Embed Size (px)
Citation preview
Phys 250 Ch12 p1
Chapter 13: ThermodynamicsThermal Equilibrium: Two systems are placed in “thermal contact” until no further changes occur.
Let A and B come into thermal contact with C, but not with each other. -A is in thermal equilibrium with C. -B is in thermal equilibrium with C.
(simultaneously)
Now, let A and B come into thermal contact with each other. -No further changes occur. => A and B are in thermal Equilibrium
The Zeroeth Law of Thermodynamics: Two systems in thermal equilibrium with a third system are in thermal equilibrium with each other. One parameter (temperature) determines equilibrium.
System A System B
System A System B
System C
System A System B
System C
Phys 250 Ch12 p2
Internal Energy U :
Sum of microscopic kinetic and potential energies
Changes in response to heat addition (Q) to the system
Changes in response to Work done (W) by the system
The First Law of Thermodynamics:
U = Q W or Q = U + W
= Conservation of energy
=> U is independent of path!
U is a function of the state of the system (i.e. a function of the state variables).
U = U(p,V,T) for an ideal gas.
Phys 250 Ch12 p3
Thermodynamic Processes: Expansion of a Gas
U = Q WIsothermal: constant temperature
generally V 0 W 0 ; Q ideal gas: U only depends upon T
U = 0 W = Q
Adiabatic: no heat transfer
Q = 0 W = U
Isochoric: constant volume (isovolumetric)
V = 0 => W = 0 => Q = U
Isobaric: constant pressure
W =Fx= F/A ·A x= p V
Polytropic processes: one generalization, not (necessarily) iso-anything.
pVr = const
p
V
Phys 250 Ch12 p4
Example: Gas in a piston expands against a constant pressure of 100 kPa. When 2E4 J of heat are absorbed, the volume of the gas expands from .15 m3 to 0.25 m3.
How much work is done by the system?
What is the change in the internal energy of the system?
Example: A heat engine undergoes a process in which its internal energy decreases by 400 J while doing 250 J worth of work. What is the net heat transferred to the system during this process?
Phys 250 Ch12 p5
Diagram of a “heat engine”
1. Heat (QH) is absorbed from a source at high temperature.
2. Mechanical work (W) is done (by converting some of the absorbed heat to mechanical work).
3. Heat (QC) is given off at a lower temperature
4. Process cycles, so that system returns to same state
W
Low temperature Reservoir
High temperature Reservoir
Q C
QH
For Cyclic Processes (Cycles):
repeating process in which the system or heat engine returns to the starting point (same thermodynamic state) each cycle.
A Cyclic Process is necessary for most practical heat engines.
Over each complete cycle
U = 0
net heat input = net work output
QH-QC = Wnet
Phys 250 Ch12 p6
Indicator Diagrams: P-V diagrams used to analyze processes which use a gas in a heat engine.
for constant pressure, W = PV (like W = Fx)
for non-constant force, W = “area under curve”
p
V
p
V
V
p
p
V
Work done by system
Work done on system
Net work done by system equals enclose area
Phys 250 Ch12 p7
p
V
ab
dc
The most efficient engine cycle operating between two specified temperatures: Carnot Cyclea-b : Isothermal Expansion at TH.
|QH| = proportional to TH
(absolute temperature!)b-c : Adiabatic Expansion to TC.c-d : Isothermal Compression at TC.
|QC| proportional to TC
d-a : Adiabatic Compression to TH.
Engine Efficiency (in general)
net mechanical work comes from net transfer of heat
W = QH QC
Efficiency is the effectiveness with which supplied heat QH is converted to work :For the
Carnot Engine only:
Q is proportional to T for both isothermal processes, so
H
C
H
CH
H Q
Q
Q
Q
WEff
1
H
Ccarnot
H
C
H
C
T
TEff
Q
Q
T
T
1
so
Phys 250 Ch12 p8
Example: What is the maximum possible thermal efficiency of a steam engine that takes in steam at 160ºC and exhausts it at 100ºC?
Example: An engine takes in 9220 J and does 1750 J of work each cycle while operating between 689C and 397C.
What is the actual efficiency?
What is the maximum theoretic efficiency?
Phys 250 Ch12 p9
Low temperature Reservoir
High temperature Reservoir
W
Q C
QH
Refrigerators and Heat Pumps
getting heat to flow from cold to hot requires work!
1. Heat (QC) is absorbed from a source at low temperature.
2. Mechanical work (W) is done on the system (work is input).
3. Heat (QH) is given off to the higher temperature reservoir.
First Law!: QH = Wnet+QC
Coefficient of Performance
CH
HH
CH
CC
TT
T
W
QTT
T
W
Qpc
pumpheat
orrefrigerat
max..ePerformanc oft Coefficien
Phys 250 Ch12 p10
Example: A refrigerator has a cp of 6.0. If the room temperature is 30 ºC, what is the lowest temperature that can be obtained inside the refrigerator?
Example: A heat pump is used to maintain an inside temperature of 20ºC when the outside temperature is 10ºC. What is the theoretical maximum cp for this heat pump? If the pump is to deliver heat at a rate of 15 kW, how much power must be supplied to run the pump?
Phys 250 Ch12 p11
The Second Law of Thermodynamics
•Heat cannot, by itself, pass from a cooler to a warmer body
•It is impossible for any system to undergo a cyclic process whose sole result is the absorption of heat from a single reservoir at a single temperature and the performance of an equivalent amount of work.
Low temperature Reservoir
High temperature Reservoir
W
QH
Low temperature Reservoir
High temperature Reservoir
Q C
Q C
Phys 250 Ch12 p12
Entropy S
A quantitative measure of disorder.
A new thermodynamic variable. (Scyclic = 0)
For reversible processes
Adiabatic Process => S = 0
for a closed system, S = 0
•In any process the entropy change of the universe increases or remains the same.
...,butT
dQS
T
dQdS
reallyT
QS
f
i
Phys 250 Ch12 p13
Example: A student takes a 2.5 kg of ice a 0C, and places it on a large rock outcropping, and watches it melt. What is the entropy change of the ice? If the rock is at an essentially constant temperature of 21 C, what is its entropy change?
Example: A student mixes 0.100 kg of water at 60C with 0.200 kg of water at 40C. What is the entropy change of the system? (Note: use the average temperature of each sample to calculate its entropy change)
