PHY 113 C General Physics I
11 AM - 12:15 pM MWF Olin 101
Plan for Lecture 21:
Chapter 20: Thermodynamics
1. Heat and internal energy
2. Specific heat; latent heat
3. Thermodynamic work
4. First “law” of thermodynamics
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PHY 113 C Fall 2013 -- Lecture 21
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PHY 113 C Fall 2013 -- Lecture 21
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In this chapter T ≡ temperature
in Kelvin or Celsius units
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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withdrawn from the tank to lower the pressure of the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.
P1V1  n1 RT1
P1V1 n1 RT1

P2V2 n2 RT2
P2V2  n2 RT2
P1 n1

P2 n2
P2
5.1
n2  n1
 1.5moles 
 0.2684moles
P1
28.5
n1  n2  1.2316 moles removed
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PHY 113 C Fall 2013 -- Lecture 21
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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 T1 < T2
Q>0 if T1 > T2
T1
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T2
Q
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Heat withdrawn
from system
Thermal equilibrium
– no heat transfer
Heat added to
system
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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
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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?
A. 4186
B. 8372
C. 41860
D. 83720
E. None of these
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Heat capacity: C = amount of heat which must be
added to the “system” to raise its temperature by 1K
(or 1o C).
Q = C DT
Heat capacity per mass: C=mc
Heat capacity per mole (for ideal gas): C=nCv
C=nCp
More generally :
Tf
Qi  f   C (T )dT
Ti
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Some typical specific heats
Material
Water (15oC)
Ice (-10oC)
Steam (100oC)
Wood
Aluminum
Iron
Gold
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J/(kg·oC)
4186
2220
2010
1700
900
448
129
PHY 113 C Fall 2013 -- Lecture 21
cal/(g·oC)
1.00
0.53
0.48
0.41
0.22
0.11
0.03
10
iclicker question:
Suppose you have 0.3 kg of hot coffee (at 100 oC).
How much heat do you need to remove so that the
temperature is reduced to 40 oC? (Note: for water,
c=1000 kilocalories/(kg oC))
A. 300 kilocalories (1.256 x 106J)
B. 12000 kilocalories (5.023 x 107J)
C. 18000 kilocalories (7.535 x 107J)
D. 30000 kilocalories (1.256 x 108J)
E. None of these
Q=m c DT = (0.3)(1000)(100-40) = 18000 kilocalories
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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=-30oC)
Q=333J
Q=2260J
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m=0.001 kg
Ti
Tf
C/L
Q
A
-30 oC
0 oC
2.09 J/oC
62.7 J
B
0 oC
0 oC
333 J
333 J
C
0 oC
100 oC
4.186 J/oC
418.6 J
D
100 oC
100 oC
2260 J
2260 J
E
100 oC
120 oC
2.01 J/oC
40.2 J
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Some typical latent heats
Material
Ice Water (0oC)
Water Steam (100oC)
Solid N Liquid N (63 K)
Liquid N Gaseous N2 (77 K)
Solid Al Liquid Al (660oC)
Liquid Al Gaseous Al (2450oC)
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PHY 113 C Fall 2013 -- Lecture 21
J/kg
333000
2260000
25500
201000
397000
11400000
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iclicker question:
Suppose you have 0.3 kg of hot coffee (at 100 oC)
which you would like to convert to ice coffee (at 0 oC).
What is the minimum amount of ice (at 0 oC) you must
add? (Note: for water, c=4186 J/(kg oC) and for ice, the
latent heat of melting is 333000 J/kg. )
A. 0.2 kg
B. 0.4 kg
C. 0.6 kg
D. 1 kg
E. more
mwater c(T f  Ti )  mice L  0 mice  (.3)(4186)(100) / 333000
 0.377 kg
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Review
Suppose you have a well-insulated cup of hot coffee
(m=0.3kg, T=100oC) to which you add 0.3 kg of ice (at
0oC). When your cup comes to equilibrium, what will be
the temperature of the coffee?
Q  mwater cwater (T f  100)  mice Lice  mice cwater (T f  0)  0
(mwater  mice )cwaterT f  mwater cwater 100  mice Lice
mwater cwater 100  mice Lice
Tf 
(mwater  mice )cwater
cwater  4186 J/(kg  o C)
mwater  mice  0.3kg
Lice  333000 J/kg
T f  10.22 o C
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Work defined for thermodynamic process
General definition of work :
rf
Wi  f   F  dr
ri
In most text books,
thermodynamic work is work
done on the “system”
the “system”
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yf
yf
Vf
F
W   Fdy   Ady    PdV
A
yi
yi
Vi
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Thermodynamic work
Vf
W    PdV
Vi
Sign convention:
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W>0
W<0
for compression
for expansion
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Work done on system --
P (1.013 x 105) Pa
“Isobaric” (constant pressure process)
Po
W= -Po(Vf - Vi)
Vi
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Vf
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Work done on system -“Isovolumetric” (constant volume process)
P (1.013 x 105) Pa
Pf
W=0
Pi
Vi
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Work done on system which is an ideal gas
“Isothermal” (constant temperature process)
P (1.013 x 105) Pa
For ideal gas:
PV = nRT
Pi
Vf
nRT
dV   nRT ln
V
 Vi
Vi
Vf
Wi  f    PdV   
Vi
Vf
  PiVi ln
 Vi
Vi
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Vf



Vf
PHY 113 C Fall 2013 -- Lecture 21
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


Work done on a system which is an ideal gas:
“Adiabatic” (no heat flow in the process process)
P (1.013 x 105) Pa
For ideal gas:
PV = PiVi
Pi
PiVi
PiVi  V f

   PdV     dV  
V
  1  Vi
Vi
Vi

Vf
Wi  f
Vf




 1
PiVi   Vi

1
  1   V f

Vi
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Qif = 0




Vf
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
 1

 1




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Thermodynamic processes:
Q:
W:
heat added to system (Q>0 if TE>TS)
work done on system (W<0 if
system expands
Tf
Q   C (T )dT
Ti
Vf
W    PdV
Vi
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iclicker question:
What happens to the “system” when Q and W are
applied?
A. Its energy increases
B. Its energy decrease
C. Its energy remains the same
D. Insufficient information to answer question
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Internal energy of a system
Eint(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.
DEint  Eint (T f ,V f , Pf )  Eint (Ti ,Vi , Pi )
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First law of thermodynamics:
DEint  Q  W
Ei
Ef
Q
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W
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Applications of first law of thermodyamics
System  ideal gas
8.314 J/(mol K)
PV  nRT
temperature in K
volume in m3 # of moles
pressure in Pascals
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Ideal gas -- continued
Equation of state : PV  nRT
1
1
Internal energy :
Eint 
nRT 
PV
 1
 1
  parameter depending on type of ideal gas
 53 for monoatomic
7
 5
for diatomic
..............................

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iclicker question:
We are about to see several P-V diagrams. Why is
this helpful?
A. It will help us analyze the thermodynamic
work
B. Physicists like nice graphs
C. It is not actually helpful
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Constant volume process on an ideal gas
Equation of state : PV  nRT
1
1
Internal energy :
Eint 
nRT 
PV
 1
 1
P
Pf Vf=Vi
W 0
DEint  Q

P

f
Pi Vi
 Pi Vi
 1
V
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Constant pressure process on an ideal gas
Equation of state : PV  nRT
1
1
Internal energy :
Eint 
nRT 
PV
 1
 1
W   Pi V f  Vi 
P
DEint 
Pi Vi
Pf = Pi Vf
V
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Pi V f  Vi 
 1
Pi V f  Vi 
Q
 1
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Constant temperature process on an ideal gas
Equation of state : PV  nRT
1
1
Internal energy :
Eint 
nRT 
PV
 1
 1
DEint  0
Vf
Q  W   PdV
Vi
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PHY 113 C Fall 2013 -- Lecture 21
Vf 
 nRT ln 
 Vi 
 Pi 
 PiVi ln 
P 
 f 
34
Consider the process described by ABCA
iclicker exercise:
What is the net change in
total energy?
A. 0
B. 12000 J
C. Who knows?
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Consider the process described by ABCA
iclicker exercise:
What is the net heat
added to the system?
A. 0
B. 12000 J
C. Who knows?
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Consider the process described by ABCA
iclicker exercise:
What is the net work done
on the system?
A. 0
B. -12000 J
C. Who knows?
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Webassign problem from Assignment #19
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