Thermal
Physics
Temperature
and Heat
Phases of Matter
Four Phases of Matter:
Solid
Liquid
Gas
Plasma
Temperature is one
thermodynamic variable
that determines the phase.
Common Temperature
Scales
Temperatures are reported in degrees
Celsius or degrees Fahrenheit.
Temperatures changed, on the
other hand, are reported in Celsius
degrees or Fahrenheit degrees:
9
1C  F
5
Converting from a Fahrenheit to a Celsius Temperature
A healthy person has an oral temperature of 98.6oF. What would this
reading be on the Celsius scale?
degrees above ice point
98.6 F  32 F  66.6 F
 1 C 
66 .6 F  9    37 .0 C 
5F 



0 C  37.0 C  37.0 C
ice point
Converting from a Celsius to a Fahrenheit Temperature
A time and temperature sign on a bank indicates that the outdoor
temperature is -20.0oC. Find the corresponding temperature on
the Fahrenheit scale.
degrees below ice point
 95 F 
20 .0 C     36 .0 F
1C 



32.0 F  36.0 F  4.0 F
ice point
Daniel Fahrenheit
1686 - 1736
Fahrenheit
According to a journal article Fahrenheit wrote
in 1724, he based his scale on three reference
points of temperature. The zero point is
determined by placing the thermometer in
Later,
other scientists
observed
brine:
hework
usedby
a mixture
of ice, water,
and
that waterchloride.
boils about
degrees
higher
ammonium
The180
mixture
automatically
than theits
freezing
point and
to
stabilizes
temperature
at 0 decided
°F.
degree slightly
tothe
make
it
Heredefine
then putthe
a thermometer
into
mixture
and
180indegrees
higher. It descend
is for thisto its
letexactly
the liquid
the thermometer
reason
thatThe
normal
bodypoint
temperature
lowest
point.
second
is the 32 is
degree
98.6by
onputting
the revised
scale (whereas
it was
found
the thermometer
in still
water
original
scale).
as96
iceon
is Fahrenheit's
just forming on
the surface.
The third point, the 96 degree, was the level of
the liquid in the thermometer when held in the
mouth or under the armpit.
Unsatisfied with the Celsius and Fahrenheit temperature scales, you decide to
create your own. On your temperature scale, the ice point is 77 M and the
steam point is at 437 M, where “M” stands for “my scale.” What
temperature on your scale corresponds to 68 F?
1.
2.
3.
4.
5.
154 M
168 M
140 M
136 M
149 M
William Thomson,
1st Baron Kelvin (Lord Kelvin)
1824 - 1907
Kelvin
Temperature Scale
T  Tc  273.15
Temperatures have an
extremely large range,
both on Earth and
throughout the
Universe.
Thermometers
Thermometers make use of the change in some physical property with
temperature. A property that changes with temperature is called a
thermometric property. All thermometers require calibration.
Zeroth Law of Thermodynamics
If bodies A and B are each in thermal equilibrium
with a third body C, then they are in thermal
equilibrium with each other.
Translation: Every body has a property called
temperature. When two bodies (one might be a
thermometer) are found to be in thermal
equilibrium, then their temperatures are the
same. This may be used to determine the
temperature of a third body (through calibration).
DEFINITION OF HEAT
Heat is energy that flows from a highertemperature object to a lower-temperature
object because of a difference in
temperatures.
SI Unit of Heat: joule (J)
OTHER UNITS
1 kcal = 4186 joules
1 cal = 4.186 joules
1 BTU = 1055 J
The heat that flows from
hot to cold originates in
the internal energy of
the hot substance.
It is not correct to say
that a substance
contains heat.
Heat and Temperature Change: Specific Heat Capacity
The heat that must be supplied or removed to
change the temperature of a substance is
Q  m cT
specific heat
capacity
Common Unit for Specific
Heat Capacity: J/(kg·Co)
1 kcal = 4186 J
1 BTU = 1055 J
When you drink cold water, your
body must expend metabolic
energy to maintain normal body
temperature of 37 oC by warming
up the water in your stomach.
Could drinking ice water substitute
for exercise as a way to “burn
calories?” Suppose you expend
430 kilocalories during a brisk onehour walk. How many liters of ice
water would you have to drink in
order to use 430 kilocalories of
metabolic energy? Note: the
stomach can hold about one liter.
In a half-hour, a 65-kg jogger can generate 8.0x105J of heat. This heat
is removed from the body by a variety of means, including the body’s
own temperature-regulating mechanisms. If the heat were not
removed, how much would the body temperature increase?
Q  m cT
Q
8.0 105 J

T 


3
.
5
C
mc 65 kg 3500J kg  C



Heat and Phase Change: Latent Heat
Saving Energy
69%
Suppose you are cooking spaghetti for dinner, and the
instructions say “boil pasta in water for 10 minutes.” To cook
spaghetti in an open pot with the least amount of energy,
31%
should you turn up the burner to its fullest so the water
vigorously boils, or should you turn down
the burner so the water barely boils?
A.
B.
turn up the burner to its fullest so
the water vigorously boils, or
should you turn down the burner
so the water barely boils?
1
2
Answer



We need water at 100 oC for 10 minutes.
Adding additional energy only produces
steam, which doesn’t help cook the
spaghetti.
Keep the flame to just enough to boil and
save energy and money.
A calorimeter consisting of a thin copper cup of mass 150 g containing
500 g of water is at a temperature of 20.0 oC. A 225-g sample of an
unknown material at 508 oC is lowered into the bath, and the device is
sealed. After a few minutes, the system reaches a constant
temperature of 40.0 oC. Determine the specific heat of the unknown
material. Ignore any losses due to the thermometer.
The heat transferred out of the sample (which is a
negative number) equals the heat transferred into the
water and the cup.
Qu  Qw  QCu
mucu Tu  mwcwTw  mCucCu TCu
cu 
mwcw Tw  mCu cCu TCu
mu T
 0.500 kg  4186 J   40 C  20 C    0.150 kg  390 J   40 C  20 C 

  0.225 kg   40 C  508 C 
 409 J/(kg  C )
You find that you have let a 2.5 kg stainless steel barbeque grate become
too hot for cooking. You Decide to cool the grate from 296 oC to 185 oC by
spraying water onto the grate. The specific heat of steel = 445 J/kg - K.
Calculate the mass of water at 20 oC you will need, assuming all the water
evaporates to steam at 100 oC.
heat lost by plate = heat gained by water in raising its own temp + heat to
transform water into steam
mscs Ts  mwcwTw  mwLv
ms cs Ts  mwcw Tw  mw Lv  mw  cw Tw  Lv 
ms cs Ts
mw 
cw Tw  Lv


 185 C  296 C 

 100 C  20 C  22.6 10
  2.5 kg  445
4186
J
kg  C
J
kg  C
 0.048 kg or 48 g
5 J
kg
Heat = transfer of energy




From someplace hot to someplace cold
Convection
Conduction
Radiation
Convection
Convection is the process in which heat is carried from one place
to another by the bulk movement of a fluid.
Concepts in action
Hot water baseboard heating units are mounted on the wall
next to the floor. The cooling coil in a refrigerator is mounted
near the top of the refrigerator. Each location is designed to
maximize the production of convection currents.
Forced Convection
CONDUCTION
Conduction is the process whereby heat is transferred directly through
a material, with any bulk motion of the material playing no role in the
transfer.
One mechanism for conduction occurs when the atoms or molecules
in a hotter part of the material vibrate or move with greater energy than
those in a cooler part. These vibrations have specific values and are called
phonons.
By means of collisions, the more energetic molecules pass on some of
their energy to their less energetic neighbors.
Another mechanism is free electrons within a metal conduct heat.
Materials that conduct heat well are called thermal conductors, and those
that conduct heat poorly are called thermal insulators.
Conduction
The
The
heat
amount
Q conducted
of heat Q
during
that isa conducted
time t through
through
a barthe
of length
bar depends on
L and
a number
cross-sectional
of factors:area A is
kAT  t

Q
1. The time during which conduction takes place.
2. The temperature difference between the ends of the bar.
3. The cross sectional area of the bar.
The of
length
of theConductivity:
bar.
SI4.Units
Thermal
J/(s·m·Co)
L
Thermal
Conductivity
What thickness of concrete, with a thermal
conductivity of 1.1 J/(smK) will conduct heat at the
same rate as 0.25 m of air, which has a thermal
conductivity of 0.0256 J/(smK), if all other conditions
are the same?
kAT  t

Q
L

Q kAT

t
L
kconcrete AT kair AT

Lconcrete
Lair
kconcrete kair

Lconcrete Lair
Lconcrete
 1.1 smJK 
kconcrete

Lair  
0
.
25
m
=
11
m

J
kair
0
.
0256
smK 

Insulation
Materials with dead air
spaces are usually
excellent thermal
insulators.
Example: Thermal Conduction
One wall of a house consists of
plywood backed by insulation.
The thermal conductivities of
the insulation and plywood are,
respectively, 0.030 and 0.080
J/(s·m·Co), and the area of the
wall is 35 m2.
Find the amount of heat
conducted through the wall in one
hour.
Example: Thermal Conduction
Q  Qinsulation  Qplywood
But first we must solve for the interface
temperature.
  kAT  t 
  kAT  t 




L
L

insulation 
 plywood
0.030J s  m  C A25.0 C  T t  0.080J s  m  C AT  4.0 Ct



0.076 m
0.019 m
T  5.8 C

Example: Thermal Conduction
0.030 J  s  m  C    35 m 2  25.0 C  5.8 C  3600 s 

Qinsulation  
0.076 m
 9.5 105 J
Three building materials, plasterboard,
brick, and wood, are sandwiched
together as the drawing illustrates. The
temperatures at the inside and outside
surfaces are 27 °C and 0 °C,
respectively. Each material has the
same thickness and cross-sectional
area. Find the temperature (a) at the
plasterboard–brick interface and (b) at
the brick–wood interface.
The rate of heat transfer is the same for all three
materials so
Let Ti be the inside temperature, T1 be the
temperature at the plasterboard-brick
interface, T2 be the temperature at the brickwood interface, and To be the outside
temperature.








 0.30 J/ s  m  C    0.10 J/ s  m  C  

 1  
  27 C
 0.60 J/ s  m  C    0.60 J/ s  m  C  

 




 



kb
 0.10 J/ s  m  C 
 0C

kw
 0.60 J/ s  m  C 


T1 
  0.10 J/ s  m  C     0.30 J/ s  m  C  
1  
  1  
  1
  0.60 J/ s  m  C     0.60 J/ s  m  C  
  

 
 21 C








 0.60 J/  s  m  C 
 0.10 J/  s  m  C 
k p  0.30 J/ s  m  C


 0.60 J/  s  m  C 
 0.10 J/  s  m  C 
k p  0.30 J/ s  m  C
kb
kw





0.30 J/ s  m  C  0.60 J/ s  m  C  21 C 0.30 J/ s  m  C

T2  

0.60 J/ s  m  C
0.60 J/ s  m 
 18 C



 27 C
C
Radiation
Radiation is the process in which
energy is transferred by means of
electromagnetic waves.
A material that is a good absorber
is also a good emitter.
A material that absorbs completely
is called a perfect blackbody.
Black Robe Mystery
Why do Bedouins wear black robes?
Scientists
Thechecked
extra warm
intoair
thetrapped
matter by the
having
black
a
man color
standexperiences
in the hot desert
a buoyancy
sun firstforce
in a white
equal
robe and
to the
thenweight
in a black
of the
one.
coolThey
air itfound
displaces.
that
the black
The warm
robe absorbed
air rises up
2.5and
times
outmore
of thesolar
robe
oC) hotter
radiation top.
and This,
was 11
in oturn,
F (6 creates
a breeze
than the
as
white one.
cooler air is drawn in at the bottom. The
breeze evaporates sweat, cooling the robe
Thewearer,
skin temperature
same in air out
and moveswas
thethe
sweat-laden
either robe, but the man felt coolerthe
in the
robe top.
black one.
The loose-fitting burnoose is a solar airconditioning system.
Blackbody Radiation
The maximum of the
intensity shifts to shorter
wavelengths as the blackbody temperature increases.
THE STEFAN-BOLTZMANN LAW OF RADIATION
The radiant energy Q, emitted in a time t by an object that has a
Kelvin temperature T, a surface area A, and an emissivity e, is given by
Q  e T At
4
Stefan-Boltzmann constant

  5.67  108 J s  m2  K 4

The emissivity e is a
dimensionless number between
zero and one. It is the ratio of
what an object radiates to what
the object would radiate if
it were a perfect emitter.
Rate of Heat Transfer
Q
4
 e T A
t
A cube, 10 cm on a side, of rough steel is heated in a furnace to a
temperature of 400.0 oC. If its total emissivity is 0.97, determine the rate
at which it radiates energy from each face.
Q
 e T 4 A
t
4
2
8
2
4
  0.97  5.67  10 J (s  m  K )  400  273 K   0.10 m 
 113 J/s = 113 W


Example: Radiation
The supergiant star Betelgeuse has a
surface temperature of
about 2900 K and emits a power of
approximately 4 x1030 W.
Assuming that Betelgeuse is a
perfect emitter and spherical,
find its radius.
Q  e T At
4
4 r 2
Q  e T (4 r )t
4
2
Q t
4 1030 W
r

4
4
4  e T
4  1 5.67 108 J  s  m2  K 4   2900 K 
 3 1011 m
Application: Thermos Bottle
A thermos bottle minimizes heat
transfer via conduction,
convection, and radiation.