PHY 712 Electrodynamics 10-10:50 AM MWF Olin 107 Plan for Lecture 31:
PHY 113 C General Physics I 11 AM - 12:15 p M MWF Olin 101 Plan for Lecture 21: Chapter 20:...
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Transcript of PHY 113 C General Physics I 11 AM - 12:15 p M MWF Olin 101 Plan for Lecture 21: Chapter 20:...
PHY 113 C Fall 2013 -- Lecture 21 111/12/2013
PHY 113 C General Physics I11 AM - 12:15 pM MWF Olin 101
Plan for Lecture 21:Chapter 20: Thermodynamics
1. Heat and internal energy2. Specific heat; latent heat3. Thermodynamic work4. First “law” of thermodynamics
PHY 113 C Fall 2013 -- Lecture 21 211/12/2013
PHY 113 C Fall 2013 -- Lecture 21 311/12/2013
In this chapter T ≡ temperature in Kelvin or Celsius units
PHY 113 C Fall 2013 -- Lecture 21 411/12/2013
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 Determine the number of moles of gas that must be withdrawn from the tank to lower the pressure of the gas from 28.5 atm to 5.10 atm.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.
removed moles 2316.1
moles2684.028.55.1moles5.1
21
1
212
2
1
2
1
22
11
22
11
22221111
nnPPnn
nn
PP
RTnRTn
VPVP
RTnVPRTnVP
PHY 113 C Fall 2013 -- Lecture 21 511/12/2013
T1T2 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 T1 < T2
Q>0 if T1 > T2
PHY 113 C Fall 2013 -- Lecture 21 611/12/2013
Heat withdrawn from system
Thermal equilibrium – no heat transfer
Heat added to system
PHY 113 C Fall 2013 -- Lecture 21 711/12/2013
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
PHY 113 C Fall 2013 -- Lecture 21 811/12/2013
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. 4186B. 8372C. 41860D. 83720E. None of these
PHY 113 C Fall 2013 -- Lecture 21 911/12/2013
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
f
i
T
Tfi dTTCQ )(
:generally More
PHY 113 C Fall 2013 -- Lecture 21 1011/12/2013
Some typical specific heats
Material J/(kg·oC) cal/(g·oC)Water (15oC) 4186 1.00Ice (-10oC) 2220 0.53Steam (100oC) 2010 0.48Wood 1700 0.41Aluminum 900 0.22Iron 448 0.11Gold 129 0.03
PHY 113 C Fall 2013 -- Lecture 21 1111/12/2013
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
PHY 113 C Fall 2013 -- Lecture 21 1211/12/2013
Q=3
33J
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=2260J
PHY 113 C Fall 2013 -- Lecture 21 1311/12/2013
Ti Tf C/L QA -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
m=0.001 kg
PHY 113 C Fall 2013 -- Lecture 21 1411/12/2013
Some typical latent heats
Material J/kgIce ÞWater (0oC) 333000Water Þ Steam (100oC) 2260000Solid N Þ Liquid N (63 K) 25500Liquid N Þ Gaseous N2 (77 K) 201000
Solid Al Þ Liquid Al (660oC) 397000Liquid Al Þ Gaseous Al (2450oC) 11400000
PHY 113 C Fall 2013 -- Lecture 21 1511/12/2013
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 kgB. 0.4 kgC. 0.6 kgD. 1 kgE. more
kg 0.377
333000/)100)(4186)(3(. 0)(
iceiceifwater mLmTTcm
PHY 113 C Fall 2013 -- Lecture 21 1611/12/2013
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?
C10.22
J/kg 333000 C) J/(kg 4186
3.0 )(
100
100)(
0)0()100(
o
o
f
icewater
icewaterwatericewater
iceicewaterwaterf
iceicewaterwaterfwatericewater
fwatericeiceicefwaterwater
T
Lc
kgmmcmm
LmcmT
LmcmTcmm
TcmLmTcmQ
PHY 113 C Fall 2013 -- Lecture 21 1711/12/2013
Work defined for thermodynamic process
: workof definition General
f
i
dW fi
r
r
rF
the “system”
In most text books, thermodynamic work is work done on the “system”
PHY 113 C Fall 2013 -- Lecture 21 1811/12/2013
f
i
f
i
f
i
V
V
y
y
y
y
PdVAdyAFFdyW
PHY 113 C Fall 2013 -- Lecture 21 1911/12/2013
Thermodynamic work
f
i
V
V
PdVW
Sign convention: W > 0 for compression W < 0 for expansion
PHY 113 C Fall 2013 -- Lecture 21 2011/12/2013
Work done on system --
“Isobaric” (constant pressure process)P
(1.0
13 x
105 )
Pa
Po
Vi Vf
W= -Po(Vf - Vi)
PHY 113 C Fall 2013 -- Lecture 21 2111/12/2013
Work done on system --
“Isovolumetric” (constant volume process)P
(1.0
13 x
105 )
Pa
Vi
Pi
Pf
W=0
PHY 113 C Fall 2013 -- Lecture 21 2211/12/2013
Work done on system which is an ideal gas
“Isothermal” (constant temperature process)P
(1.0
13 x
105 )
Pa
Pi
Vi Vf
i
fii
i
fV
V
V
Vfi
VV
VP
VV
nRTdVV
nRTPdVWf
i
f
i
ln
ln
For ideal gas:
PV = nRT
PHY 113 C Fall 2013 -- Lecture 21 2311/12/2013
Work done on a system which is an ideal gas:
“Adiabatic” (no heat flow in the process process)P
(1.0
13 x
105 )
Pa
Pi
Vi Vf
Qif = 0
11
1
i
fiiV
V
iiV
Vfi V
VVPdVVVPPdVW
f
i
f
i
For ideal gas:
PV = PiVi
1
11
f
iii
VVVP
PHY 113 C Fall 2013 -- Lecture 2111/12/201324
PHY 113 C Fall 2013 -- Lecture 21 2511/12/2013
Thermodynamic processes: Q: heat added to system (Q>0 if TE>TS) W: work done on system (W<0 if
system expands
f
i
f
i
V
V
T
T
PdVW
dTTCQ )(
PHY 113 C Fall 2013 -- Lecture 21 2611/12/2013
iclicker question:What happens to the “system” when Q and W are applied?
A. Its energy increasesB. Its energy decreaseC. Its energy remains the sameD. Insufficient information to answer question
PHY 113 C Fall 2013 -- Lecture 21 2711/12/2013
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.
),,(),,( intintint iiifff PVTEPVTEE D
PHY 113 C Fall 2013 -- Lecture 21 2811/12/2013
First law of thermodynamics:
WQE D int
Ei Ef
Q W
PHY 113 C Fall 2013 -- Lecture 21 2911/12/2013
Applications of first law of thermodyamics
System ideal gas
nRTPV
pressure in Pascals
volume in m3 # of moles
temperature in K
8.314 J/(mol K)
PHY 113 C Fall 2013 -- Lecture 21 3011/12/2013
Ideal gas -- continued
..............................diatomicfor
monoatomicfor
gas ideal of on type dependingparameter 1
11
1 :energy Internal
:state ofEquation
5735
int
PVnRTE
nRTPV
PHY 113 C Fall 2013 -- Lecture 21 3111/12/2013
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 graphsC. It is not actually helpful
PHY 113 C Fall 2013 -- Lecture 21 3211/12/2013
Constant volume process on an ideal gas
PVnRTE
nRTPV
11
11 :energy Internal
:state ofEquation
int
P
V
Pf Vf=Vi
Pi Vi
1
0
int
D
iif VPP
QEW
PHY 113 C Fall 2013 -- Lecture 21 3311/12/2013
Constant pressure process on an ideal gas
PVnRTE
nRTPV
11
11 :energy Internal
:state ofEquation
int
P
V
Pf = Pi VfPi Vi
1
1int
D
ifi
ifi
ifi
VVPQ
VVPE
VVPW
PHY 113 C Fall 2013 -- Lecture 21 3411/12/2013
Constant temperature process on an ideal gas
PVnRTE
nRTPV
11
11 :energy Internal
:state ofEquation
int
D
f
iii
i
f
V
V
PPVP
VV
nRT
PdVWQ
Ef
i
ln
ln
0int
PHY 113 C Fall 2013 -- Lecture 21 3511/12/2013
Consider the process described by ABCA
iclicker exercise:What is the net change in total energy?
A. 0B. 12000 JC. Who knows?
PHY 113 C Fall 2013 -- Lecture 21 3611/12/2013
Consider the process described by ABCA
iclicker exercise:What is the net heat added to the system?
A. 0B. 12000 JC. Who knows?
PHY 113 C Fall 2013 -- Lecture 21 3711/12/2013
Consider the process described by ABCA
iclicker exercise:What is the net work done on the system?
A. 0B. -12000 JC. Who knows?
PHY 113 C Fall 2013 -- Lecture 21 3811/12/2013
Webassign problem from Assignment #19