Chapter 16: Temperature and Heat by DWt7kRZs

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									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       Boiling
       equilibrium at atmospheric pressure.
                                                           point
    2) a mixture of water and steam in thermal
       equilibrium at atmospheric pressure.
         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)
  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
                                                      const. volume
                                        T1 p1
                pressure




                                      The absolute or Kelvin
                                      temperature scale:
                                        T(K) = T(ºC) + 273.15

                                       In fact it is also true that:

                                       T2 V2
                                                     const. pressure
                                       T1 V1
         Temperature and its scales
                                                      Thank you Mr.
 Temperature    scales                               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  L f  Li  Li T or L  L0 T   ( Li  L0 , T  T f  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  3L2L0dT  3L3 dT  V0dT
                                 0             0
             dL
                         dL  L0dT


         3
               Thermal expansion (Ch.17) (cont’d)
 Thermal              expansion of water
                                              • Water contracts when
             Density of Water                   heated from 0ºC to 4ºC, then
                                                expands when heated from
                                                4 ºC to 100 ºC.
              1                               • Just above the freezing
                                                point, the coldest (and least
            0.99                                dense) water rises to the
g/(cm**3)




                                                surface, and lakes freeze
            0.98                                from the surface downward.
            0.97                              • This unusual property
                                                permits aquatic life on earth
            0.96                                to survive winter!

            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  mcT
 For an infinitesimally small change in temperature:

                    dQ  mcdT
                       1 dQ
                    c                  specific heat
                       m dT
              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        molar heat capacity
                             n dT
  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  mciceT , T  30.0C
      62.7 J cice  2090 J/(kg C)

B:Q  mL f , L f  3.33 10 J / kg
                           5


     333 J
C: Q  mcwater T , T  100C
      4.19 J cwater  4190 J/(kg C)


D: Q  mLv , Lv  2.26 10 J / kg       E: Q  mcsteamT , T  20.0C
                          6


      2.26 103 J                            40.2 J csteam  2010 J/(kg C)
                          Calorimetry            “measuring heat”
 Isolated   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                            Tf common to all objects in equilibrium.
     Q
      k 1
             k    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 c g (T  Tg )  0


       mwcwTw  mal calTal  mg cgTg
  T                                      37.9C
           mwcw  mal cal  mg cg
                  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        m(kg) c(J/(kgoC)) L (J/kg) Tf   Ti    Exp.
Qice  Qmelt  Qice water  Qwater  0
                                          Qice     6.00 2090                  0    -5.00 mcT
                                          Qmelt    6.00             3.33x105 0      0    mLf
 T  3.03C
                                          Qice-water 6.00 4190                T     0   mcT
                                          Qwater   30.0   4190                T    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
                                                     dQ      T1  T2
                                        Heat
                                        current
                                                  H     kA
                                                     dt         l
                                                   Thermal Insulation
                                                   R = l/k ,R=R /A

                                              dQ     T1  T2 T1  T2
                                                  A          
Heat flows in the dir. of decreasing temp.    dt                 R
        dQ       dT                           For Layers : eff   i
     H     kA
        dt       dx                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                            420
         copper                             380
          steel                              40
          glass                            0.84
          water                            0.56
       fiberglass                          0.048
       styrofoam                           0.024
           air                             0.023
       Mechanism of heat transfer (cont’d)
 Conduction         (cont’d)
Example 17.13
                        2.00cm
                     steel     copper
 TH   =100oC                                         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.
            ksteal A(100C  T )              kcopper A(T  0C )
  H steel                        H copper 
                     Lsteal                           Lcopper
  ksteel  50.2 W /(m  K ), kcopper  385 W /(m  K )
   T  20.7C
            Mechanism of heat transfer (cont’d)
     Conduction
     • Example : Two rods cases
                                                             k1,L,,A1
                  k1,L1         k2,L2


                           Tm
                                                              k2,L,,A2

   Q         T T        T T                    Q1        T  T Q2        T T
       k1 A h m   k2 A m c 
             L           L                           k1 A1 h c      k2 A2 h c
   t           1          2                     t          L   t          L
                Q
Th  Tm 
          A(k1 / L1 )t               Q  L1 L2     Q Q1  Q2              T T
                           Th  Tc                         (k1  k 2 ) h c
                                     At  k1 k2 
                Q                                    t   t
Tm  Tc                                                                    L
          A(k 2 / L2 )t

     Q A(Th  Tc )
        
     t   L1 L2
            
          k1 k 2
           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                Stefan's Law
            dt
           : 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:

                  dQ
                       A(T1  T2 )
                               4    4
        H net                                 Heat transfer by radiation
                  dt
                                         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 cross-
sectional area of the cup.
     Voil  V0,oil  oil T  (9.9cm ) A oil T   Vglass  V0, glass  glass T  (10 .0cm ) A glass 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)mv 2    v2        (7700 m / s ) 2
                                                       54 .3
      Q   cmT        2cT 2(910 J /(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             (4190 J /(kg  K ))( 42 .0 K )
                                                                       0.0696
      mw cw Ts  Lv (4190 J /(kg  K ))( 65 .0 K )  2256 10 3 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 )))  0.15 m 2  K / W  (( 7.0  10 3 m ) /(0.024 W /(m  K )))]
                                                                                                             2.9.
                             ( 4.2  10 3 m ) /(0.80W /(m  K ))  0.15 m 2  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.

								
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