∫∫)(

==

f

i

T

T

VR

T

dTTC

TdQ

S

Entropy Change (at Constant Volume)

For an ideal gas, CV (and CP) are constant with T.

But in the general case, CV (and CP) are functions of T.

Then dQR = CV(T)dT, and

The case of a system at Ti in contact with a reservoir at Tf.

Thermodynamic Universe

System

Res. 1

Q1

Res. 2

Q2

Res. 3

Q3

Adiabatic Wall

Not necessarily the same as THE Universe

Entropy Change, S, in a Thermodynamic Universe

Suniv = Ssyst + Sres1 + Sres2 + ….

Entropy can decrease or increase within the various parts of the thermodynamic universe.

For an irreversible process within the universe, Suniv 0.

For a reversible process within the universe, Suniv =0.

Thus, S goes to a maximum within a thermodynamic universe (i.e., a thermally-isolated system). (2nd Law!)

Implication: In a thermodynamic universe, a higher entropy state must follow a lower entropy state.

Entropy gives us the direction of time!

Example of an Entropy Change in a Thermodynamic Universe

Question: What is the entropy change of the thermodynamic universe (Suniverse) consisting of 0.5 kg water and a hotplate (reservoir), when the water is heated from an initial temperature of 293 K to a final temperature of 373 K?

Hot Reservoir, Tf = 373 K

Ti = 293 K

Suniverse = Swater + Shotplate

Irreversible process: Therefore expect that

Suniverse 0.

Suniverse = Ssystem + Sreservoir(s)

Previously, we calculated Swater for this same problem:

)ln(===i

fP

PRwater

TT

CTdTC

TdQ

S i

f Tf

Ti

For 0.5 kg of water, Swater = +0.241 CP = +506 J/K

What is Shot plate ?We note that heat flows out of it, so we expect that Shot plate < 0.

But we note that at the initial point, the heat flow is irreversible because of the large temperature difference between the “system” and the “reservoir”.

We imagine an equivalent process in which the water is placed in contact with a series of hot plates, each at a temperature that is dT greater than the previous one:

Ti = T1T2 = T1+dT T3 = T2+dT

The total heat flow will be equivalent to an experiment in which water at Ti is placed into contact with a reservoir at Tf.

…Tf

i

We can apply our “usual” equation for S:

fR

f

Rhotplate

TQ

dQTT

dQS =

1==

i

f f

What is Q, the total heat flow out of the reservoir (hotplate)?

If no heat is “wasted”, e.g. flows into the atmosphere, etc., then the total heat out of the reservoir is equal to the heat required to heat the water.

Let Cp be the heat capacity of the water (system):

)(== ifPP TTCTCQ

f

ifP

f

hotplate

TTTC

TQ

S)(

==

What is the entropy change of the thermodynamic Universe?

Suniverse = Swater + Shotplate

PPuniverse CCS 027.0+=]

37380

)293373

[ln(=

Entropy change is +ve as expected for an irreversible process!

Something to ponder: What if the system is cooling when in contact with the reservoir? What is Suniverse?

)373

293373()

293373

ln(+= PPuniverse CCS

Plug in some numbers:

TR = Reservoir T

TS = System T

C = System’s heat capacity

General Result

R

RS

S

RUniv

T

TT

T

TCS ln

Reservoir, TR

System, TS

SUniv is always ≥0

+

-

What is the entropy change of an equivalent two-stage process?

Two reservoirs - rather than one - are used to heat the water.

Thp1 = 323 K

Ti = 293 K

Hotplate 1

Thp2 = 373 K

Hotplate 2

Suniverse = Swater + Shp1 + Shp2

We observe that the “reversible” heat flow into the water is the same for two hotplates as for one - approximated as a series of reversible steps. Therefore Swater is as before! The S for the water is path independent.

What is the entropy change of each reservoir?

Hotplate 1: Thp1 = 323 K

Hotplate 2: Thp2 = 373 K

Q = - CP(Thp1 - Ti) = - CP(323 - 293)

)32330

(==1

1P

hp

hp CTQ

S

Flow out of hotplate:

)37350

(==2

2P

hp

hp CT

QS

Q = - CP(Thp2 - Thp1) = - CP(373 - 323)Flow out of hotplate:

What is the entropy change of the t.d. Universe?

Suniverse = Swater + Shp1 + Shp2

PPuniverse CCS 014.0+=]

37350

32330

)293373

[ln(=

Compare this result to that for heating the water on a single hotplate: Suniverse = +0.027 CP

How would S change if a third hotplate was added to the process? Try it and see!

Entropy change of a reversible process

As more and more hotplates are added, Suniverse will decrease.

Adding more and more hotplates is causing the process to approach a reversible process.

As the number of hotplates increases towards , Suniverse will approach 0, just as expected for a reversible process!