Physics 231 Lecture 29
• Main points of today’s lecture:
• Temperature
5
Tcelcius  (Tfarenheit  320 )
9
Tcelcius  Tkelvin  273.15
• Thermal expansion
L  TL; A  A0 T ;
V  V0 T
• Ideal gas properties:
PV  nRT  Nk BT
2  N  1
2N 
2 
m
v

 

  KE
3  V  2
 3V 
3
KE  K E  k BT
2
P
Zeroth Law of Thermodynamics
• If objects A and B are in thermal equilibrium with a third object, C, then
A and B are in thermal equilibrium with each other.
• Allows a definition of temperature
Different Temperature Scales
• Three different temperature scales are
commonly in use:
– Fahrenheit
– Celsius
– Kelvin
• Celsius define such that the freezing
point of water is at a temperature of
about 0o C and boiling temperature is
at about 100o C.
• The conversions from Celcius to • At TK=0, molecules in gas are nearly at rest
and many material have unusual properties.
Farenheit and to Kelvin are as
What is this temperature in Fahrenheit?
follows:
0


273.15
C
T

T

273.15
0
C
K
a) 0 F
9
TF  TC  32
9
b) -2730 F
5
TF  TC  32
5
c) – 4600 F
TK  TC  273.15
9
0

 273.15  32  459.670 F
d) -923 F
5
Thermometers
• While we know what is hot or cold. Much of what we think is simply
perception:
– Wind makes day “feel” colder because moving air strips away the
insulating layer of warm air next to our skin.
– Cold metal feels colder than equally cold wood.
– Humid winter days feel colder than dry winter days. How do we
construct a reliable thermometer?
• Need a quantity that has a unique dependence on
temperature. Example the length of a column of mercury in
a thermometer grows linearly with temperature.
• Many solids and liquids expand linearly with T: L  L0 T
• Example: A aluminum rod which is 1.0000 m long at 0C.
What is its length at 100 C? (Al=23x10-6 / C)
6 o
o
a) 0.9989 m L   L0 T  23x10 / C 1.0000m  100 C
b) 1.0011 m L  0.0023m
c) 1.0023 m
L  L0  L  1.0023m
d) 1.0045 m




Temperature and Thermal Expansion
Slide 12-42
Area and Volume expansion
• The areas and volumes of many objects also expand with temperature.
• Example: Calculation of the area expansion coefficient  for a
rectangular aluminum plate.
Af  xy   x0  x  y0  y ; x  x0 T ; y  y0 T
 Af   x0  x0 T  y0  y0 T   x0 y0 1  T 1  T 
 Af  A0 1  2T   2 T 2   A  Af  A0  2TA0
y
x
A  TA0 where   2
• By analogy, the volume expansion coefficient =3 for a uniform
material, where:
V  TV0
• Quiz: A swimming pool contains 110 m3 ( about 30,000 gal of water). The
sun heats the water from 17 to 27C. What is the change in the volume of
the water? =2.07x10-4 / C.
– a) 0.08 m3
  2.07x10-4 / o C
V0  110m3 V  V0 T
– b) 0.13 m3
V  110m3 2.07x10-4 / o C 10o C  0.228m3
3
– c) 0.17 m
– d) 0.23 m3



Example
• A brass ring of diameter 10.00 cm at 20.0°C is heated and
slipped over an aluminum rod of diameter 10.01 cm at 20.0°C.
Assuming the average coefficients of linear expansion are
constant, to what temperature must this combination be cooled
to separate them? (Al=24x10-6 /oC, Br=19x10-6 /oC)
Want D Br  D Al
D Br  D Br,0  D Br ; D Al  D Al,0  D Al
D Br,0  D Br  D Al,0  D Al
 D Br,0  D Al,0  D Al  D Br
10.00  10.01   Al D Al T   Br D Br T
 0.01    A D Al   Br D Br  T


 0.01  (24x10 6 10.01  19x10 6 10.00) T  5.0x10 5 T
0.01 0
0


199
C  T

C  T
5
5.0x10
 Tf  200 C  T
 1790 C
The special case of Ice
• Many material contract when changing from liquid to solid. Ice is
an exception. The lowest energy configuration of the solid
occupies a larger volume than the same mass of liquid.
• The fact is key to spread of life across the planet because it means
that ice is on the surface of large bodies of water where it will
melt during summer rather than on the bottom.
Phases of Matter
In the solid form, ice, the molecules are farther apart than they
are in the liquid form, water.
Slide 12-16
Reading Quiz
2. A sample of nitrogen gas is in a sealed container with a constant
volume. Heat is added to the gas. The pressure
A.
B.
C.
D.
increases
stays the same
decreases
can’t be determined with the information given
Slide 12-8
Ideal gas
• Ideal gas pressure depends linearly on temperature.
nR
TK
V
If T is measured in Kelvin
P  aT  b 
• Ideal gas pressure depends linearly on temperature.
• Here n=number of moles of the gas. There are NA = 6.02x1023
molecules or atoms (if the atoms don’t combine into molecules) per
mole. One mole = .0224 m3 of gas at T=0oC and P=1 atm.
• R=8.31 J/(moleK)
• Example: A molecular gas is contained in an 8.0-L vessel at a
temperature of 20°C and a pressure of 9.0 atm. (a) Determine
the number of moles of gas in the vessel. (b) How many
molecules are in the vessel?



PV 9 1.01x105 Pa .008m3
a) n 
 3.0 moles
RT  8.31J / K  293K 
b) N  nN A  3  6.02x1023  1.8x1024 molecules
Example
• Gas is confined in a tank at a pressure of 10.0 atm and a temperature of
15.0°C. If half of the gas is withdrawn and the temperature is raised to
65.0°C, what is the ratio of the final density over the initial density?
N0m
0 
V
1
N0m
Nf m
f 
 2
V
V

1
0
2
• What is the new pressure in the tank?
N
N0
Pf  f k BTf
k BT0
V
V
Nf
k BTf
N f Tf
Pf
1 273  65 1 338
V




 0.59
N 0 T0
P0 N 0 k T
2 273  15 2 288
V B 0
P0 
Pf  0.59 10atm
 5.9atm
Physics 231
Lecture 30
• Main points of today’s lecture:
PV  nRT  Nk BT
• Ideal gas law:
2  N  1
2  N
3
2 
P     m v     KE ; KE  KE  k BT
 3  V
3  V  2
2
• Heat and heat capacity: Q  cmT
• Work in thermodynamic processes: Wsystem  PV
• First Law of Thermodynamics: U  Q  PV
Checking Understanding: Pressure and Forces
The two identical cylinders contain
samples of gas. Each cylinder has
a lightweight piston on top that is
free to move, so the pressure inside
each cylinder is equal to atmospheric
pressure. One cylinder contains
hydrogen, the other nitrogen. Both
gases are at the same temperature.
The number of moles of hydrogen is
A. greater than the number of moles of
nitrogen.
B. equal to the number of moles of nitrogen.
C. less than the number of moles of nitrogen.
PV  nRT

n
TH =TN
VH =VN
PH =PN
nH/nN=?
PV
RT
Slide 12-24
Checking Understanding: Pressure and Forces
The two identical cylinders contain
samples of gas. Each cylinder has
a lightweight piston on top that is
free to move, so the pressure inside
each cylinder is equal to atmospheric
pressure. One cylinder contains
hydrogen, the other nitrogen. The
mass of gas in each cylinder is the same.
The temperature of the hydrogen gas is
A. greater than the temperature of the
nitrogen.
B. equal to the temperature of the nitrogen.
C. less than the temperature of the nitrogen.
TH /TN=?
VH =VN=V
PH =PN=P
mH=mN=m
PV
1
PV

T
PV  nRT
n
T
n R n
1
nR
 H  H  H  N 
m
PV
1
TN
n H 14
m  n H 2g  n N 28g n H  m & n N 
28g
n NR n N
Slide 12-26
2g
Reading Quiz
2. When the temperature of an ideal gas is increased, which of the
following also increases? (1) The thermal energy of the gas; (2) the
average kinetic energy of the gas; (3) the average potential energy
of the gas; (4) the mass of the gas atoms; (5) the number of gas
atoms.
A.
B.
C.
D.
E.
1, 2, and 3
1 and 2
4 and 5
2 and 3
All of 1–5
Slide 11-8
Reading Quiz
What is the mass, in u, of a molecule of carbon dioxide, CO2?
A.
B.
C.
D.
E.
12
24
32
36
44
Each nucleon has a mass of about 1 u.
Most carbons has 6 protons and 6 neutrons
for a total of 12 nucleons or m=12u.
Most Oxygens have 8 protons and 8 neutrons
for a total of 16 nucleons or m=16u.
The total mass is 12+2*16 u = 44 u
Slide 12-12
Kinetic properties of an ideal gas
• Consider the gas volume at shown on the right. A
molecule with mass m is elastically scattered by the
wall which supplies an impulse force F1.
F1t  mv x  m( v x )  2mv x  F1 
wall
2mv x
t
wall
• After a time t=2d/(vx), the mass bounce off the far
wall and return to bounce of this wall again. If we
use this value for t in the expression above, we
can get the average force for much large time
intervals.
2mv
F1 
x
2d / vx 
d
A
 mv x2 / d
• Multiplying by the number of molecules in the gas and dividing by the area A,
we get the pressure on the wall. P  N mv 2  N mv 2
Ad
x
V
x
• This would be correct if each molecule was moving the x direction with
exactly velocity vx. To correct for the differences in velocities, we use the
average <vx2> instead. Also <v2>= <vx2>+ <vy2>+ <vz2>=3 <vx2>. Thus”
P
N
2 N 1
2N
2 
m v2 / 3 
m
v

 KE 


V
3 V 2
 3V
Average kinetic energy of gas molecules
• If we combine the last expression with the ideal gas law equation of
state, we get a useful expression for the means kinetic energy of gas
molecules:
2N
R
n
N
P
 KE  P  RT N  nN A k B 
 P  k BT
3V
NA
V
V
2N
N
1
1
3

 KE  k BT   KE  m v 2  mv 2rms  k BT  v rms 
3V
V
2
2
2
3k BT
m
note: increasing temperature increases thermal (KE) energy
(adding heat energy Q)
• Example: If the translational rms speed of water molecules (H2O) in a
volume of air is 648 m/s, what is the translational rms speed of carbon
dioxide (CO2)? Both gases are at the same temperature.
(vrms[<v2>]1/2)
v rms,H2O
 v rms,CO2 
3k BT

m H2O
m H2O
m CO2
v rms,H2O 
v rms,CO2
3k BT


m CO2
N A m H2O
N A m CO2
v rms,H2O 
m H2O 3k BT

m CO2 m H2O
m H2O
m CO2
3k BT
m H2O
2  16
 648m / s  414m / s
12  2 16
quiz
• Consider a gas mixture of the two monoatomic gases: 50% Helium
(Mmole=4.0 g; Mmole is the mass of one mole of the gas.) and 50%
Argon (Mmole=36.0 g). The ratio vrms,He / vrms,Ar of the rms speed for
helium atoms divided by the rms speed for Argon atoms is:
– a) .06
3k BT
3k BT
– b) .17
v rms,Ar 
v rms,He 
m Ar
m He
– c) .67
– d) 1.3
– e) 3.0

v rms,He
v rms,Ar

3k BT
m He
3k BT
m Ar
1
m He m He m Ar
1 m He m Ar
m Ar


m Ar
36

3
m He
4
Speed and Kinetic Energy of Gas Molecules
1 2
Kavg = mv rms
<KE>
2
<KE>
2K
avg
T
3 kB
2
K avg  123kmv
T
vrms = B rms
K avg  23 kmBT
kB  1.38  10 23 J/K
PV  NkBT
 m 
F(v)  4 
 2 k BT 
3/2
 mv 2 
v exp  
 2k T 
2
B
Slide 12-17
The Ideal Gas Model
2 <KE>
Kavg
T
3 kB
3
E th  N  KE  Nk BT
2
Transfer of heat at constant
volume:
Slide 11-18
Checking Understanding
Two containers of the
same gas (ideal) have
these masses and
temperatures:
• Which gas has
atoms with the
largest average
thermal
energy?
• Which container of gas has the largest thermal energy?
A. P, Q
1
3
2
 mv  k BT, which is independent of mass
B. P, P
2
2
C. Q, P
Q has the highest T  highest thermal
D. Q, Q
energy per atom
3
E th  Nk BT  E th,P
2
E th,Q
3
N P k BTP
 2
3
N Q k BTQ
2

N P TP 100 273  0

1
NQ TQ 20 273  50
Slide 11-21
Physics 231 Lecture 31
• Main points of today’s lecture:
• Heat and heat capacity:
Q  cmT
• Phase transitions and latent heat:
Q  Lm
• Mechanisms of heat flow.
• Conductive heat flow
Q kAT2  T1 

t
L
• Examples of heat conductivity, R
values for insulators
H
Ri  Li / ki
Heat and heat capacity
• We saw that for atoms or molecules in an ideal gas <KE>=3/2kBT. In
general, the atoms or molecules in matter increase in energy if the
object is heated. This is thermal energy, which can be transfer to and
from the gas.
• We define the heat energy Q to be the energy that flows from a hot
object to a cold object solely because of the difference in T.
• When the heat flow is sufficient, the two objects will reach the same
temperature and we say they are in thermal equilibrium at the same T.
• If we put Q into a mono-atomic ideal gas at constant volume:
Q  E th  E th,f  E th,i 
Q
3
3
3
Nk b Tf  Nk b Ti  Nk b  Tf  Ti 
2
2
2
3
3
Nk b T  nRT  nc molar,V T cmolar,V=3/2R is the heat capacity per mole at
2
2
constant volume for mono-atomic ideal gas
• For many other materials, the relationship between heat transferred and
temperature change is given by: (usually transferred at constant V or P)
Q  cmT
c is the specific heat capacity per unit mass
m is the mass and cm the total heat capacity of the object.
Example
• Helium (He), a monoatomic gas, fills a 0.010 m3 container. The
pressure of the gas is 6.2x105Pa. How long would a 0.25 hp engine
have to run (1 hp=746 W) to produce an amount of work equal to the
thermal energy of this gas?
3
3
E th  N KE  Nk BT  nRT
2
2
PV  nRT



3
3
 E th  Ppressure V  6.2x105 Pa 0.010m3  9300J
2
2
E th  Work  0.25  746W  t
E
9300J
 t 

 50s
0.25  746W  187W
Quiz
• Two moles of the monatomic gas helium (CV=3/2R) are initially at a
temperature of 300K. The gas is cooled at constant volume. How much
heat must be removed to decrease the temperature (in Kelvin) by a
factor of two? (Note: R=NAkB=8.31 J/(molek))
– a) 1.7 kJ
3
3
E th,f  nRTf
E th,i  nRTi
– b) 2.7 kJ
2
2
E th,f  E th,i  Q
– c) 3.7 kJ
– d) 4.7 kJ
3
3
1
Q

E

E

nRT

nRT
th,f
th,i
Ti
f
i  TF 
– e) 5.7 kJ
2
2
2
3
3

  2moles   8.31J / mole  K    150K 
Q  nR  Tf  Ti 
2
2
Q  3.7kJ
removed energy = 3.7 kJ
Conservation of heat energy
• Imagine two objects A and B, with thermal energies Eth,A.0 and Eth,B,0
and TA>TB.
• When they are put in thermal contact, they will come to equilibrium
and heat Q will be transferred from A to B. Then, we can write
E th,A,f  E th,A,0  QA
E th,B,f  E th,B,0  QB
From conservation of energy: QB  QA  0
QB  QA  Q
E th,A,f  E th,A,0  Q
E th,B,f  E th,B,0  Q
Example of equilibration and energy conservation
• At a fabrication plant, a hot metal forging has a mass of 75 kg and a specific
heat capacity of 430 J/(kg °C). To harden it, the forging is quenched by
immersion in 710 kg of oil that has a temperature of 32°C and a specific heat
capacity of 645 cal/(kg °C). The final temperature of the oil and forging at
thermal equilibrium is 47°C. Assuming that heat flows only between the
forging and the oil, determine the initial temperature of the forging.
Qoil  coil m oil  Tf  Toil 
Q metal  c metal m metal  Tf  Tmetal 
Energy conservation:
Q metal  Qoil  0
 Q metal  Qoil
4.19J
coil  645cal / (kg  C) 
 2700kJ / (kg  o C)
cal
o
c metal m metal  Tf  Tmetal   coil m oil  Tf  Toil 
Tf  Tmetal 
coil m oil  Tf  Toil 
 Tmetal  Tf 
c metal m metal
coil m oil  Tf  Toil 
c metal m metal
mmetal
75 kg
moil
710 kg
cmetal
430J/kg/ oC
coil
645cal/kg/ oC
Tf
47 oC
Toil,0
32 oC
Tmetal,0
?
o
o
2700J
/
(kg

C)
710kg
47

32
C



o
 47 C 
 939 o C
o
430J / (kg  C)  75kg 
Example
• Example: A 100 kg mass of water at a temperature of 30°C is dropped
into a thermally isolated vessel containing 50 kg of water which is a
temperature of 10°C. After system comes to thermal equilibrium the
final temperature is (useful information: cwater=4186 J/(kg• °C))
– a) 11°C
Q100  c water 100kg Tf  30o C
– b) 14°C
– c) 19°C
Q50  c water 50kg Tf  10o C
– d) 23°C




Q100  Q50  0




c water 100kg Tf  30o C  c water 50kg Tf  10o C  0


2 Tf  30o C  Tf  10o C  0
Tf  23o C
3Tf  60o C  10o C
Phase transitions
T
Q
•
Many substances display a plateaus in the caloric curve that describes T vs. Q.
Regions with a linear increase correspond to a constant heat capacity where
Q  cmT  T  Q /(cm)
• The flat regions occur when the system has a phase transition. For water,
there is one where ice changes to water and other where water changes to
steam. If the pressure remains constant during the phase change, the
temperatures will remain constant. The heat required to change a mass
m of the matter is given by the latent heat L for the phase change.
Q  Lm L fusion  33.5 x104 J / kg
Lvaporization  22.6 x105 J / kg
Example
• When it rains, water vapor in the air condenses into liquid water, and
the energy is released. (a) How much energy is released when 0.0254
m (one inch) of rain falls over an area of 2.59x106 m2 (one square
mile)? (b) If the average energy needed to heat one home for a year is
1.5x1011 J, how many homes can be heated with the energy determined
in part (a)?


a) m water  water Vwater  1000kg / m3 2.59x106 m 2 .0254m   6.58x107 kg



Q released  L vaporization m water  22.6x105 J / kg 6.58x107 kg  1.5x1014 J
b) n houses 1.5x1011 J  1.5x1014 J  n houses  1000
Heat flow
• There are three major processes that
transfer heat from one point to
another.
– Conduction
– Convection
– Radiation
• Convection results from the fact that
hot objects generally expand. This
decreases their density. If this occurs
in a fluid, the less dense hot fluid
rises and the colder denser fluid falls.
Conduction
• Conduction concerns the transfer of heat through materials without
convection.
• Consider an concrete wall of a heated garage. The outside of the
garage is at temperature T1 and the interior of the garage is a
temperature T2. The conductive heat flow through a portion of the wall
with area A is given by:
H
Q kAT2  T1  kAT


t
L
L
• H is the heat flow through the wall, k is the thermal conductivity of
concrete, and L is the thickness of the concrete wall.
• Example: Calculate the heat flow through a 2 m2 section of a 20 cm
thick concrete wall when the outside temperature is 0 oC and the inside
temperature is 20 oC. Assume the thermal conductive of the concrete is
1.3 J/(s m  oC)



1.3J / (s  m  o C) 2m 2 20 o C
Q kAT
H


 260W
t
L
0.2m
Physics 231 Lecture 32
• Main points of today’s lecture:
• Heat flow
Q kAT2  T1 
H

t
L
• Examples of heat conductivity, R
values for insulators
Ri  Li / ki
• Convection
• Radiation
P  AeT 4
where   5.67x10-8W /( A  K 4 )
• Work in thermal processes:
Wsystem  PV
Quiz
• Two materials have the same insulating value if the same amount of
heat per second per square meter flows through each due to the same
temperature difference. Ignoring air convection, what thickness of
body fat is required to give the same insulating value as a 0.010 inch
thickness of air? (kfat=0.2 J/(sm oC), kair=0.023 J/(sm oC))
–
–
–
–
a) 0.09 inches
b) 0.7 inches
c). 2.3 inches
d) 4.2 inches
want H fat  H air
k fat AT k air AT

Lfat
Lair
 Lfat  Lair

k fat k air

Lfat Lair
k fat
0.2
  0.01inch 
 .09inch
k air
0.023
Conceptual quiz
• Two large metal blocks made of the same material and at the
same temperature are brought together and placed into thermal
contact. Block A contains twice as much heat as Block B. Which
of the following statements is true
– a) There will be net flow of heat from A to B
– b) Net flow of heat from B to A
– c) No net heat flow will occur
– d) Not enough information to tell
H
Q kAT2  T1  kAT


t
L
L
Layered materials – R values
Li
• Consider the layered insulating structure at the right. The
area of each layer is the same. Each layer can have a
A
different thickness Li and heat conductivity ki.
• Here the important thing to remember is that we are
concerned with a steady state solution. There is no build H
up of heat in any of these layers. For each layer one has:
Li
Q k i ATi
H Li H
H

 Ti     R i ; R i 
t
Li
A ki A
ki
• Here Ri it the “R value” of the ith insulating layer. The total temperature
difference is given by the sum.
i.e. T 
H
T   Ti    R i
A i
i
H
Q AT AT


t  R i
R tot
i
H
  Ri
A i
Example
R Cu
• Two rods, one of aluminum and the other of copper are joined end to
end. The cross-sectional area of each is 4.0x10-4 m2, and the length of
each is 0.04 m. The free end of the aluminum rod is kept at 302 oC,
while the free end of the copper rod is kept at 25 oC. The loss of heat
through the sides of the rods may be ignored. (a) What is the
temperature at the aluminum-copper interface? (b) How much heat is
conducted through the unit in 2.0 s? (kalum = 238 J/(s  m oC), kCu =
397 J/(s  m oC)) Do b) first.
Q AT
R tot  R Cu  R Al
b) H 

t
R tot
LCu
.04m


 1.01x10 4 s m 2 C / J
k Cu 397J / (s mC)
R Al 
L Al
.04m

 1.68x10 4 s m 2 C / J
k Al 238J / (s mC)
}
R tot  2.69x10 4 s m 2 C / J
4
2
AT 4x10 m  302  25 C
H

 412W
 Q  Ht  412W 2s  824J
4
2
R tot
2.69x10 s m C / J
4
2
R Cu
H LCu
1.01x10
s
m
C / J
a) TCu  
=H 
TCu  412W 
 103.7C
4
2
A k Cu
A
4x10 m
Tinterface  25 o C  103.7 o C  128.7 o C
Radiation
• The Earth is warmed by the
radiation from the sun.
• The rate at which an object
radiates energy is given by
Stefan’s Law which provides the
power radiated from a object with
surface area A at temperature T.
P  AeT 4 where   5.67x10-8W /( A  K 4 )
• Here e is emissivity (0e1) which describes how the radiation or
absorption can be reduced below that for a “black” surface that absorbs
light completely.
• Example: The filament of a light bulb has a temperature of 3.0x 103 oC
and radiates sixty watts of power. The emissivity of the filament is 0.36.
Find the surface area of the filament.
– a) 3.4 m2
60W
P

A
4
4
– b).063 m2
 eT
5.67x108 W / (m 2  K 4 )(.36)  3273K 
– c) 2.6x10-5 m2
5 2

A

2.6x10
m
-6
2
– d) 9.8x10 m
Example
• A solid aluminum sphere is coated with lampblack (emissivity=0.97) and
hung inside an evacuated container. The sphere has a radius of 0.020 m
and is initially at 20.0 oC. The container is maintained at a temperature
of 70.0 oC. (a) Assuming that the temperature of the sphere does not
change very much, what is the net energy gained by the sphere in 10.0 s?
(b) Estimate the change in temperature of the sphere. (Note: all objects
radiate and absorb radiation at the same time.)
a) Detailed balance: In equilibrium power emitted  power absorbed.
P20   Ae  (293K) 4  power emitted
P70   Ae  (343K) 4  power absorbed
E   P70  P20  t is energy gained by sphere
E   Ae   P70  P20  t  5.67x10-8 4 (0.02) 2 (.97)[ 343   293 )10s  18J
4
4
b) Have E  Q (no work), need heat capacity to get temperature change
ΔE


T

ΔE  calum m alum T  calum alum Valum T
3
calum alum 4 .02m  / 3
18J
0
T 

0.22
C
3
3
(900J / (kg  K)) 2700kg / m 4 .02m  / 3


Temperature of the Planets
• The surface temperature of the Sun is about
Planet
5800 K. a) Taking the Sun’s radius to be 6.96 x
Mercury
108 m, calculate the total energy radiated by the
Venus
Sun each second. (Assume e = 0.965.) b)
Earth
Calculate the equilibrium temperature of the
Earth, Mercury, Venus and Mars. Assume for
Mars
simplicity that the emissivity of the Earth,
Mercury, Venus and Mars are all also 0.965.
The orbital radii of these planets are given in the
table.
Orbital radius
~5.8x1010 m
1.1x1011 m
1.5x1011 m
2.3x1011 m
First calculate the radiant power emitted by the sun

4
a) Psun   AeTsun
 5.67x10 8 W / (m 2  K 4 )4 6.96x108 m
 Psun  3.76x1026 W

2
.9655800K  4
Temperature of planets continued
• Let’s calculate for the earth first. At a distance equal to the earth’s
radius, rearth, Psun is spread over a spherical area of 4r2earth.. The earth
covers a circular area of R2earth where Rearth is the radius of the earth.
• The power absorbed on earth therefore is:
2
eearthRearth
b) Psun _ on _ earth  Psun
2
4rearth
• The earth is at temperature Tearth and radiates in all directions with a
power: P  A e T 4   4R 2 e T 4
earth
earth earth earth
earth earth earth
In steady state
Psun _ on _ earth
T
4
earth
2Re
2
eearthRearth
2
4
 Psun

P


4

R
e
T
earth
earth
earth
earth
2
4rearth
 Psun
re
2
eearthRearth
Psun
9
4


5
.
88
x
10
K
2
2
2
4rearth
 4Rearth
eearth 16rearth
 Tearth  277 K 00C
4
Note : Tmerc
. 
Psun
Psun
4
T

venus
2
2
16rmerc
16rvenus
.
4
Tmars

Psun
2
16rmars
 Tmerc.  445 K  1720C Tvenus  323K  500C Tmars  223K  500C
Sometimes
Planet
Orbital radius Calculated T
Measured <T>
Mercury 5.8x1010 m
445 K
443K
Venus
1.1x1011 m
323 K
723 K
Earth
1.5x1011 m
277 K
287 K
Mars
2.3x1011 m
223 K
230 K
• Even though Mercury is much closer to the Sun than Venus is, Venus
is the hottest planet in our Solar System. Temperatures on the plains of
Venus often reach 850 degrees Fahrenheit (450 degrees Celsius). This
is because of the thick layer of carbon dioxide clouds that surround
Venus. Much of the heat from the Sun that hits Mercury escapes back
into space because Mercury does not have a large atmosphere with
clouds to trap in the warmth. But the carbon dioxide clouds on Venus
are so thick they create what scientists call a Greenhouse Effect.
• We cannot calculate the Greenhouse effect using Stefan’s equation. It
occurs because the emissivity of the Venusian atmosphere is much
larger for long wavelength infrared light and smaller for shorter
wavelength light.
Physics 231 Lecture 33
• Main points of today’s lecture:
• Work in thermodynamic
processes:
Wsystem  PV
• First Law of Thermodynamics:
U  Q  PV
• Specific heat at constant pressure
• Processes
P  V  T  0
– cyclic:
– isobaric:
P  0
– isovolumetric: V  0
T  0
– isothermal:
Q  0
– adiabatic:
Conceptual question
• Which gives the largest average radiation absorbed on a 1 m2
area at that distance?
– 1. a 50-W source at a distance R.
– 2. a 100-W source at a distance 2R.
– 3. a 200-W source at a distance 4R.
Pabsorbed  fraction of area covered e emitted power
Pabsorbed
1m2

eP
2
4 r
1m2
P1 
e 50
2
4 R
1m2
P2 
e 100
2
4 (2R)
1m2
P2 
e 200
2
4 (4R)
Work and the 1st Law of thermodynamics
•
•
•
Consider the ideal gas contained in the volume under the
cylinder at right. The piston compresses the gas very
slowly, moving downwards with a constant velocity.
The gas exerts a force Fgas=PA upward on the piston and
conversely, the piston exerts a force Fpiston=-PA downward
on the gas.
The work done by the piston on the gas is:
W piston  Fpiston y  PA y   PV
•
The work done by the gas on the piston is:
Wgas   Fgas y   PA y  PV
•
If U denotes the internal energy of the system, the
conservation of energy dictates:
U  E th  Q  Wpiston  Q  Wgas  Q  PV 1st Law of Thermodynamics
Here I have changed notation to U, because other systems,
besides the ideal gas, can have potential energy.
Why does work increase thermal energy
• Recall that the thermal energy U is given by U=3/2NkBT
• If work done on the gas increases U it must increase T.
• Positive work done on the gas increases U because the collisions with
the piston moving downward increases the velocity of the molecules.
• Negative work done on the gas decreases U because the collisions with
the piston moving upward decreases the velocity of the molecules.
Example
• The work done to compress one mole of a monatomic ideal gas is 6200
J. The temperature of the gas changes from 350 K to 550 K. How
much heat flows between the gas and its surroundings? Determine
whether the heat flows into or out of the gas.
Work done by the gas is : Wgas  6200J
3
Internal energy change in the gas is: U  nR  T 
2
3
3
Q  U  Wgas  nR  T   6200J  8.31J / K  550  350K   6200J
2
2
Q  3700J
Computation of work:
• What is the work done by the gas going from (Pi ,Vi) to (Pf ,Vf) ?
– a) Pi(Vi-Vf)
– b) Pi(Vf-Vi)
– c) 0
– d) Vi(Pi-Pf)
– e) Vf(Pi-Pf)
• The work done by the gas is the mathematical area under the curve Pi(Vf-Vi). It
is negative if the volume is decreasing. It is pushing to increase the volume, but
the volume is being decreased.
• No work is done on the isovolumetric (constant volume) part of the path!
Computation of work:
• What is the work done on the gas going from (Pi ,Vi) to (Pf ,Vf) ?
– a) Pi(Vi-Vf)
– b) Pi(Vf-Vi)
– c) 0
– d) Vi(Pi-Pf)
– e) Vf(Pi-Pf)
• Note that the work done on the gas is (-1) times the mathematical area under the
curve, -Pi(Vf -Vi). The mathematical area is negative if the volume is decreasing.
The work done on the gas is positive, you are compressing it and thereby
exerting a force in the direction of that it is being compressed.
• No work is done on the isovolumetric (constant volume) part of the path!
Conceptual quiz
• A gas is taken from an initial state of pressure and volume, PA
VA,to a final, different, state, PB VB. As it changes work is done
on the gas.
– a) The amount of work depends on the path taken from A to
B
– b) The amount of work does not depend on the path taken
from A to B
quiz:
• What is the work done on the gas going from (Pi ,Vi) to (Pf ,Vf) ?
– a) Pf(Vi-Vf)
– b) Pf(Vf-Vi)
– c) 0
– d) Vi(Pi-Pf)
– e) Vf(Pi-Pf)
• Note that the work done on the gas is the negative of the mathematical area
under the curve. The mathematical area is negative if the volume is
decreasing. We are compressing the gas and the volume is decreasing and we
are pushing in the direction to decrease it.
• No work is done on the isovolumetric part of the path!
Computation of work:
• What is the work done on the gas going from (Pi ,Vi) to (Pf ,Vf) ?
– a) Pi(Vi-Vf)
– b) Pf (Vi-Vf)
– c) 0
– d) (Pi+ Pf)(Vi-Vf)/2
– e) -(Pi+ Pf)(Vi-Vf)/2
Heat capacity at constant pressure
• Consider a cylinder filled with a n=2 moles of
monatomic gas and plugged at the top by a frictionless
piston. Above the piston is atmospheric pressure. This
ensures the piston will be at a constant pressure
P=299kPa. The bottom of the cylinder piston is a heat
source that can heat the gas. The sides of the cylinder
and the piston have zero thermal conductivity. What is
the heat Q required to raise the temperature of the gas
from initial temperature Ti =300 K to final temperature
Tf =400K?
Heat source
Q  U  W
3
U  n RT
2
W  pV
3
 n R  Tf  Ti 
2
 p  Vf  Vi   pVf  pVi
 nRTf  nRTi  nR  Tf  Ti 
3
5
Q  n R  Tf  Ti   nR  Tf  Ti   n R  Tf  Ti   2 8.31*(400  300)J  1.66kJ
2
2
3
5
c p  c v  R  R  R  R is the molar heat capacity at constant pressure
2
2
Example
• A person takes in a breath of 0°C air and holds it until it warms to
37.0°C. The air has an initial volume of 0.600 L and a mass of 7.70 x
10–4 kg. Determine (a) the work done by the air on the lungs if the
pressure remains constant at atmospheric pressure, (b) the change in
internal energy of the air, and (c) the energy added to the air by heat.
Model the air as if it were a diatomic gas: cV =5/2R., cp=7/2R.
Pf  Pi  Patm  1.01x105 Pa
Wgas  Patm  Vf  Vi   nR  Tf  Ti 
 nRTf 
 Tf 
 Patm Vf 
 1  Patm Vi   1
 Patm Vi 
 1  Patm Vi 
 nRTi 
 Ti 
 Patm Vi 
 310 
Wgas  1.01x105 Pa .0006m3 
 1  8.2J
 273 
Assume diatomic gas:
T

5
5
5
U  nR  Tf  Ti   nRTi  f  1  Wgas  20.5J
2
2
2
 Ti 7 
Q  U  Wgas  nR  Tf  Ti   28.7J
2



Physic 231 Lecture 34
• Main points of today’s lecture:
• Cycles
• Reversible and irreversible
processes.
• Carnot cycle and Carnot engine.
e
Weng
Qh

Th  Tc
Th
 1
– T is in Kelvin.
• Engines and refrigerators.
• Entropy:
Q reversible
S 
T
Tc
Th
Thermal systems
• In thermal systems, there are a set of “state variables” that are sufficient
to completely describe the macroscopic state of the system.
– example: ideal gas (P,T,N), (V,T,N), (P,U,N) and (U,V,N) are
examples of a set of state variables. U=3/2NkBT is the internal
energy.
• Zeroeth law of thermodynamics: Two macroscopic systems are in
thermal equilibrium if and only if they are at the same temperature.
• In a thermal process, a macroscopic system changes its state variables in
a smooth and controlled manner. Examples of some common processes
are:
– Isobaric process: P=0; the pressure remains the same
– Isovolumetric process: V=0; the volume remains the same.
– Adiabatic process: Q=0; the system is thermally isolated.
– Isothermal process: T=0; the temperature remains the same.
Additional Questions
When I do work on a gas in an adiabatic process, compressing it, I add
energy to the gas. Where does this energy go?
A. The energy is transferred as heat to the environment.
B.
The energy is converted to thermal energy of the gas.
C.
The energy converts the phase of the gas.
Slide 12-56
Cyclic process:
• A thermal path which returns to its initial
condition is called a cycle.
• The work done by the gas on a clockwise cycle
is the area contained in the path.
• The work done by the gas on a
counterclockwise cycle is the negative of the
area in the path.
• The work done on the gas is the negative of the
work done by the gas.
Example of Cyclic process
•
•
•
•
The drawing refers to one mole of monatomic ideal
gas and shows a process that has four steps, two
isobaric (A to B and C to D) and two isovolumetric
(B to C and D to A).
a) Complete a table by calculating U, W and Q
(including the algebraic signs) for each of the four
steps.
b)What is the net heat/cycle absorbed by the system ?
c) What is the net work/cycle done by the system? n  1
3
3
3
RTA = (8.31)400J  4986J U B  RTB  9972J
2
2
2
3
3
U C  RTC  4986J U D  RTD =2493J
2
2
U A  B  U B  U A UA B  9972  4986  5.0kJ
UA 
Path
A-B
U
W
W
Q
Q
5.0kJ
3.3kJ
8.3kJ
B-C
C-D
- 5.kJ
0
- 5.kJ
D-A
2.5kJ
0
2.5kJ
similarly get U BC , UC D , & U D A
Net
0
1.6 kJ
1.6 kJ
WD A  WBC  0 because isovolumetric PV  nRT
-2.5kJ -1.7kJ -4.2kJ
WAB  PA  VB  VA   PB VB  PA VA  RTB  RTA  3.3kJ
WC D  RTD  RTC  1662J
Q  U  W
Important Cyclic Processes: Engines
• In a heat engine, thermal energy Qh is used to do
work, Weng. Some of the original thermal energy Qc
escapes and ends up heating something else
• A heat engine involves some working substance in a
cyclical process.
• In many cases heat comes from reservoir at TH and
is exhausted to the invironment at TC.
• Thermal efficiency is defined as the ratio of the
work done by the engine to the energy absorbed at
the higher temperature. For simplicity both can be
computed over one cycle:
e
Weng
Qh
Qh  Qc
Qc

 1
Qh
Qh
• e = 1 (100% efficiency) only if Qc = 0
– No energy expelled to cold reservoir,
which is theoretically possible for
Tc=0, but practically impossible.
Example
• The energy absorbed by an engine is three times as large as
the work it performs. (a) What is its thermal efficiency? (b) What
fraction of the energy absorbed is expelled to the cold reservoir?
a) Q h  3W
e
W
W 1


Q h 3W 3
b) Q h  W  Qc
 Qc  Q h  W  2W 
Qc 2


Qh 3
2
Qh
3
Quiz
• A heat engine performs 200 J of work in each cycle and has an
efficiency of 30%. For each cycle of operation, (1) how much
energy is taken in from the hot reservoir (Qh) and (2) how much
energy is expelled to the cold reservoir (QC)?
• Answers below are in the form Qh, Qc.
– a) 667J, 467J
– b) 467J, 667J
– c) 60J, 140J
– d) 140J, 60J
a) e 
W
W 200 J
 Qh 

 667 J
Qh
e
0.3
b) QH  W  Qc  Qc  Qh  W  667 J  200 J  467 J
Reading Quiz
1. A sample of nitrogen gas is inside a sealed container. The container
is slowly compressed, while the temperature is kept constant. This is
a ________ process.
A.
B.
C.
D.
constant-volume
isobaric
isothermal
adiabatic
Slide 12-6
Reversible and Irreversible Processes
• reversible process is one in which every state along some path is an
equilibrium state.
– And one for which the system can be returned to its initial state by
going along the same path in the (p,V) diagram but in the opposite
direction.
– Volume and pressure changes are “slow”.
– When objects are brought into thermal contact, they are at the same
temperatures.
– Carnot cycle is an example of a reversible process.
• An irreversible process does not meet these requirements
– Most natural processes are irreversible
• Burning fueling in an automobile engine
• Dropping ice into warm water
• Heating water on a range
• Reversible process are an idealization, but some real processes are good
approximations.
Constant-Volume Process: Isovolumetric
– Isovolumetric process: V=0; the volume remains the same, internal
energy changes, heat can be exchanged with environment, but no
work is done.
Slide 12-29
Constant-Pressure Process
– Isobaric process: P=0; the pressure remains the same, work can be
done, heat can be exchanged and internal energy can change.
Constant-Temperature Process: isothermal
– Isothermal process: T=0; the temperature remains the same.
Internal energy of a gas remains the same, heat can be exchanged
and work can be done.
nRT
P
V
Adiabatic process: no exchange of heat with environment
– Adiabatic process: Q=0; the system is thermally isolated. Heat
cannot come in or out of the system, internal energy can change and
work can be done. .
.
PV  const an t
  5 / 3 for monatomic ideal gas
When a gas expands, it does work, when a gas is compressed,
work is done on it. In an adiabatic process, the work done by
a gas while expanding decreases its thermal energy and
temperature. Conversely, the work done on the gas while
compressing it increases its thermal energy and temperature.
Operation of a Heat Engine
Slide 11-26
Carnot Engine
• A Carnot engine is the most
efficient possible engine that takes
heat from a hot reservoir at
temperature Th and expells heat into
a cold reservoir at temperature Tc .
– It uses gas as the working
substance.
– It absorbs heat Qh during an
isothermal expansion while in
contact with the hot reservoir.
– It expands adiabatically a little
further.
– It compresses isothermally
while depositing QC into the
cold reservoir.
– It compresses adiabatically a
little further.
Carnot Cycle, A to B
• A to B is an isothermal
expansion
• The gas is placed in contact with
the high temperature reservoir
• The gas absorbs heat Qh
• The gas does work WAB in
raising the piston
• What happens to the internal
energy of the gas?
– a) It increases
– b) It decreases
– c) It stays the same
• What happens to the pressure?
– a) It increases
– b) It decreases
– c) It stays the same
nRT
P
V
Fig. 12.13, p. 374
Slide 22
Carnot Cycle, B to C
• B to C is an adiabatic expansion
• The base of the cylinder is
replaced by a thermally insulating
wall
• No heat enters or leaves the system
• The temperature falls from Th to
Tc
PV  const an t
• The gas does work WBC
  5 / 3 for monatomic ideal gas
• What happens to the pressure?
– a) increases
– b) decreases
– c) stays the same
• What happens to the internal
energy?
Q  U  pV  0
– a) increases
– b) decreases
– c) stays the same
Fig. 12.13, p. 374
Slide 22
Carnot cycle efficiency
• The efficiency of the Carnot cycle depends only on the
temperatures of the hot and cold reservoirs:
e
Weng
Qh

Th  Tc
T
 1 c
Th
Th
– No engine operating between these two temperatures is more
efficient than the Carnot engine
• A heat engine operates between two reservoirs at temperatures
of 20°C and 300°C. What is the maximum efficiency possible for
this engine?
• a) .24
TC
293
e

e

1


1

 0.49
• b) .49
max
carnot
TH
573
• c) .56
• d) .73
Carnot Cycle, C to D
• The gas is placed in contact with the
cold temperature reservoir
• C to D is an isothermal compression
• The gas expels energy QC
• Work WCD is done on the gas
• What happens to the internal energy
of the gas?
– a) It increases
– b) It decreases
– c) It stays the same
• What happens to the pressure?
– a) It increases
– b) It decreases
– c) It stays the same
Fig. 12.13, p. 374
Slide 22
Carnot Cycle, D to A
• D to A is an adiabatic
compression
• The gas is again placed against a
thermally nonconducting wall
– So no heat is exchanged with
the surroundings
• The temperature of the gas
increases from Tc to Th
• The work done on the gas is WCD
• What happens to the pressure?
– a) increases
– b) decreases
– c) stays the same
• What happens to the internal
energy?
– a) increases
– b) decreases
– c) stays the same
Fig. 12.13, p. 374
Slide 22
Example
• An engine does 20900 J of work and rejects 7330 J of heat into a cold
reservoir at 298K. What is the smallest possible temperature of the hot
reservoir?
a) Q h  W  Qc
 20900J  7330J  28230J
W 20900J
e

 0.74
Q h 28230J
 ecarnot
Tc
 1
Th
Tc
T
 c  1  .74  .26
Th
Th
Tc
Tc
298K
 Th

 .26 
 Th
.26
Th
.26
0.74  1 
 1146K  Th
If Th were less than 1146, then the calculated efficciency 0.74
is larger than that of a Carnot engine running at that temperaure,
which is impossible.
Additional Questions
Suppose you have a sample of gas at 10°C that you need to warm up to
20°C. Which will take more heat energy: raising the temperature while
keeping the pressure constant or raising the temperature while keeping
the volume constant?
A. It takes more energy to raise the temperature while keeping the
volume constant.
B.
It takes more energy to raise the temperature while keeping the
pressure constant.
C.
The heat energy is the same in both cases.
If you raise the temperature and keep the pressure
constant, the gas will expand and do positive work.
Q  U  W
Slide 12-54
Additional Questions
When I do work on a gas in an adiabatic process, compressing it, I add
energy to the gas. Where does this energy go?
A. The energy is transferred as heat to the environment.
B.
The energy is converted to thermal energy of the gas.
C.
The energy converts the phase of the gas.
Q  U  pV  0
Slide 12-56
Quiz
• A heat engine operating between a hot reservoir at 500 K and a cold
reservoir at 200 K has an efficiency that is 70% of its maximum
possible value. If it receives lx106 J heat energy from the hot reservoir
in 25 minutes, it can do a quantity of work equal to
– a)6.3x105J.
Tc
200
– b)4.2x105J.
a) e max  ecarnot  1 
 1
 0.6
Th
500
– c)3.lxl05J.
– d)2.5x105J.
 e  0.7e max  0.7(0.6)  .42
5
– e)1.7x10 J.
W
e
Qh


 W  e  Q h  .42 1x106 J  4.2x105 J
Heat pumps and refrigerators
• Heat engines can run in reverse
– Send in energy
– Energy is extracted from the cold reservoir
– Energy is transferred to the hot reservoir
• This process means the heat engine is running
as a heat pump
– A refrigerator is a common type of heat
pump
– An air conditioner is another example of a
heat pump
– In the south, people often use heat pumps
to heat homes
• A standard measure of the performance of a
refrigerator is its “coefficient of performance”
coef. of performance =
QC
W
heat removed per cycle
work required to remove it
A Carnot refrigerator is a Carnot engine run in
reverse
•
A Carnot refrigerator maintains the food inside it at 276 K while the
temperature of the kitchen is 298 K. The refrigerator removes 3.00x
104 J of heat from the food. How much heat is delivered to the
kitchen?
As a Carnot heat engine, we know that
Qc
Tc  Qc  Tc
ecarnot  1 
 1
Q h Th
Qh
Th
The Carnot engine run in reverse takes mechanical
energy W to move Qc from the inside of the refrigerator
and deposit Q h in the kitchen.
Qc Tc
Th
4 298

 Q h  Qc
 3x10 J
 3.23x104 J
Q h Th
Tc
276
Note:
Carnot coefficient of performance=
QC
QC
TC


W Qh  Qc Th  TC
Checking Understanding: Increasing Efficiency of
a Heat Pump
Which of the following changes would allow your refrigerator to use
less energy to run? (1) Increasing the temperature inside the
refrigerator; (2) increasing the temperature of the kitchen; (3)
decreasing the temperature inside the refrigerator; (4) decreasing the
temperature of the kitchen.
A. All of the above
B. 1 and 4
C. 2 and 3
Assume the temperature dependence is similar to that of a
carnot refrigerator, where
Q h Th

QC Tc
Carnot Coefficent of performance =
Tc
Th  Tc
This is largest when Th and TC are closer to each other in
temperature.
Slide 11-31
Qreversible
S 
T
• Entropy can only be calculated from a reversible path, and must be done that
that way even if the system actually follows an irreversible path
– To calculate the entropy for an irreversible process, model it as a
reversible process
• When heat energy is absorbed, Q is positive and entropy increases
• When heat energy is expelled, Q is negative and entropy decreases
• In an adiabatic process Q=0 and entropy remains the same.
• S  ln(probability).
• A disordered state with energy and matter spread out everywhere is more
probable than having all of the energy stored in an organized way that can be
used to do work.
Example
• The surface of the Sun is approximately at 5700 K, and the
temperature of Earth’s surface is approximately 290 K. What
entropy change occurs when 1000 J of energy is transferred by
heat from the Sun to Earth?
The sun loses Q of heat and therefore decreases its entropy by the amount
Q
Ssun 
Tsun
The earth gains Q of heat and therefore increases its entropy by the amount
Q
Searth 
Tearth
The total entropy change is:
S  Ssun  Searth  Q  1  1   1000 J  1  1   3.27J / K


 T

290
K
5700
K
T
earth
sun
S 
T

Example
• What is the change in entropy of 1.00 kg of liquid water at 100°C
as it changes to steam at 100°C?
Q
S  
T
L vap m
T
22.6x10 J / kg  1kg 


 6x10 J / K
5
3
373K
Entropy
Higher entropy states
are more likely.
Systems naturally
evolve to states of
higher entropy.
Slide 11-33
Second Law of Thermodynamics
Slide 11-34
Example
• A power plant has been proposed that would make use of the
temperature gradient in the ocean. The system is to operate
between 20.0°C (surface-water temperature) and 5.00°C (water
temperature at a depth of about 1 km). (1) What is the maximum
efficiency of such a system? (2) If the useful power output of the
plant is 75.0 MW, how much energy is absorbed per hour? (3) In
view of your answer to (1), do you think such a system is
worthwhile (considering that there is no charge for fuel)?
• Answers for 1 and 2:
278
Tc
– a) .025, 3.2x106 J a) e  e
 1
 0.051
max
carnot  1 
12
293
Th
– b).051, 5.3x10 J
– c) . 25, 3.2x107 J b) e  W  Q  W   75MW  3600s   5.3x1012 J
h
e
0.051
Qh
– d) 051, 5.3x1013 J
c) What is the energy required to pump the water?
Story of Hawaiian deep water project
•
•
Keahole sits at a point where underwater •
land slopes sharply down into the sea, it
was a place where warm water can be
piped from the surface of the sea and cold
water can be piped from depths of about a
half-mile.
•
A process called ocean thermal energy
conversion, or OTEC, used the
temperature difference between hot and
cold sea water to produce 50 KW of
electricity at Keahole in 1993. The
process worked but it was uneconomical.
KAILUA, HAWAI'I — Koyo USA
Corp., a company selling deep-sea
water from Keahole Hawai'i, is
expanding its plant and has applied to
sell the water in the United States
The company is producing more than
200,000 bottles a day and says it
can't keep up with demand in Japan,
where it sells 1.5 liter bottles of its
MaHaLo brand for $4 to $6 each.
Physics 231 Lecture 23
• Main points of today’s lecture:
• Springs and masses
• Simple harmonic motion of a
spring:
x  A cos(t   0 )
v x  A sin(t   0 )
a x   2 A cos(t   0 )
 2  k / m  0 and A are constants
• Pendulum:
L
T  2
g
   max cos2ft  0 
Final Exam
•
•
•
•
A common final exam time is scheduled for all sections of Physics 231
Time: 3pm-5pm, Thursday, May 5 .
Location : E100 Veterinary Medical Center.
This information can also be found on our course schedule page
• An alternate exam time will be scheduled for students who have
conflicts with the regular time.
– Three students have confirmed conflicts with me and will take the
exam then.
– You must contact me by email and obtain permission from me to
take the exam at the alternate time.
• Alternate time: TBA (probably sometime Wed. May 4th)
• Location: TBA
Example
• The surface of the Sun is approximately at 5700 K, and the
temperature of Earth’s surface is approximately 290 K. What
entropy change occurs when 1000 J of energy is transferred by
heat from the Sun to Earth?
The sun loses Q of heat and therefore decreases its entropy by the amount
Q
Ssun 
Tsun
The earth gains Q of heat and therefore increases its entropy by the amount
Q
Searth 
Tearth
The total entropy change is:
S  Ssun  Searth  Q  1  1   1000 J  1  1   3.27J / K


 T

290
K
5700
K
T
earth
sun
Qreversible
S 
T
Second Law of Thermodynamics
Slide 11-34
Reading Quiz
1. The type of function that describes simple harmonic motion is
A.
B.
C.
D.
E.
linear
exponential
quadratic
sinusoidal
inverse
Slide 14-6
Reading Quiz
2. A mass is bobbing up and down on a spring. If you increase the
amplitude of the motion, how does this affect the time for one
oscillation?
A. The time increases.
B. The time decreases.
C. The time does not change.
Slide 14-8