Physics 101: Lecture 31
Thermodynamics, part 2
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Review of 1st law of thermodynamics
2nd Law of Thermodynamics
Engines and Refrigerators
The Carnot Cycle
Physics 101: Lecture 31, Pg 1
Quick Review
1st Law of Thermodynamics:
energy conservation
Q = DU + W
Work done by (or on) system
Increase (or decrease) in internal
energy of system
P
Heat flow
into (or out of) system
U depends only on T (U = 3nRT/2 = 3PV/2)
 Point on P-V plot completely specifies
state of system (PV = nRT)
 work done is area under curve
 for complete cycle

V
DU=0  Q=W
Physics 101: Lecture 31, Pg 2
Second Law of Thermodynamics
Not all processes that are allowed by energy conservation occur in
nature. Why ?
Example:
Stone falls from height h:
mgh -> ½ m v2 (just before impact) -> heat (contact with floor)
This process is consistent with energy conservation.
The reversed process:
Stone lying on floor cools down and moves upward to height h,
has never been observed in nature, although it is also allowed by
energy conservation: Q->1/2 mv2->mgh
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Or: Ice melts but water does not spontaneously freeze,
heat flows from hot to cold but never from cold to hot.
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We need a new concept which makes these (reversed) processes
highly unlikely.
Physics 101: Lecture 31, Pg 3
New concept: Entropy (S)
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A measure of “disorder” or probability of state of a system.
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A property of a system (=state function, just like P, V, T, U)
related to number of different “states” of system
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Examples of increasing entropy:
ice cube melts
gases expand into vacuum
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Change in entropy:
DS = Q/T (T in K !)
SI unit: [J/K]
» >0 if heat flows into system (Q>0)
» <0 if heat flows out of system (Q<0)
Physics 101: Lecture 31, Pg 4
Reversible vs. Irreversible changes in a thermodynamic system:
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Reversible changes are conceived to be those that would occur very
slowly, giving all the molecules in the system time to 'adjust' to new
conditions, and all state variables time to adjust while still remaining
uniform throughout a system. Theoretically you could imagine
stopping at any point and reversing the change slowly, recovering the
previous thermodynamic state.
Definition given by Fermi (1936), in Thermodynamics:
"A transformation is said to be reversible when the successive states
of the transformation differ by infinitesimals from equilibrium
states.”
DSrev = 0
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Irreversible: Processes in which new entropy is “created”. A system
spontaneously changes, or energy is transformed in a way that creates
new entropy. This does not allow complete recovery of all aspects
of previous thermodynamic states.
DSirrev > 0
Processes that happen spontaneously are irreversible.
Physics 101: Lecture 31, Pg 5
1st and 2nd Law of Thermodynamics:
A Perpetuum Mobile (perpetual motion) of 1st
and 2nd kind is impossible.
Or:
The energy of the universe is constant, the
entropy of the universe seeks to be maximal.
R.Clausius
(1822-1888)
Perpetuum Mobile of 1st kind :
A machine that is able to provide useful work
without input of external energy (e.g. heat)
and without change of the physical or chemical
status of its parts does not exist (or a machine
that creates energy continuously does not
exist).
M.C. Escher
“Waterfall” (1961)
Perpetuum Mobile of 2nd kind:
A machine undergoing a cyclic process which
does nothing more than convert heat into
mechanical (or other) work does not exist.
Physics 101: Lecture 31, Pg 6
Second Law of Thermodynamics
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The entropy change (Q/T) of the system+environment  0
never < 0
order to disorder
The entropy of the universe increases whenever an irreversible
process occurs. All real processes in nature are irreversible.
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Consequences:
A “disordered” state cannot spontaneously transform into an
more “ordered” state.
No engine operating between two reservoirs can be more
efficient than one that produces zero change in entropy. The
latter is called a “Carnot engine” (no real engine can ever be
perfectly reversible but Carnot is a useful idealization, since
it represents the limiting case) .
 Heat cannot be transferred spontaneously from cold to hot.
Physics 101: Lecture 31, Pg 7
Engines and Refrigerators
HEAT ENGINE
REFRIGERATOR
TH
TH
QH
system
QH
W
QC
TC
W
QC
TC
System taken in closed cycle  DUsystem = 0
 Therefore, net heat absorbed = work done
QH - QC = W (engine)
QC - QH = -W (refrigerator)
energy going into “green blob” = energy leaving “green blob”
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Physics 101: Lecture 31, Pg 8
HEAT ENGINE
The objective: turn heat from hot
reservoir (QH) into work
The cost: “heat is wasted”
TH
QH
1st Law: QH -QC = W
efficiency
e  W/QH
=W/QH
W
QC
TC
= (QH-QC)/QH
= 1-QC/QH
Physics 101: Lecture 31, Pg 9
REFRIGERATOR
The objective: remove heat from
cold reservoir (QC)
The cost: work needs to be done
TH
QH
1st Law: QH = W + QC
coefficient of performance
CPr  QC/W
= QC/W
W
QC
TC
= QC/(QH - QC)
Physics 101: Lecture 31, Pg 10
Engines and the 2nd Law
The objective: turn heat from hot
reservoir into work.
The cost: “heat is wasted”
HEAT ENGINE
TH
QH
1st Law: QH -QC = W
efficiency e  W/QH =W/QH = 1-QC/QH
W
QC
TC
DS = QC/TC - QH/TH  0
DS = 0 for Carnot
Therefore, QC/QH  TC/ TH
QC/QH = TC/ TH for Carnot
Therefore e = 1 - QC/QH  1 - TC/ TH
e = 1 - TC/ TH for Carnot
=> efficiency of
a realistic engine can never be larger than eCarnot !
e largest if TC << TH
Physics 101: Lecture 31, Pg 11
Concept Question
Consider a hypothetical device that takes 1000 J of heat from a hot
reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and
produces 800 J of work.
Does this device violate the first law of thermodynamics ?
1. Yes
2. No
correct
This device doesn't violate the first law of
thermodynamics because no energy is being created
nor destroyed. All the energy is conserved.
W (800) = Qhot (1000) - Qcold (200)
 Efficiency = W/Qhot = 800/1000 = 80%
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Physics 101: Lecture 31, Pg 12
Concept Question
Consider a hypothetical device that takes 1000 J of heat from a hot
reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and
produces 800 J of work.
Does this device violate the second law of thermodynamics ?
correct
1. Yes
2. No
W (800) = Qhot (1000) - Qcold (200)
 Efficiency = W/Qhot = 800/1000 = 80%
 Max eff = 1 - 100/300 = 67% = eCarnot
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.
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e > eCarnot is forbidden by second law :
D S = DSH+DSC=200/100 J/K– 1000/300 J/K < 0
Physics 101: Lecture 31, Pg 13
Concept Question
Consider a hypothetical refrigerator that takes 1000 J of heat from a cold
reservoir at 100K and ejects 1200 J of heat to a hot reservoir at
300K.
How much work does the refrigerator do?
2. What happens to the entropy of the universe?
3. Does this violate the 2nd law of thermodynamics?
1.
TH
QC = 1000 J
QH = 1200 J
QH
W
QC
TC
Answers:
200 J
Decreases
yes
Since QC + W = QH, W = 200 J
DSH = QH/TH = (1200 J) / (300 K) = 4 J/K
DSC = -QC/TC = (-1000 J) / (100 K) = -10 J/K
DSTOTAL = DSH + DSC = -6 J/K  decreases (violates 2nd law)
Physics 101: Lecture 31, Pg 14
Heat Capacities of an Ideal Gas
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As discussed in Chapter 12, the heat needed to raise the
temperature of a solid or liquid is given by: Q=cm DT
where c is the heat capacity of the material.
Gases: Volume and/or pressure change when temperature
changes (this effect can be safely neglected in case of
solids and liquids).
 Heat capacity of a gas depends on if T changes
at constant V, cV, or constant P, cP,:
V=const.: DU=Q=cV m DT=3/2 n R DT => CV=cVm/n = 3/2 R
P=const.: DU=Q-PDV=cP m DT-n R DT=3/2 n R DT
=> CP=cP m/n = 5/2 R
CV and CP are the molar specific heat capacities of an ideal
monatomic gas.
Physics 101: Lecture 31, Pg 15