1、Chapter 16: Temperature and Heat,Temperature and thermal equilibrium,Temperature,is a measure of how hot or cold an object isis 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
2、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.,Thermometer,Thermometers are devices used to measure the temperature ofan object or a system.,When a thermometer is in thermal contact with a syst
3、em, energy isexchanged until the thermometer and the system are in thermalequilibrium with each other.,All the thermometers use some physical properties that depend onthe temperature. Some of these properties are:1) the volume of a fluid2) the length of a solid3) the pressure of a gas held at consta
4、nt volume4) the volume of a gas held at constant pressure5) electric resistance of a conductor6) the color of very hot object.,Temperature and thermal equilibrium (contd),Thermometer (contd),One common thermometer consists of a mass of liquid: mercury oralcohol. The fluid expands into a glass capill
5、ary tube when itstemperature rises.,When the cross-sectional area of the tubeis constant, the change in volume of theliquid varies linearly with its length alongthe tube.,The thermometer can be calibrated byplacing it in thermal contact withenvironments that remain at constant temp.,Two of such envi
6、ronments are:1) a mixture of water and ice in thermalequilibrium at atmospheric pressure.2) a mixture of water and steam in thermalequilibrium at atmospheric pressure.,0oC (Celsius) 100oC,Freezing point,Boiling point,Temperature and thermal equilibrium (contd),Constant-volume gas thermometer and the
7、 Kelvin scale,A constant-volume gas thermometer measures the pressure ofthe gas contained in the flask immersed in the bath. The volumeof the gas in the flask is kept constant by raising or loweringreservoir B to keep the mercurylevel constant in reservoir A.,Temperature and its scales,Constant-volu
8、me gas thermometer and the Kelvin scale,It has been experimentallyobserved that the pressure varieslinearly with temperature ofa fixed volume of gas, which doesnot depend on what gas is used.,It has been experimentallyobserved that these straight linesmerge at a single point at temp.-273.15oC at pre
9、ssure = 0.This temperature is called absolutezero, which is the base of the Kelvin temperature scale T=TC-273.15measured in kelvin (K) where TC is temperature in Celsius,0 K = -273.15oC,Temperature and its scales (contd),Gas thermometer and absolute (Kelvin) scale (contd),The pressure of any gas at
10、constant volume is a linear function of temperature, which always extrapolates to zero at 273.15 C.,The absolute or Kelvin temperature scale:T(K) = T(C) + 273.15,pressure,const. volume,Temperature and its scales (contd),In fact it is also true that:,const. pressure,Temperature and its scales,Tempera
11、ture scales,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 Earths surface at sea level.,( 0o C : the freezing point 100o C : the boiling point ),Thank you M
12、r. Fahrenheit!,Temperature scales,The common temperature scale in US isFahrenheit:,Temperature and its scales,Linear expansion,Thermal expansion (Ch.17),Most materials expand when heated: The average distance between atoms increases as the temperature is raised. The increase is proportional to the c
13、hange 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 = coefficient of linear expansion C-1( is a property of the material),Coefficients of linear expansion,Thermal expansion (Ch.17)(contd),Volume expansion,Th
14、ermal expansion (Ch.17)(contd),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
15、 and the initial volume:,Relation between a and b,For solid materials there is a simple relation between a and b as V=L3:,Thermal expansion of water,Thermal expansion (Ch.17) (contd),Water contracts when heated from 0C to 4C, then expands when heated from 4 C to 100 C. Just above the freezing point,
16、 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!,Heat,Quantity of heat,When two objects of different temperatures are in thermal contact, their temperature eventually reach th
17、e 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 temperatureof 1 g of water from 14.5 oC to 15.5 oC Btu : the amount of heat required to r
18、aise the temperatureof 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,Specific heat,Quantity of heat (contd),The quantity of heat Q required to increase the temperature of a mass m of a certain material from T1 to T2 is fo
19、und to be approximately proportional to the temperature change DT=T2-T1 and to mass m.,For an infinitesimally small change in temperature:,specific heat,Molar heat capacity,Quantity of heat (contd),Often it is more convenient to describe a quantity of substancein terms of moles n rather than the mas
20、s m of material.A mole of any pure substance contains the same number ofmolecules.The molar mass of any substance M is the mass per mole.,molar heat capacity,For water C=(0.0180 kg/mol)4190 J/(kg T) = 75.4 J/(mol K),Phase changes,Phase transition,Phases of matter : solid, liquid, gas,A change of pha
21、se : phase transition,For any given pressure a phase change takes place at adefinite temperature, usually accompanied by absorptionor emission of heat and a change of volume and density,Latent heat (see Table 17.1),Phase transition (Ch.17),heat of fusion,heat of vaporization,phase equilibrium,Heat o
22、f fusion Lf : heat needed to change from liquid to gas per kgof material,Heat of vaporization Lv : heat needed to change from solid to liquid per kgof material,3.34 x 105 J/kg = 79.6 cal/g =143 Btu/lb,Latent heat and phase change,Water,Consider an addition of energy to a 1.00-g cube of ice at -30.0o
23、C in a container held at constant pressure. Suppose this input energy turns ice to steam (water vapor) at 120.0oC.,A:,B:,C:,D:,E:,Phase transition (Ch.17),Calorimetry,Isolated system,A system whose energy does not leave out of the system iscalled isolated system.,The principle of energy conservation
24、 for an isolated systemrequires that the net result of all the energy transfer is zero.If one part of the system loses energy, another part has togain the energy.,Calorimeter and calorimetry,Imagine a vessel made of good insulating material and containingcold water of known mass and temperature and
25、the temperature ofthe water can be measure. Such a system of the vessel and wateris called calorimeter. If the object is heated to a higher temperatureof known value before it is put into the water in the vessel, the specificheat of the object can be measured by measuring the change in temperature o
26、f the water when the system (the object, vessel, andwater) reaches thermal equilibrium. This measuring process is calledcalorimetry.,“measuring heat”,Calorimetry,Calorimeter and calorimetry,When a warm object is put into a calorimeter with cooler waterdescribed in the previous page, it becomes coole
27、r while the waterbecomes warmer.Qcold (0 ) is the heat transferred (energy change) to the coolerobject and Qhot (0) is the heat transferred (energy change) to thewarmer object.,In general, in an isolated system consisting of n objects :,Tf common to all objects in equilibrium.,Calorimeter and calori
28、metry,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 tempera
29、ture of the system.,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?,Q m(kg) c(J/(kgoC) L (J/kg) Tf Ti Exp.,Qice 6.00 2090
30、 0 -5.00 mcDT,Qmelt 6.00 3.33x105 0 0 mLf,Qice-water 6.00 4190 T 0 mcDT,Qwater 30.0 4190 T 20.0 mcDT,Calorimetry and phase transition,Conduction,Mechanism of heat transfer,Conduction, convection, and radiation,Conduction occurs within a body or between bodies in contactHeat transfer occurs only betw
31、een region that are at differenttemperaturesThe direction of heat flow always from higher to lower temperature,Thermal Insulation R = l/k ,R=R /A,When the temp. varies in a non-uniform way,Heat flows in the dir. of decreasing temp.,Heat current,Conduction (contd),Mechanism of heat transfer (contd),T
32、hermal Conductivities of Some Materials,Conduction (contd),Mechanism of heat transfer (contd),Example 17.13,10.0cm,20.0cm,steel,copper,2.00cm,TH=100oC,TC=0oC,The heat currents in the two bars must be equal.,What is the temperature at the junction of two bars?,Conduction,Example : Two rods cases,k1,L
33、1,k2,L2,Tm,k1,L,A1,k2,L,A2,Mechanism of heat transfer (contd),Convection,Mechanism of heat transfer (contd),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
34、Heating System for a Home,Radiation,Mechanism of heat transfer (contd),All objects radiate energy because of microscopic movements(accelerations) of charges, which increase with temperature.,Heat current in radiation (= radiated power P),s=5.670400(40) x 10-8 W/(m2 K4) :Stefan-Boltzman const.,Heat t
35、ransfer by radiation,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:,A :area,e : emissivity that depends on nature of surface (0= e =1),Exercises,Problem 1,Solution,You are making pesto for your
36、pasta and have a cylindrical measuring cup 10.0 cm high made of ordinary glass ( b=2.7 x 10-5 (Co)-1) that is filled with olive oil ( b=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 abo
37、ut 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?,Both the volume of the cup and the volume of the olive oil increase when the temperature increases, b
38、ut b is larger than for oil, so it expands more. When the oil starts to overflow, where A is the cross- sectional area of the cup.,Exercises,Problem 2,Solution,A spacecraft made of aluminum circles the Earth at a speed of 7700 m/s. Find the ratio of its kinetic energy to the energy required to raise
39、 itstemperature 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 amanned space vehicle into the Earths atmosphere.,(a),(b) Unless the kinetic energy can be converted into forms other than theincreased heat of the sa
40、tellite, the satellite cannot return intact.,Exercises,Problem 3,Solution,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 th
41、e 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?,The ratio of masses: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 t
42、he heat lost by condensed steam.,Exercises,Problem 4,Solution,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 do
43、uble-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.,The ratio will be inverse of the ratio of the total thermal resistance, as given by Eq.(17.24). With tw
44、o panes of glass with the air trapped in between, compared to the single-pane, the ratio of the heat flows is:,where R0 is the thermal resistance of the air films. Numerically, the ratio is:,Exercises,Problem 5,Solution,A physicist uses a cylindrical metal can 0.250 m high and 0.090 m in diameter to
45、 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?
46、 The emissivity of the metal can is 0.200. The only heat transfer between the metal can and the surrounding walls is by radiation.,The rate at which the helium evaporates is the heat from the surroundingsby radiation divided by the heat of vaporization. The heat gained from thesurroundings comes from both the side and the ends of the cylinder, andso the rate at which the mass is lost is:which is 5.82 g/h.,
