PHY 113 C Fall 2013 Lecture 21 1 PHY 113 C General Physics I 11 AM 1215 p M MWF Olin 101 Plan for Lecture 21 Chapter 20 Thermodynamics Heat and internal energy Specific heat latent heat ID: 623565
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Slide1
11/12/2013
PHY 113 C Fall 2013 -- Lecture 21
1
PHY 113 C General Physics I
11 AM - 12:15
p
M
MWF Olin 101
Plan for Lecture 21:
Chapter 20: Thermodynamics
Heat and internal energy
Specific heat; latent heat
Thermodynamic work
First “law” of thermodynamicsSlide2
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PHY 113 C Fall 2013 -- Lecture 21
2Slide3
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PHY 113 C Fall 2013 -- Lecture 21
3
In this chapter T
≡ temperature
in Kelvin or Celsius unitsSlide4
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PHY 113 C Fall 2013 -- Lecture 21
4
Webassign
question from Assignment #18
A rigid tank contains 1.50 moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to lower the pressure of the gas from 28.5
atm
to 5.10 atm. Assume the volume of the tank and the temperature of the gas remain constant during this operation.Slide5
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PHY 113 C Fall 2013 -- Lecture 21
5
T
1
T
2
Q
Heat – Q -- the energy that is transferred between a “system” and its “environment” because of a temperature difference that exists between them.
Sign
convention:
Q<0
if T
1
< T
2
Q>0
if T
1
> T
2Slide6
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PHY 113 C Fall 2013 -- Lecture 21
6
Heat
withdrawn
from system
Thermal equilibrium – no heat transfer
Heat
added
to systemSlide7
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PHY 113 C Fall 2013 -- Lecture 21
7
Units of heat: Joule
Other units: calorie = 4.186 J
Kilocalorie = 4186 J =
Calorie
Note: in popular usage, 1 Calorie is a measure of the heat produced in food when oxidized in the body Slide8
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PHY 113 C Fall 2013 -- Lecture 21
8
iclicker
question:
According to the first law of thermodynamics, heat and work are related through the “internal energy” of a system and generally cannot be interconverted. However, we can ask the question: How many times does a person need to lift a 500 N barbell a height of 2 m to correspond to 2000 Calories (1 Calorie = 4186 J) of work?
4186
8372
41860
83720
None of theseSlide9
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PHY 113 C Fall 2013 -- Lecture 21
9
Heat
capacity: C = amount of heat which must be added
to
the “system” to raise its
temperature
by 1K (or 1
o
C).
Q
= C
D
T
Heat capacity per mass: C=mc
Heat capacity per mole (for ideal gas): C=
nC
v
C=nCpSlide10
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PHY 113 C Fall 2013 -- Lecture 21
10
Some typical specific heats
Material
J/(kg·
o
C)
cal/(g·
o
C)
Water (15
o
C)
4186
1.00
Ice (-10
o
C)
2220
0.53
Steam (100
o
C)
2010
0.48
Wood
1700
0.41
Aluminum
900
0.22
Iron
448
0.11
Gold
129
0.03Slide11
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PHY 113 C Fall 2013 -- Lecture 21
11
iclicker
question:
Suppose you have 0.3 kg of hot coffee (at 100
o
C
). How much heat do you need to remove so that the temperature is reduced to 40
o
C
? (Note: for water, c=1000 kilocalories/(kg
o
C
))
300 kilocalories (1.256 x 10
6
J)
12000 kilocalories (5.023
x
10
7
J)18000 kilocalories (7.535 x
107J)30000 kilocalories
(1.256 x 108J)None of these
Q=m c DT = (0.3)(1000)(100-40) = 18000 kilocaloriesSlide12
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PHY 113 C Fall 2013 -- Lecture 21
12
Q=333J
Heat and changes in phase of materials
Example: A plot of temperature versus Q added to
1g = 0.001 kg of ice (initially at T=-30
o
C)
Q=2260JSlide13
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PHY 113 C Fall 2013 -- Lecture 21
13
T
i
T
f
C/L
Q
A
-30
o
C
0
o
C
2.09 J/
o
C
62.7 J
B
0 oC
0 oC333 J
333 JC0 oC
100 oC4.186 J/oC418.6 J
D100 oC
100 oC2260 J2260 JE
100 oC120 oC2.01 J/oC40.2 Jm=0.001 kgSlide14
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PHY 113 C Fall 2013 -- Lecture 21
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Some typical latent heats
Material
J/kg
Ice
Þ
Water (0
o
C)
333000
Water
Þ
Steam (100
o
C)
2260000
Solid N
Þ
Liquid N (63 K)
25500
Liquid N
Þ
Gaseous N
2
(77 K)
201000
Solid Al
Þ
Liquid Al (660
o
C)
397000
Liquid Al
Þ
Gaseous Al (2450
o
C)
11400000Slide15
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PHY 113 C Fall 2013 -- Lecture 21
15
iclicker
question:
Suppose you have 0.3 kg of hot coffee (at 100
o
C
) which you would like to convert to ice coffee (at 0
o
C
). What is the minimum amount of ice (at
0
o
C
) you must add
? (
Note: for water, c=4186 J/(kg
o
C
) and for ice, the latent heat of melting is 333000 J/kg. )
0.2 kg
0.4 kg0.6 kg1 kg
moreSlide16
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PHY 113 C Fall 2013 -- Lecture 21
16
Review
Suppose
you have a well-insulated cup of hot coffee (
m=0.3kg
,
T=100
o
C) to which you
add
0.3
kg of ice (at 0
o
C). When your cup comes to equilibrium, what will be the temperature of the
coffee?Slide17
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PHY 113 C Fall 2013 -- Lecture 21
17
Work defined for thermodynamic process
the “
system
”
In most text books, thermodynamic work is work done
on the “system”Slide18
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PHY 113 C Fall 2013 -- Lecture 21
18Slide19
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PHY 113 C Fall 2013 -- Lecture 21
19
Thermodynamic work
Sign convention: W > 0 for compression
W <
0 for
expansionSlide20
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20
Work done
on system --
“Isobaric” (constant pressure process)
P (1.013 x 10
5
) Pa
P
o
V
i
V
f
W=
-P
o
(
V
f
- V
i
)Slide21
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21
Work done
on system --
“
Isovolumetric
” (constant volume process)
P (1.013 x 10
5
) Pa
V
i
P
i
P
f
W=0Slide22
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PHY 113 C Fall 2013 -- Lecture 21
22
Work done
on system which is an ideal gas
“Isothermal” (constant temperature process)
P (1.013 x 10
5
) Pa
P
i
V
i
V
f
For ideal gas:
PV = nRTSlide23
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PHY 113 C Fall 2013 -- Lecture 21
23
Work done
on
a
system which is an ideal gas:
“Adiabatic” (no heat flow in the process process)
P (1.013 x 10
5
) Pa
P
i
V
i
V
f
Q
i
®
f
= 0
For ideal gas:
PV
g
= P
i
VigSlide24
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PHY 113 C Fall 2013 -- Lecture 21
24Slide25
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PHY 113 C Fall 2013 -- Lecture 21
25
Thermodynamic processes:
Q: heat added to system (Q>0 if T
E
>T
S
)
W: work done on system (W<0 if system expandsSlide26
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PHY 113 C Fall 2013 -- Lecture 21
26
iclicker
question:
What happens to the “system” when Q and W are applied?
Its energy increases
Its energy decrease
Its energy remains the same
Insufficient information to answer questionSlide27
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PHY 113 C Fall 2013 -- Lecture 21
27
Internal energy of a system
E
int
(T,V,P….)
The internal energy is a “state” property of the system, depending on the instantaneous parameters (such as T, P, V, etc.). By contrast, Q and W describe path-dependent processes.Slide28
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PHY 113 C Fall 2013 -- Lecture 21
28
First law of thermodynamics:
E
i
E
f
Q
WSlide29
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29
Applications of first law of
thermodyamics
System
ideal gas
pressure in
Pascals
volume in m
3
# of moles
temperature in K
8.314 J/(
mol
K)Slide30
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30
Ideal gas -- continuedSlide31
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PHY 113 C Fall 2013 -- Lecture 21
31
iclicker
question:
We are about to see several P-V diagrams. Why is this helpful?
It will help us analyze the thermodynamic work
Physicists like nice graphs
It is not actually helpfulSlide32
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32
Constant volume process on an ideal gas
P
V
P
f
V
f
=V
i
P
i
V
iSlide33
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33
Constant pressure process on an ideal gas
P
V
P
f
=
P
i
V
f
P
i
V
iSlide34
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34
Constant temperature process on an ideal gasSlide35
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35
Consider the process described by A
BCA
iclicker
exercise:
What is the net change in total energy?
0
12000 J
Who knows?Slide36
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PHY 113 C Fall 2013 -- Lecture 21
36
Consider the process described by A
BCA
iclicker
exercise:
What is the net heat added to the system?
0
12000 J
Who knows?Slide37
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PHY 113 C Fall 2013 -- Lecture 21
37
Consider the process described by A
BCA
iclicker
exercise:
What is the net work done on the system?
0
-12000 J
Who knows?Slide38
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PHY 113 C Fall 2013 -- Lecture 21
38
Webassign
problem from Assignment
#19