VIEWS: 43 PAGES: 33 POSTED ON: 2/9/2012 Public Domain
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 3L2L0dT 3L3 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 mcT 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 nMcT nCT 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 mciceT , T 30.0C 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 100C 4.19 J cwater 4190 J/(kg C) D: Q mLv , Lv 2.26 10 J / kg E: Q mcsteamT , T 20.0C 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 37.9C 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 mcT Qmelt 6.00 3.33x105 0 0 mLf T 3.03C Qice-water 6.00 4190 T 0 mcT Qwater 30.0 4190 T 20.0 mcT 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(100C T ) kcopper A(T 0C ) H steel H copper Lsteal Lcopper ksteel 50.2 W /(m K ), kcopper 385 W /(m K ) T 20.7C 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 At 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 PH 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 (1.00 103 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) Aoil T (10.0cm) A glass T (1.00 103 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 cmT 2cT 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: [hd 2 (d / 2)2 ] (Ts4 T 4 ) 1.62 106 kg / s, Lv which is 5.82 g/h.