1. PHYSICS 231 Lecture 29: Entropy
“Another way of stating the Second Law then is: the universe is constantly getting more disorderly! Viewed that way, we can see the Second Law all about us. We have to work hard to straighten a room, but left to itself, it becomes a mess again very quickly and very easily .... How difficult to maintain houses, and machinery, and our own bodies in perfect working order; how easy to let them deteriorate. In fact, all we have to do is nothing, and everything deteriorates, collapses, breaks down, wears out, all by, itself and that is what the Second Law is all about." – Isaac Asimov
Remco Zegers
Question hours:Tue 4-5
Helproom
PHY 231

2. engines and fridges engine: work was performed by the
gas, while heat was added to the gas.
Wby gas= (enclosed area)
fridge: Net work was performed on
the gas and heat extracted from the
gas.
Won gas= (enclosed area)
PHY 231

3. The 2nd law of thermodynamics
1st law: U=Q+W
In a cyclic process (U=0) Q=W: we cannot do more work
than the amount of energy (heat) that we put inside
2nd law: It is impossible to construct an engine that,
operating in a cycle produces no other effect than the
absorption of energy from a reservoir and the performance
of an equal amount of work: we cannot get 100% efficiency
What is the most efficient engine we can make
given a heat and a cold reservoir?
PHY 231

4. Carnot engine AB isothermal expansion BC adiabatic expansion W-, T-
W-, Q+
Thot
DA adiabatic compression
W+, T+
CD isothermal compression
Tcold
W+, Q-
PHY 231

5. Carnot engine General engine: W=|Qh|-|Qc| efficiency: W/|Qh|
e=1-|Qc|/|Qh|
Carnot engine
efficiency=1-Tcold/Thot
The Carnot engine is the most efficient way to operate
an engine based on hot/cold reservoirs because the
process is reversible.
PHY 231

6. A new powerplant A new powerplant is designed that makes use of the
temperature difference between sea water at 0m (200) and
at 1-km depth (50). A) what would be the maximum efficiency
of such a plant? B) If the powerplant produces 75 MW, how
much energy is absorbed per hour? C) Is this a good idea?
maximum efficiency=carnot efficiency=1-Tcold/Thot=
1-278/293= efficiency=5.1%
P=75*106 J/s W=P*t=75*106*3600=2.7x1011 J
efficiency=1-|Qcold|/|Qhot|=(|Qhot|-|Qcold|)/|Qhot|=
W/|Qhot| so |Qhot|=W/efficiency=5.3x1012 J
c) Yes! Very Cheap!! but… |Qcold|= |Qhot|-W=5.0x1012 J
every hour 5E+12 J of waste heat is produced:
Q=cmT 5E+12=4186*m*1 m=1E+9 kg of water is heated
by 10C.
PHY 231

7. A cycle Consider the cycle in the figure. what is the net work done in
one cycle?
B) What is the net energy added
to the system per cyle?
Work: area enclosed in the cycle:
W=-(2V03P0)+(2V0P0)=-4V0P0 (Negative work is done on
the gas, positive work is done by the gas)
Cycle: U=0 so Q=-W Q=4V0P0 of heat is added to
the system.
PHY 231

8. quiz (extra credit) P (Pa) 3x105 Consider this cyclic
process. Which of the following
is true?
1x105 1 3
V (m3)
clockwise: work done by the gas, so engine
W by gas=area enclosed=(3-1)x(3x105-1x105)=4x105 J
This is a heat engine and the W done by the gas is +4x105 J
This is a fridge and the W done on the gas is +4x105 J
This is a heat engine and the W done by the gas is +6x105 J
This is a fridge and the W done on the gas is +6x105 J
This is a heat engine and the W done by the gas is –4x105J
PHY 231

9. Irreversible process Thot Thot thermal work? engine contact Tcold work
e=1-Tc/Th=0
thermal
contact
engine
Tcold
e=1-Tc/Th
The transport of heat by conductance is irreversible and
the engine ceases to work.
PHY 231

10. The (loss of) ability to do work: entropy
entropy: S=QR/T R refers to a reversible process
The equation ONLY holds for
a reversible process.
example: Carnot engine:
Hot reservoir: Shot=-Qhot/Thot (entropy is decreased)
Cold reservoir: Scold=Qcold/Tcold
We saw: efficiency for a general engine: e=1-Qcold/Qhot
efficiency for a Carnot engine: e=1-Tcold/Thot
So for a Carnot engine: Tcold/Thot=Qcold/Qhot
and thus: Qhot/Thot=Qcold/Tcold
Total change in entropy: Shot+Scold=0
For a Carnot engine, there is no change in entropy
PHY 231

11. The loss of ability to do work: entropy
Now, consider the following irreversible case:
entropy: S=QR/T
This equation only holds
for reversible processes.
T=300 K
conducting
copper
wire
Qtransfer=1200 J
We cut the irreversible
process up into 2 reversible
processes
T=650 K
Shot+Scold=Qhot/Thot+Qcold/Tcold=-1200/ /300=
=+1.6 J/K
The entropy has increased!
PHY 231

12. 2nd law of thermodynamics rephrased
2nd law: It is impossible to construct an engine that,
operating in a cycle produces no other effect than the
absorption of energy from a reservoir and the performance
of an equal amount of work: we cannot get 100% efficiency
2nd law rephrased: The total entropy of the universe
increases when an irreversible process occurs.
PHY 231

13. Entropy in terms of disorder
speed n
speed n
speed
In an isolated system, disorder tends to grow and entropy
is a measure of that disorder: the larger the disorder,
the higher the entropy.
PHY 231

14. example what is the change in entropy of 1.00 kg of liquid water
at 1000C as it changes to steam at 1000C?
Lvaporization=2.26E+6 J/kg
Q=Lvaporizationm=2.26E+6 J/kg * 1 kg= 2.26E+6 J
S=Q/T=2.26E+6/(373)=6059 J/K
PHY 231

15. adiabatic process
For an adiabatic process, which of the following is true?
S<0
S=0
S>0
none of the above
Adiabatic: Q=0 so S=Q/T=0
PHY 231