Tue Dec 1
Thermodynamics
• 1st Law
ΔU
=
Q−
W
nd
SYS
• 2 Law
• Heat Engines and Refrigerators
•
•
•
•
Isobaric: W = PΔV
Isochoric: W = 0
Isothermal: ΔU = 0
Adiabatic: Q = 0
Q/A Sessions (tentative)
• Assign 13/14 Friday
• Final: Fri Dec 11 – 2:30PM
• WALTER 145
•
•
•
•
– 2 Sheets handwritten
– about 30% Thermal Physics
– 16 MC, 4 problems
– Exam Conflict? Notify me ASAP
Study: Homework, PRS, Pre-class
Examples in text
Review problems
Thursday – finish up details on
Chapter 15, then review
Textbook Question: This semester we started using the
OpenStax Textbook
(1) I purchased a hardcopy of the textbook
(2) I downloaded a copy of the textbook and referenced it
during the course
(3) I downloaded a copy of the textbook but did not use it
during the course
(4) I did not use any textbook
1st Law of Thermodynamics
ΔU = Q− WSYS
• SIGNS!
– Q: added to system
• If added, Q is +
• If removed, Q is -
– W: Work done BY system (WSYS)
• If system does (+) work on surroundings, Wsys is (+)
• If surroundings do (+) work on system, Wsys is (-)
! Sect 15.3
Transitions
• between States
• PV diagram (ideal gas)
– State is a point
– same point, same state
– same state, same P,V,n,T,U
• Cycle – loop back to initial state
– end at same U!
– For cycle: ΔU = 0
Work – Ideal Gas
• W = Force*displacement
• W = (PA) * s
• W = P (ΔV)
• If P constant, isobaric transition
ΔU = Q − ( PΔV )
• Area under Curve = Work
– Holds in general!
! Sect 15.2
Given the PV diagram to the right and the information that
ΔUAB is positive, what can you conclude about QAB and WAB?
(1) Q>0, W>0
(3) Q>0, W<0
(5) Q=0, W=0
(7) Q<0, W>0
(9) Q<0, W<0
(2) Q>0, W = 0
(4) Q=0, W>0
(6) Q=0, W<0
(8) Q<0, W=0
C
P
A
ΔU = Q − W
From the graph, W is +. The only way ΔU can be +
is if Q is + and greater than W.
B
V
Cycles
• Cycle – loop back to initial state
– end at same U!
– For cycle: ΔU = 0
– ΔUNET = ΔUAB + ΔUBC + ΔUCD + ΔUDA
• Total W – add up transitions
WNET = WAB + WBC + WCD + WDA
• Total Q – add up transitions
QNET = QAB + QBC + QCD + QDA
P
A
B
D
C
V
The NET work done by the system during the cycle shown is:
(1) positive
(2) zero
(3) negative
Positive work done from left to right. Negative work from right to left.
Larger area under curve left to right – greater magnitude of work.
What if reversed direction of arrows?
Given the cycle shown, the heat absorbed by the system is:
(1) positive
(2) zero
(3) negative
ΔU = 0;
ΔU = Q-W
Q = ΔU + W = 0 + (some positive number)
Q must be positive
In the cycle below, QAB is positive and ΔUBC is positive.
What is the sign of ΔUCA? (Can you identify the work. heat and
change in internal energy for each process? Don’t forget about the
First Law.)
(1) positive 2) zero
3) negative
ΔUAB positive (W=0, Q+)
ΔUBC is positive
ΔUCycle= ΔUAB + ΔUBC + ΔUCA =0
Some Examples of Cycles in Heat Engines
Diesel Cycle
http://www.shermanlab.com/science/
physics/thermo/engines/DieselG.php
Stirling
Cycle
http://
www.stirlingengine
.com/DisplacerAnim.adp
Otto Cycle
http://www.shermanlab.com/science/physics/
thermo/engines/OttoG.php
Carnot
Cycle
http://
www.rawbw.com/
~xmwang/myGUI/
CarnotG.html
Heat Engines
• Work
• Hot Reservoir, Cold Reservoir
• Conserve Energy
• Cycle: ΔU = 0
• Qcycle = Wcycle
• QH-QC = W
• Q H = W + QC
Eff = W/QH
Always < 1 unless in %
Example
W = QH − QC
• A heat engine has an efficiency of 64%
and produces 5500 J of work. Determine:
a) the input heat QH =W/e =5500J/0.64 = 8590 J
W
e=
QH
b) The rejected heat or heat exhausted
QC = QH -W = 8590J -5500J = 3090 J
Many "faces" of efficiency
W
e=
QH
QH − QC
e=
QH
1 − QC / QH
e=
1
What is the efficiency of a heat engine that pulls 8000J from
the hot reservoir and expels 6000J in the form of exhaust?
(1) 10%
(5) 60%
(2) 25%
(6) 75%
(3) 33%
(7) 133%
W = 8000J – 6000J
eff = W/QH = 2000J/8000J = 25%
(4) 50%
A certain heat engine operates by absorbing a fixed amount of
heat from the hot reservoir each cycle and expelling a
certain amount of heat to the cold reservoir.
The engine is redesigned such the energy absorbed from the
hot reservoir each cycle is the same, but the energy expelled
to the cold reservoir each cycle decreases.
What happens to the efficiency of the engine?
1) it increases
2) it stays the same
3) it decreases
1 − QC / QH
e=
1
QC gets smaller, QC/QH gets smaller, 1-QC/QH gets larger
Carnot Engine
• Absolute best engine if all processes reversible
• Reversible – both sys and surr returned to original state
• All real processes irreversible
• For theoretically best engine (Carnot Engine):
QC TC
=
QH TH
eff MAX, Carnot
T in KELVIN!
TC ⎞
⎛
= 1− ⎜
⎟
T
H ⎠
⎝
An engine manufacturer claims to have an engine that draws
heat from a reservoir at 375K. The engine does 5.0kJ of
work each second and expels 4.0kJ of heat each second to a
cold reservoir at 225K. Is this theoretically possible? (1) yes
(2) no Claimed eff = W/QH = W/(W+QC) = 5.0kJ / 9.0 kJ = 0.56
Carnot eff = 1- (TC/TH) = 1- (225K/375K) = 0.4
Claimed eff better than theoretical best
Clausius' Version of 2nd Law of Thermodynamics
• If two objects are placed in thermal contact:
– hot object cools down
– cool one heats up
Heat won’t flow spontaneously from colder object to hotter object
• Why?
• Doesn’t violate First Law of Thermodynamics (heat out of cold object = heat into hot object would
certainly conserve energy).
• Must be a separate law!
Versions of the 2nd Law of Thermodynamics
•
•
•
•
•
•
Clausius:
Heat won’t flow spontaneously from colder object to hotter object.
Kelvin-Planck:
Heat can’t be entirely converted into work.
Entropy:
Entropy is always increasing.
Kinetic Theory:
Disorder tends to increase.
Heat Engine:
It is impossible for a heat engine to be more efficient than a Carnot-cycle engine.
Custodian:
You can’t get something clean without getting something else dirtier.
Third Law of Thermodynamics
• For Carnot heat engine:
emax,Carnot = 1 −
TC
TH
• You can have perfect efficiency, e = 1, but only if TC = 0 K.
• The Third Law of Thermodynamics says: There’s no way to
cool something to 0 K in a finite number of steps.
Summary of the Laws of Thermodynamics
(with an attitude!)
0: There is a game.
1: You can’t win, the best you can do is break even.
2: You can only break even at absolute zero.
3: You can’t reach absolute zero.
Refrigerator
QC + W = QH
• Compress refrigerant – Heat it
• Expand – Cool it
QC
Outside
Expand
Inside
Colder
than in
Refrig
QH
Warmer than outside
Compress
Motor
W
Refrigerators
QH = QC + W
Coefficient of Performance (COP)
COPRefrig = QC / W
COPRefrig
QC / QH
=
1 − QC / QH
For Carnot Refrigerator QC/QH = TC/TH:
COPRefrig
TC / TH
=
1 − TC / TH
In cooling a warm leftovers, a refrigerator removes 60000J
of heat from the food. The COP is 3.0. We eventually want
to know the amount of heat exhausted into the room. First,
though, which quantity does the 60000J represent?
1) QC
2) QH
3) W
60000J represents the amount of
energy which must be removed
from the cold reservoir.
In cooling warm leftovers, a refrigerator removes 60000J of
heat from the food. If the COP is 3.0, what is the amount of
heat exhausted to the room (moved into the hot reservoir)?
1) 20,000J
5) 240,000J
2) 60,000J
3) 80,000J
COP = QC/W
3.0 = 60000 J/W
W = 20,000 J
QH = QC+W = 60,000 J + 20,000 J
4) 180,000J
In cooling warm leftovers, a refrigerator removes 60000J of
heat from the food. The COP is 3.0. If the power of the motor
is 200W, what is the minimum time to cool the leftovers?
1) 100s
2) 200s
3) 300 s
COP = QC/W
W = 20,000J
P = Energy/Time
Time = 100 s
200W = 20,000J/time
Heating Your Kitchen …
How long would a 3.40 kW space heater have to run to put into a kitchen the same
amount of heat as a refrigerator (coefficient of performance = 3.29) does when it
freezes 1.30 kg of water at 21.4°C into ice at 0°C?
Another way …
One way …
Refrig
Warms
Kitchen
QC
QH
Q
Same
W
Warms
Kitchen
Electrical Heater
Elect Energy = P*t
QH (Refrig) = Power (of elect heater) * time (elect heater run)
Each drawing shows a hypothetical heat engine or a hypothetical heat
pump and shows the corresponding heats and works. Only one is
allowed in nature. Which is it?
1)
2)
3)
4)
5)
Heat Pump
• Coefficient of Performance (COP) a bit different
COPHeat pump
QH
=
W
Heat Engines, Refrigerators, Heat Pumps
• Heat Engine
– Natural flow of heat (hot to cold)
– can do work on surroundings
• Refrig or Heat Pump
– Unnatural flow of heat (cold to hot)
– requires work from surroundings
• Both cases: QH = QC + W
• What is related to what????
– Think about Reservoirs, not the temp
of something in the reservoir
– Which reservoir is hotter? Colder?
– QC inside refrigerator; QH Kitchen
Heat Engine, Heat Pump, and Refrigerators Summary
Device
Heat engine
Carnot
Heat pump
Carnot
Refrigerator
Carnot
What we What
want
we buy
Wnet
QH
QC
QH
Wnet
Wnet
Q H = W + QC
Efficiency, e, or Coefficient of
Performance (“COP”)
0 < e = Wnet/ QH = 1 - (QC/QH) < 1
e = 1 - (TC/TH) = (ΔΤ/ΤΗ)
COP = QH/Wnet = 1 + [QC/(QH - QC)] > 1
COP = 1/[1 - (TC/TH)]
COP = QC/Wnet
COP = 1/[(TH/TC) - 1]
(which may be more or less than 1)