Carnot engine,

Δ S tot =– Q h T h + Q c T c = 0 . size 12{DS rSub { size 8{"tot"} } "=-" { {Q rSub { size 8{h} } } over {T rSub { size 8{h} } } } + { {Q rSub { size 8{c} } } over {T rSub { size 8{c} } } } =0 "." } {}

This result, which has general validity, means that the total change in entropy for a system in any reversible process is zero.

The entropy of various parts of the system may change, but the total change is zero. Furthermore, the system does not affect the entropy of its surroundings, since heat transfer between them does not occur. Thus the reversible process changes neither the total entropy of the system nor the entropy of its surroundings. Sometimes this is stated as follows: Reversible processes do not affect the total entropy of the universe. Real processes are not reversible, though, and they do change total entropy. We can, however, use hypothetical reversible processes to determine the value of entropy in real, irreversible processes. The following example illustrates this point.

Entropy increases in an irreversible (real) process

Spontaneous heat transfer from hot to cold is an irreversible process. Calculate the total change in entropy if 4000 J of heat transfer occurs from a hot reservoir at T h = 600 K 327º C size 12{T rSub { size 8{h} } ="600"" K " left ("327"°C right )} {} to a cold reservoir at T c = 250 K 23º C size 12{T rSub { size 8{c} } ="250"" K " left (-"23" "." 0°C right )} {} , assuming there is no temperature change in either reservoir. (See [link] .)

Strategy

How can we calculate the change in entropy for an irreversible process when Δ S tot = Δ S h + Δ S c size 12{ΔS rSub { size 8{"tot"} } =ΔS rSub { size 8{h} } +ΔS rSub { size 8{c} } } {} is valid only for reversible processes? Remember that the total change in entropy of the hot and cold reservoirs will be the same whether a reversible or irreversible process is involved in heat transfer from hot to cold. So we can calculate the change in entropy of the hot reservoir for a hypothetical reversible process in which 4000 J of heat transfer occurs from it; then we do the same for a hypothetical reversible process in which 4000 J of heat transfer occurs to the cold reservoir. This produces the same changes in the hot and cold reservoirs that would occur if the heat transfer were allowed to occur irreversibly between them, and so it also produces the same changes in entropy.

Solution

We now calculate the two changes in entropy using Δ S tot = Δ S h + Δ S c size 12{DS rSub { size 8{"tot"} } =DS rSub { size 8{h} } +DS rSub { size 8{c} } } {} . First, for the heat transfer from the hot reservoir,

Δ S h = Q h T h = 4000 J 600 K = 6 . 67 J/K . size 12{DS rSub { size 8{h} } = { {-Q rSub { size 8{h} } } over {T rSub { size 8{h} } } } = { {-"4000"" J"} over {"600 K"} } "=-"6 "." "67"" J/K"} {}

And for the cold reservoir,

Δ S c = Q c T c = 4000 J 250 K = 16 . 0 J/K . size 12{DS rSub { size 8{c} } = { {-Q rSub { size 8{c} } } over {T rSub { size 8{c} } } } = { {"4000"" J"} over {"250 K"} } ="16" "." 0" J/K"} {}

Thus the total is

Δ S tot = Δ S h + Δ S c = ( 6 . 67 +16 . 0 ) J/K = 9.33 J/K. alignl { stack { size 12{DS rSub { size 8{"tot"} } =DS rSub { size 8{h} } +DS rSub { size 8{c} } } {} #" =" \( +- 6 "." "67 +16" "." 0 \) " J/K" {} # " =9" "." "33 J/K" "." {}} } {}

Discussion

There is an increase in entropy for the system of two heat reservoirs undergoing this irreversible heat transfer. We will see that this means there is a loss of ability to do work with this transferred energy. Entropy has increased, and energy has become unavailable to do work.

Questions & Answers

How is the de Broglie wavelength of electrons related to the quantization of their orbits in atoms and molecules?
Larissa Reply
How do you convert 0.0045kgcm³ to the si unit?
EDYKING Reply
how many state of matter do we really have like I mean... is there any newly discovered state of matter?
Falana Reply
I only know 5: •Solids •Liquids •Gases •Plasma •Bose-Einstein condensate
Thapelo
Alright Thank you
Falana
Which one is the Bose-Einstein
James
can you explain what plasma and the I her one you mentioned
Olatunde
u can say sun or stars are just the state of plasma
Mohit
but the are more than seven
Issa
what the meaning of continuum
Akhigbe Reply
What state of matter is fire
Thapelo Reply
fire is not in any state of matter...fire is rather a form of energy produced from an oxidising reaction.
Xenda
Isn`t fire the plasma state of matter?
Walter
all this while I taught it was plasma
Victor
How can you define time?
Thapelo Reply
Time can be defined as a continuous , dynamic , irreversible , unpredictable quantity .
Tanaya
unpredictable? but I can say after one o'clock its going to be two o'clock predictably!
Victor
what is the relativity of physics
Paul Reply
How do you convert 0.0045kgcm³ to the si unit?
flint
What is the formula for motion
Anthony Reply
V=u+at V²=u²-2as
flint
S=ut+½at
flint
they are eqns of linear motion
King
S=Vt
Thapelo
v=u+at s=ut+at^\2 v^=u^+2as where ^=2
King
hi
Mehadi
hello
King
Explain dopplers effect
Jennifer Reply
Not yet learnt
Bob
Explain motion with types
Bob
Acceleration is the change in velocity over time. Given this information, is acceleration a vector or a scalar quantity? Explain.
Alabi Reply
Scalar quantity Because acceleration has only magnitude
Bob
acleration is vectr quatity it is found in a spefied direction and it is product of displcemnt
bhat
its a scalar quantity
Paul
velocity is speed and direction. since velocity is a part of acceleration that makes acceleration a vector quantity. an example of this is centripetal acceleration. when you're moving in a circular patter at a constant speed, you are still accelerating because your direction is constantly changing.
Josh
acceleration is a vector quantity. As explained by Josh Thompson, even in circular motion, bodies undergoing circular motion only accelerate because on the constantly changing direction of their constant speed. also retardation and acceleration are differentiated by virtue of their direction in
fitzgerald
respect to prevailing force
fitzgerald
What is the difference between impulse and momentum?
Manyo
Momentum is the product of the mass of a body and the change in velocity of its motion. ie P=m(v-u)/t (SI unit is kgm/s). it is literally the impact of collision from a moving body. While Impulse is the product of momentum and time. I = Pt (SI unit is kgm) or it is literally the change in momentum
fitzgerald
Or I = m(v-u)
fitzgerald
the tendency of a body to maintain it's inertia motion is called momentum( I believe you know what inertia means) so for a body to be in momentum it will be really hard to stop such body or object..... this is where impulse comes in.. the force applied to stop the momentum of such body is impulse..
Pelumi
Calculation of kinetic and potential energy
dion Reply
K.e=mv² P.e=mgh
Malia
K is actually 1/2 mv^2
Josh
what impulse is given to an a-particle of mass 6.7*10^-27 kg if it is ejected from a stationary nucleus at a speed of 3.2*10^-6ms²? what average force is needed if it is ejected in approximately 10^-8 s?
John
speed=velocity÷time velocity=speed×time=3.2×10^-6×10^-8=32×10^-14m/s impulse [I]=∆momentum[P]=mass×velocity=6.7×10^-27×32×10^-14=214.4×10^-41kg/ms force=impulse÷time=214.4×10^-41÷10^-8=214.4×10^-33N. dats how I solved it.if wrong pls correct me.
Melody
what is sound wave
Nworu Reply
sound wave is a mechanical longitudinal wave that transfers energy from one point to another
Ogor
its a longitudnal wave which is associted wth compresion nad rearfractions
bhat
what is power
PROMISE Reply
it's also a capability to do something or act in a particular way.
Kayode
Newton laws of motion
Mike
power also known as the rate of ability to do work
Slim
power means capabilty to do work p=w/t its unit is watt or j/s it also represents how much work is done fr evry second
bhat
what does fluorine do?
Cheyanne Reply
strengthen and whiten teeth.
Gia
a simple pendulum make 50 oscillation in 1minute, what is it period of oscillation?
Nansing Reply
length of pendulm?
bhat
Practice Key Terms 3

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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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