Transcript Lecture 1: Rotation of Rigid Body

Chapter 16: Temperature and Heat
Temperature and thermal equilibrium
 Temperature
• is a measure of how hot or cold an object is
• is measured by a thermometer
 Thermal
equilibrium
Objects placed in contact will eventually reach the same temperature.
When this happens, they are in thermal equilibrium.
Zero’th law of thermodynamics
If an object C is in thermal equilibrium with both objects A and B,
Then A and B are in thermal equilibrium with each other too.
Temperature and thermal equilibrium (cont’d)
 Thermometer
• Thermometers are devices used to measure the temperature of
an object or a system.
• When a thermometer is in thermal contact with a system, energy is
exchanged until the thermometer and the system are in thermal
equilibrium with each other.
• All the thermometers use some physical properties that depend on
the temperature. Some of these properties are:
1) the volume of a fluid
2) the length of a solid
3) the pressure of a gas held at constant volume
4) the volume of a gas held at constant pressure
5) electric resistance of a conductor
6) the color of very hot object.
Temperature and thermal equilibrium (cont’d)
 Thermometer
(cont’d)
• One common thermometer consists of a mass of liquid: mercury or
alcohol. The fluid expands into a glass capillary tube when its
temperature rises.
o
o
• When the cross-sectional area of the tube 0 C (Celsius) 100 C
is constant, the change in volume of the
liquid varies linearly with its length along
the tube.
• The thermometer can be calibrated by
placing it in thermal contact with
environments that remain at constant temp.
• Two of such environments are:
Freezing
1) a mixture of water and ice in thermal
point
equilibrium at atmospheric pressure.
2) a mixture of water and steam in thermal
equilibrium at atmospheric pressure.
Boiling
point
Temperature and its scales
 Constant-volume
gas thermometer and the Kelvin scale
• A constant-volume gas thermometer measures the pressure of
the gas contained in the flask immersed in the bath. The volume
of the gas in the flask is kept
constant by raising or lowering
reservoir B to keep the mercury
level constant in reservoir A.
Temperature and its scales (cont’d)
 Constant-volume
gas thermometer and the Kelvin scale
• It has been experimentally
observed that the pressure varies
linearly with temperature of
a fixed volume of gas, which does
not depend on what gas is used.
• It has been experimentally
observed that these straight lines
merge at a single point at temp.
-273.15oC at pressure = 0.
This temperature is called absolute
zero, which is the base of the Kelvin temperature scale T=TC-273.15
measured in kelvin (K) where TC is temperature in Celsius..
0 K = -273.15oC
Temperature and its scales (cont’d)
 Gas
thermometer and absolute (Kelvin) scale (cont’d)
pressure
The pressure of any gas at constant volume is a linear function of
temperature, which always extrapolates to zero at –273.15 ºC.
T2 p2

T1 p1
const. volume
The absolute or Kelvin
temperature scale:
T(K) = T(ºC) + 273.15
In fact it is also true that:
T2 V2

T1 V1
const. pressure
Temperature and its scales
 Temperature
scales
Thank you Mr.
Fahrenheit!
Fahrenheit
Based on the ability of farm animals to survive without attention!
( 0o F : the coldest 100o F : the hottest )
Celisius/Centigrade
Based on the physical properties of water on the Earth’s surface
at sea level.
( 0o C : the freezing point 100o C : the boiling point )
9
5
TF  TC  32 ; TC  (TF  32)
5
9
9
TF  TC
5
Temperature and its scales
 Temperature
scales
• The common temperature scale in US is
Fahrenheit:
Thermal expansion (Ch.17)
 Linear
expansion
Most materials expand when heated:
• The average distance between atoms increases as the
temperature is raised.
• The increase is proportional to the change in temperature (over a
small range).
Consider an object of length Li at temperature Ti:
If the object is heated or cooled to temperature Tf
L  Lf  Li  Li T or L  L0T
( Li  L0 , T  Tf  Ti )
α = coefficient of linear expansion [ºC-1]
(α is a property of the material)
L  L0 (1  T )
Thermal expansion (Ch.17)(cont’d)
 Coefficients
of linear expansion
Material
α (ºC -1)
Glass
9 x 10-6
Concrete
12 x 10-6
Copper
17 x 10-6
Lead
29 x 10-6
Mercury
1.8x 10-4
Gasoline
3.2 x 10-4
Thermal expansion (Ch.17)(cont’d)
 Volume
expansion
Increasing temperature usually causes increases in volume
for both solid and liquid materials. Experiments show that if
the temperature change is not too great (less than 100 Co
or so), the increase in volume is approximately proportional
to both the temperature change and the initial volume:
V  V0T
 Relation
between  and 
For solid materials there is a simple relation between  and 
as V=L3:
dV
dV 
dL  3L2dL  3L20L0dT  3L30dT  V0dT
dL
dL  L0dT
  3
Thermal expansion (Ch.17) (cont’d)
 Thermal
expansion of water
• Water contracts when
heated from 0ºC to 4ºC, then
expands when heated from
4 ºC to 100 ºC.
• Just above the freezing
point, the coldest (and least
dense) water rises to the
surface, and lakes freeze
from the surface downward.
• This unusual property
permits aquatic life on earth
to survive winter!
Density of Water
g/(cm**3)
1
0.99
0.98
0.97
0.96
0.95
0
4
12
20
50 100
Temperature in Celsius
Quantity of heat
 Heat
When two objects of different temperatures are in thermal contact,
their temperature eventually reach the thermal equilibrium. The
change in temperature to reach the thermal equilibrium is achieved
by an interaction that transfers energy called heat.
 Unit
of heat
calorie : the amount of heat required to raise the temperature
of 1 g of water from 14.5 oC to 15.5 oC
Btu
: the amount of heat required to raise the temperature
of 1 lb (weight) of water from 1 oF from 63 oF.
1 cal = 4.186 J
1 kcal = 1000 cal = 4186 J
1 Btu = 778 ft lb = 252 cal = 1055 J
Quantity of heat (cont’d)
 Specific
heat
The quantity of heat Q required to increase the temperature of a
mass m of a certain material from T1 to T2 is found to be approximately
proportional to the temperature change T=T2-T1 and to mass m.
Q  m cT
For an infinitesimally small change in temperature:
dQ  m cdT
1 dQ
c
m dT
specific heat
Quantity of heat (cont’d)
 Molar
heat capacity
• Often it is more convenient to describe a quantity of substance
in terms of moles n rather than the mass m of material.
• A mole of any pure substance contains the same number of
molecules.
• The molar mass of any substance M is the mass per mole.
m  nM
Q  nMcT  nCT
1 dQ
C
 Mc
n dT
molar heat capacity
For water C=(0.0180 kg/mol)[4190 J/(kg T)] = 75.4 J/(mol K)
Phase transition
 Phase
changes
• Phases of matter : solid, liquid, gas
• A change of phase : phase transition
• For any given pressure a phase change takes place at a
definite temperature, usually accompanied by absorption
or emission of heat and a change of volume and density
Phase transition (Ch.17)
 Latent
heat (see Table 17.1)
Heat of fusion Lf
: heat needed to change from liquid to gas per kg
of material 3.34 x 105 J/kg = 79.6 cal/g =143 Btu/lb
Heat of vaporization Lv : heat needed to change from solid to liquid per kg
of material
heat of vaporization
phase equilibrium
heat of fusion
Phase transition (Ch.17)
 Latent
heat and phase change
• Water
Consider an addition of energy to a 1.00-g cube of ice at -30.0oC in
a container held at constant pressure. Suppose this input energy
turns ice to steam (water vapor) at 120.0oC.
A: Q  m ciceT , T  30.0C
 62.7 J cice  2090J/(kg C)
5
B:Q  m Lf , L f  3.3310 J / kg
 333J
C: Q  mcwater T , T  100C
 4.19 J cwater  4190J/(kg C)
6
D: Q  m Lv , Lv  2.2610 J / kg
 2.26103 J
E: Q  mcsteam T , T  20.0C
 40.2 J csteam  2010J/(kg C)
Calorimetry
 Isolated
“measuring heat”
system
• A system whose energy does not leave out of the system is
called isolated system.
• The principle of energy conservation for an isolated system
requires that the net result of all the energy transfer is zero.
If one part of the system loses energy, another part has to
gain the energy.
 Calorimeter
and calorimetry
• Imagine a vessel made of good insulating material and containing
cold water of known mass and temperature and the temperature of
the water can be measure. Such a system of the vessel and water
is called calorimeter. If the object is heated to a higher temperature
of known value before it is put into the water in the vessel, the specific
heat of the object can be measured by measuring the change in
temperature of the water when the system (the object, vessel, and
water) reaches thermal equilibrium. This measuring process is called
calorimetry.
Calorimetry
 Calorimeter
and calorimetry
• When a warm object is put into a calorimeter with cooler water
described in the previous page, it becomes cooler while the water
becomes warmer.
Qcold  Qhot
Qcold (>0 ) is the heat transferred (energy change) to the cooler
object and Qhot (<0) is the heat transferred (energy change) to the
warmer object.
• In general, in an isolated system consisting of n objects :
n
Q
k 1
k
Tf common to all objects in equilibrium.
 0 Qk  mk ck [(Tk ) f  (Tk )i ]
Calorimetry and phase transition
 Calorimeter
and calorimetry
• Example: Calculate an equilibrium temperature
Suppose 0.400 kg of water initially at 40.0oC is poured into a 0.300-kg
glass beaker having a temperature of 25.0oC. A 0.500-kg block of
aluminum at 37.0oC is placed in the water, and the system is insulated.
Calculate the final equilibrium temperature of the system.
Qw  Qal  Qg  0
mwcw (T  Tw )  mal cal (T  Tal )  mg cg (T  Tg )  0
T
mwcwTw  mal calTal  mg cgTg
mwcw  mal cal  mg cg
 37.9C
Calorimetry and phase transition
 Latent
heat and phase change
• Example : Ice water
6.00 kg of ice at -5.00oC is added to a cooler holding 30 liters of water
at 20.0oC. What temperature of the water when it comes to equilibrium?
mwater  waterV  30.0 kg
Q
Qice  Qmelt  Qicewater  Qwater  0 Q
ice
T  3.03C
m(kg) c(J/(kgoC)) L (J/kg) Tf
6.00 2090
0
Ti
Exp.
-5.00 mcT
3.33x105 0
0
mLf
Qice-water 6.00 4190
T
0
mcT
Qwater
T
Qmelt
6.00
30.0
4190
20.0 mcT
Mechanism of heat transfer
Conduction, convection, and radiation
 Conduction
• Conduction occurs within a body or between bodies in contact
• Heat transfer occurs only between region that are at different
temperatures
• The direction of heat flow always from higher to lower temperature
Heat
current
dQ
T1  T2
H
 kA
dt
l
Thermal Insulation
R = l/k ,R=R /A
Heat flows in the dir. of decreasing temp.
dQ
dT
H
 kA
dt
dx
dQ
T1  T2 T1  T2
A

dt

R
For Layers: eff   i
When the temp. varies in a non-uniform way
Mechanism of heat transfer
(cont’d)
 Conduction
(cont’d)
Thermal Conductivities of Some Materials
Material
k (J/(s m °C)
silver
copper
steel
420
380
40
glass
water
fiberglass
0.84
0.56
0.048
styrofoam
air
0.024
0.023
Mechanism of heat transfer (cont’d)
 Conduction
(cont’d)
Example 17.13
2.00cm
TH
=100oC
steel
copper
TC=0oC
10.0cm
20.0cm
What is the temperature at the junction of two bars?
The heat currents in the two bars must be equal.
H steel
kcopper A(T  0C )
ksteal A(100C  T )

 H copper 
Lsteal
Lcopper
ksteel  50.2 W /(m  K ), kcopper  385W /(m  K )
 T  20.7C
Mechanism of heat transfer (cont’d)
 Conduction
• Example : Two rods cases
k1,L,,A1
k2,L2
k1,L1
Tm
k2,L,,A2
 Th  Tm 
 Tm  Tc 
Q



 k1 A
 k2 A

t
 L1 
 L2 
Q
A(k1 / L1 )t
Q
Tm  Tc 
A(k2 / L2 )t
Th  Tm 
Q A(Th  Tc )

L1 L2
t

k1 k 2
Th  Tc 
Q  L1 L2 
  
At  k1 k 2 
T T
Q1
 k1 A1 h c
t
L
T T
Q2
 k 2 A2 h c
t
L
T T
Q Q1  Q2

 (k1  k 2 ) h c
t
t
L
Mechanism of heat transfer
(cont’d)
 Convection
Convection is the transfer of heat due to the net movement
of the medium by gravitational forces.
e.g. warm air is less dense than cold air and rises under the
influence of gravity.
Convection Heating
System for a Home
Mechanism of heat transfer
(cont’d)
 Radiation
• All objects radiate energy because of microscopic movements
(accelerations) of charges, which increase with temperature.
Heat current in radiation (= radiated power P)
dQ
PH 
 A  T 4
dt
Stefan' s Law
 : emissivity that depends on nature of surface (0=<  =<1)
A :area
=5.670400(40) x 10-8 W/(m2 K4) :Stefan-Boltzman const.
• If an object is at temperature T1 and its surroundings are at
temperature T2, the net flow of heat radiation between the
object and its surroundings is:
H net
dQ
4
4

   A(T1  T2 )
dt
Heat transfer by radiation
Exercises
Problem 1
You are making pesto for your pasta and have a cylindrical measuring cup
10.0 cm high made of ordinary glass ( =2.7 x 10-5 (Co)-1) that is filled with
olive oil ( =6.8 x 10-4 (Co)-1) to a height of 1.00 mm below the top of the cup.
Initially the cup and oil are at room temperature (22.0 oC). You get a phone
call and forget about the olive oil, which you inadvertently leave on the hot
stove. The cup and oil heat up slowly, and have a common temperature. At
what temperature will the olive oil start to spill out of the cup?
Solution
Both the volume of the cup and the volume of the olive oil increase when the
temperature increases, but  is larger than for oil, so it expands more. When
the oil starts to overflow, Voil  Vglass  (1.00103 m) A, where A is the crosssectional area of the cup.
Vglass  V0,glass  glass T  (10.0cm) A glass T
Voil  V0,oil oil T  (9.9cm) Aoil T
(9.9cm) Aoil T  (10.0cm) Aglass T  (1.00103 m) A
T  15.5 C  T2  T1  T  37.5 C
Exercises
Problem 2
A spacecraft made of aluminum circles the Earth at a speed of 7700 m/s.
(a) Find the ratio of its kinetic energy to the energy required to raise its
temperature from 0 oC to 600 oC. The melting point of aluminum is 660 oC.
(b) Discuss the bearing of your answer on the problem of the reentry of a
manned space vehicle into the Earth’s atmosphere.
Solution
(a)
K (1 / 2)m v2
v2
(7700m / s)2



 54.3
Q
cmT
2cT 2(910J /(kg  K ))(600C )
(b) Unless the kinetic energy can be converted into forms other than the
increased heat of the satellite, the satellite cannot return intact.
Exercises
Problem 3
In a household hot water heating system, water is delivered to the
radiator at 70.0 oC and leaves 28.0 oC. The system is to be replaced
by a steam system in which steam at atmospheric pressure condenses
in the radiators and the condensed steam leaves the radiator at 35.0 oC.
How many kilograms of steam will supply the same heat as was supplied
by 1.00 kg of hot water in the first system?
Solution
The ratio of masses:
ms
cw Tw
(4190J /(kg  K ))(42.0 K )


 0.0696
3
mw cw Ts  Lv (4190J /(kg  K ))(65.0 K )  225610 J / kg
so 0.0696 kg of steam supplies the same heat as 1.00 kg of water. Note
the heat capacity of water is used to find the heat lost by condensed steam.
Exercises
Problem 4
Calculate the ratio of the rate of heat loss through a single-pane window
with area 0.15 m2 to that for a double-pane window with the same area.
The glass of a single-pane is 4.2 mm thick, and the air space between
the two panes of the double-pane window is 7.0 mm thick. The glass has
thermal conductivity 0.80 W/(m K). The air films on the room and outdoor
surfaces of either window have a combined thermal resistance of 0.15
m2K/W.
Solution
The ratio will be inverse of the ratio of the total thermal resistance, as given
by Eq.(17.24). With two panes of glass with the air trapped in between,
compared to the single-pane, the ratio of the heat flows is:
[2( Lglass / k glass )  R0  ( Lair / kair )]
( Lglass / k glass )  R0
,
where R0 is the thermal resistance of the air films. Numerically, the ratio is:
[2((4.2  103 m) /(0.80W /(m  K )))  0.15m2  K / W  ((7.0  103 m) /(0.024W /(m  K )))]
 2.9.
(4.2  103 m) /(0.80W /(m  K ))  0.15m2  K / W
Exercises
Problem 5
A physicist uses a cylindrical metal can 0.250 m high and 0.090 m in
diameter to store liquid helium at 4.22 K; at that temperature the heat
of vaporization of helium is 2.09 x 104 J/kg. Completely surrounding the
metal can are walls maintained at the temperature of liquid nitrogen,
77.3 K, with vacuum between the can and the surrounding walls. How
much helium is lost per hour? The emissivity of the metal can is 0.200. The
only heat transfer between the metal can and the surrounding walls is by
radiation.
Solution
The rate at which the helium evaporates is the heat from the surroundings
by radiation divided by the heat of vaporization. The heat gained from the
surroundings comes from both the side and the ends of the cylinder, and
so the rate at which the mass is lost is:
[hd  2 (d / 2)2 ] (Ts4  T 4 )
 1.62  106 kg / s,
Lv
which is 5.82 g/h.