2. What is thermodynamics?
• The study of thermodynamics is concerned with ways energy is stored within a
body and how energy transformations, which involve heat and work, may take
place.
• Approaches to studying thermodynamics
– Macroscopic (Classical thermodynamics)
• study large number of particles (molecules) that make up the
substance in question
• does not require knowledge of the behavior of individual molecules
– Microscopic (Statistical thermodynamics)
• concerned within behavior of individual particles (molecules)
• study average behavior of large groups of individual particles
2
4. Thermodynamic Systems
•Thermodynamic System
– quantity of matter or a region of
space chosen for study
•Boundary
– real or imaginary layer that
separates the system from its
surroundings
•Surroundings
– physical space outside the system
boundary
• Types of Systems
– Closed
– Open
4
5. A thermodynamic system is that portion of the
Universe that we have selected for investigation
•The surroundings are everything outside
the system.
5
8. Properties
• Any characteristic of a system in equilibrium is called a
property.
• Types of properties
– Extensive properties - vary directly with the size of the system
Examples: volume, mass, total energy
– Intensive properties - are independent of the size of the system
Examples: temperature, pressure, color
• Extensive properties per unit mass are intensive properties.
specific volume v = Volume/Mass = V/m
density ρ = Mass/Volume = m/V
8
9. 9
State & Equilibrium
• State of a system
– system that is not undergoing any change
– all properties of system are known & are not changing
– if one property changes then the state of the system
changes
• Thermodynamic equilibrium
– “equilibrium” - state of balance
– A system is in equilibrium if it maintains thermal (uniform
temperature), mechanical (uniform pressure), phase (mass
of two phases), and chemical equilibrium
10. 10
Processes & Paths
• Process
– when a system changes from one equilibrium state to another one
– some special processes:
• isobaric process - constant pressure process
• isothermal process - constant temperature process
• isochoric process - constant volume process
• isentropic process (adiabatic process) - constant entropy process
• Path
– series of states which a system passes through during a
process
11. 11
Thermodynamic equilibrium (Zeorth Law)
• If two systems are in thermal equilibrium with a third
system, they are also in thermal equilibrium with each
other.
• If A and C are each in thermal equilibrium with B, A is
also in equilibrium with C.
• Temperature as a quality of heat, by Galileo and Newton
• The temperatures are equal for all systems in thermal
equilibrium.
• Temperature scale
• Thermometers
12. 12
First Law of thermodynamics
• The first law of thermodynamics is an extension
of the law of conservation of energy
• The change in internal energy of a system is equal
to the heat added to the system minus the work
done by the system
ΔU = Q - W
14. 14
• An adiabatic process transfers no heat
– therefore Q = 0
• ΔU = Q – W
• When a system expands adiabatically, W is
positive (the system does work) so ΔU is
negative.
• When a system compresses adiabatically, W is
negative (work is done on the system) so ΔU is
positive.
Adiabatic process
15. 15
• An isothermal process is a constant temperature
process. Any heat flow into or out of the system
must be slow enough to maintain thermal
equilibrium
• For ideal gases, if ΔT is zero, ΔU = 0
• Therefore, Q = W
– Any energy entering the system (Q) must leave as
work (W)
Isothermal process
16. 16
Isobaric process
• An isobaric process is a constant pressure process. ΔU, W,
and Q are generally non-zero, but calculating the work done
by an ideal gas is straightforward
W = P·ΔV
• Water boiling in a saucepan is an example of an isobar
process
17. 17
Isochoric process
• An isochoric process is a constant volume
process. When the volume of a system doesn’t
change, it will do no work on its surroundings.
W = 0
ΔU = Q
• Heating gas in a closed container is an isochoric
process
18. 18
• The amount of heat required to raise a certain
mass of a material by a certain temperature is
called heat capacity
Q = mcxΔT
• The constant cx is called the specific heat of
substance x, (SI units of J/kg·K)
Heat Capacity
19. 19
( ) ( )
V T
U U
dU dT dV
T V
( )
V V
U
C
T
The heat capacity at
constant volume
dU=dQ=Qv, at constant volume, no
additional work
QV=CvdT
Heat Capacity
20. 20
• CV = heat capacity at constant volume
CV = 3/2 R
• CP = heat capacity at constant pressure
CP = 5/2 R
• For constant volume
Q = nCVΔT = ΔU
• The universal gas constant R = 8.314 J/mol·K
Heat Capacity for of ideal gas
21. 21
Latent Heat
The word “latent” comes from a Latin word that means “to lie
hidden.” When a substance changes phases (liquid solid or
gas liquid) energy is transferred without a change in
temperature. This “hidden energy” is called latent heat. For
example, to turn water ice into liquid water, energy must be
added to bring the water to its melting point, 0 ºC. This is not
enough, however, since water can exist at 0 ºC in either the
liquid or solid state. Additional energy is required to change 0 ºC
ice into 0 ºC water. The energy increases the internal energy of
the water but does not raise its temp. When frozen, water
molecules are in a crystalline structure, and energy is needed to
break this structure. The energy needed is called the latent heat
of fusion. Additional energy is also needed to change water at
100 ºC to steam at 100 ºC, and this is called the latent heat of
vaporization.
22. 22
Latent Heat Formula
L is the energy per unit mass needed to change the state
of a substance from solid to liquid or from liquid to gas.
Ex: Lf (the latent heat of fusion) for gold is 6440 J/kg.
Gold melts at 1063 ºC. 5 grams of solid gold at this
temp will not become liquid until additional heat is
added. The amount of heat needed is:
(6440 J/kg)(0.005 kg) = 32 J. The liquid gold will still be at
1063 ºC.
Q = mLf or Q = mLv
Q = thermal energy
m = mass
L = heat of fusion or vaporization
23. 23
Enthalpy – heat of transformation
• △U= Qv (Closed system, only expansion work exchanging,
constant volume)
• Qv , heat transactions at constant volume
• Qp, heat transactions at constant pressure (heat of
transformation = (heat absorbed / (m or n)
ΔU= Q-W=Qp- PΔV
ΔU= Qp – P2 V2 + P1V1= U2-U1
QP=（U2+ P2 V2）-（U1+ P1V1）
H≡U+PV，∴Qp=H2-H1=ΔH
•State function, extensive
• the heat of transformation in any change of phase is equal to the
difference between the enthalpies of the system in the two phases.
Enthalpy
24. 24
Enthalpy – heat of transformation
Let l12 = lS-L : fusion, l23 = lL-V : vaporization, l13 = lS-V : sublimation
In a cyclic process, Δh = 0. S –L—V—S
Therefore: Δh1 + Δh2 + Δh3 = 0. l13 = l12 + l23
e.g. latent heat of vaporization of water:
26. 26
Heat capacity
• Heat capacity at constant pressure
• Heat capacity at constant volume
• Closed system, in equilibrium, PV work only
( )
d
p
p p
Q H
C
T
T
d
p p
H Q C T
D
( )
d
V
V V
Q U
C
T
T
d
V V
U Q C T
D
27. 27
Second law of thermodynamics
• The second law of thermodynamics introduces the
notion of entropy (S), a measure of system disorder
• The 2nd Law can also be stated that heat flows
spontaneously from a hot object to a cold object
(spontaneously means without the assistance of external work)
• The 2nd Law helps determine the preferred direction
of a process
• A reversible process is one which can change state
and then return to the original state
• This is an idealized condition – all real processes are
irreversible
28. 28
Heat engines
• A device which transforms heat into work is called a
heat engine
• This happens in a cyclic process
• Heat engines require a hot reservoir
to supply energy (QH) and a cold
reservoir to take in the excess energy
(QC)
- QH is defined as positive, QC is negative
Cold Reservoir, TC
Engine
Hot Reservoir, TH
QH
QC
W
Real engine. QH = QC + W
29. 29
Thermal Efficiency of a Heat Engine
eng
1
h c c
h h h
W Q Q Q
e
Q Q Q
Second Law: Kelvin-Planck Form
• It is impossible to construct a heat 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
– Means that Qc cannot equal 0
Some Qc must be expelled to the environment
– Means that e cannot equal 100%
Cold Reservoir, TC
Engine
Hot Reservoir, TH
QH
QC = 0
W
Impossible engine. QH = W
30. Refrigerators
Cold Reservoir, TC
Engine
Hot Reservoir, TH
QH
QC
W
Real fridge. QC + W = QH
A refrigerator forces heat from a cold region into a warmer one. It takes
work to do this, otherwise the 2nd Law would be violated. Can a fridge
be left open in the summer to provide a make shift air conditioner?
Nope, since all heat pumped out of the fridge is pumped back into the
kitchen. Since QH > QC because of the work done, leaving the
refrigerator open would actually make your house hotter!
31. 31
Impossible fridge. QC = QH
Cold Reservoir, TC
Engine
Hot Reservoir, TH
QH
QC
W = 0
Second Law – Clausius Form
• It is impossible to construct a cyclical
machine whose sole effect is to transfer energy
continuously by heat from one object to another
object at a higher temperature without the input
of energy by work
• Or – energy does not transfer spontaneously by
heat from a cold object to a hot object.
32. 32
COP c
Q
W
• In cooling mode,
• A good refrigerator should have a high COP
– Typical values are 5 or 6
• The effectiveness of a heat pump is described by a
number called the coefficient of performance (COP).
• In heating mode:
energy transferred at high temp
COP =
work done by heat pump
h
Q
W
33. 33
Carnot Cycle
• AB and CD are isothermal processes
• BC and DA are adiabatic processes
• The work done by the engine is
shown by the area enclosed by
the curve, Weng
• The net work is equal to |Qh| –
|Qc|
• DEint = 0 for the entire cycle
h
c
c
h
c
h
c
T
T
e
and
T
T
Q
Q
1
• Temperatures must be in Kelvins
• All Carnot engines operating between the same two temperatures will
have the same efficiency
34. 34
Entropy and second Law of thermodynamics
• The processes in which the entropy of an isolated system would decrease do
not occur. In every process taking place in an isolated system, the entropy of
the system either increase or remains constant.
• Entropy of Carnot cycle (ΔS =0)
• for all cycles ΔS =0. therefore
• dS is exact differential . S: extensive properties.
Calculation of entropy changes in Reversible Process.
1. reversible adiabatic process: dQ= 0 and dS=0 (S=cte) : isentropic
process.
2. reversible isothermal process: from a to b
35. 35
E.g. : Change in phase at cte pressure and T:
3. reversible isochoric process: no phase change
4. reversible isobaric process: no phase change
Since System+ surrounding =universe
Suniverse =cte
36. 36
T-S diagram for Carnot cycle:
Qr : Area of the cycle = net heat flow into the system.
37. 37
Calculation of entropy changes in irreversible Process.
for reversible process only
• Consider a body at temperature T1 in contact with heat reservoir
at T2 (T2 > T1)
heat flow into the boby , therefore ΔS> 0
How does Sreservoir changes?
Q =cp (T2 –T1) into the body
