1、Chapter 23 The first law of thermodynamics23-1 Heat:energy in transit(传输)1.“Heat is energy that flows between a system and its environment because of a temperature difference between them”We choose our sign convention so that Q is positive in the case that the internal energy of the system tends to
2、be increased.Like other forms of energy,heat can be expressed in the SI unit of Joules(J).2.Misconceptions(误解)about heatNeither heat nor work is an intrinsic property of a system.We cannot say that a system“contains”a certain amount of heat or work.They are not state functions.Both heat and work ass
3、ociated with a“thermodynamic process”3.The understanding of heat in historySee动画库力学夹5-03作功与传热23-2 The transfer of heat1.Thermal conductionConsider a thin slab of a homogeneous material of thickness and area A(Fig 23-2).One face is held at T and the other at a somewhat higher constant temperature.QAT
4、xxTT+TT+Fig 23-2Experiment shows thatis(23-1)H is the“rate of heat transfer”k is the“thermal conductivity”of the materialA is the area of the slabis the thickness of the slab)(tQH=xTkAH=xConsidering the direction of H and infinitesimal thicknessof the slab,we have:dxdTkAH=(23-4)Sample Problem 23-2A
5、thin,cylindrical metal pipe is carrying steam at a temperature of Ts=1000c.The pipe has a diameter of 5.4 cm and is wrapped with a thickness of 5.2 cm of fiberglass(玻璃丝)insulation.A length D=6.2 m of the pipe passes through a room in which the temperature is TR=110c.At what rate does heat energy pas
6、s through the insulation?2.Convection(对流)3.RadiationProblem Two identical rectangular rods of metal are welded end to end as shown in Fig.a),and 10 J of heat flows through the rods in 2.0 min.How long would it take for 30 J to flow throught the rods if they are welded as shown in Fig.b?a)b)00C1000C1
7、000C00C23-3 The first law of thermodynamics 1.For a thermodynamic system,internal energy is the only type of energy the system may have.The law of conservation of energy of the system can be expressed as(First law of thermodynamics)(23-6)(i)Q is the energy transferred(as heat)between the system and
8、its environment because of a temperature different.intEWQ=+(ii)W is the work done on(or by)the system by forces that act through the system boundary.(a)(b)(c)initial stateprocessfinal stateFig 23-9EnvironmentQWEint,iEint,fWQEEEif+=int,int,intBoundary(iii)In any thermodynamic process between equilibr
9、ium state i and f,the quantity Q+W has the same value.This quantity is equal to the change in the internal energy .(iv)The first law of the thermodynamics is a general result that is thought to apply to every process in nature that proceeds between equilibrium states.intE23-4 Heat capacity(热容)and sp
10、ecific heat（比热）1.Heat capacity C：(23-7)2.Specific heat：The heat capacity per unit mass of a body(23-8)The heat capacity(C)is characteristic of a particular object,but the specific heat(c)characterizes a kind of substance.TQC=TmQmCc=Usually both C and c depend on the temperature and conditionunder wh
11、ich the heat Q is added to the material.3.Molar heat capacityIf we multiply the specific heat by the molar mass M,we obtain the“molar heat capacity”.=dTTcmdTTcmQ)()(23-10)TnQTmQMcmol=n-the molar number*4.Heats of Transformation(Latent heat(潜热)When heat enters a sample,the sample may change from one
12、phase or state to another.In this case,the temperature of the sample does not rise.Vice versa.Larger3R*Measured at room temperature and atmospheric pressure.A cube of copper of mass mc=75 g is placed in an oven at a temperature of T0=3120C until it comes to thermal equilibrium.The cube is then dropp
13、ed quickly into an insulated beaker(烧杯)containing a quantity of water of mass mw=220 g.The heat capacity of the beaker alone is Cb=190 J/K.Initially the water and the beaker are at a temperature of Ti=12.00c.What is the final equilibrium temperature Tfof the system consisting of the copper+water+bea
14、ker?Sample Problem 23-323-5 Work done on or by an ideal gasFig 23-13pv(b)W1v2v1.Work done on an ideal gasVPW=(a)xideal gasxFdVVPdW=)(=fivvdVVPW)(Work done on gas in a more general form:(23-15)=PdVW(a)If PV relationship is known,the work done on the gas is equal to the area under the curverepresentin
15、g the process.(b)The pressure force is not a conservative force.ABCDPVFig 23-14ivfvTwo paths:A B DA C D(a).Work done at constant volume(V is const.)(23-16)0=W2.Several typical thermal processes=PdVW(b).Work done at constant pressure(P is const.)(23-17)=dVPdVVPW)(constantPV=(c).Work done at constant
16、temperature(T is const.)namely“isothermal(等温)process”,pvW=0ivfvW 0pvfivv=VP)vP(vif=vivfvpP(V)curve is hyperbolic(双曲线).The curve of PV=const.is called an“isotherm(等温线)”.PVWFig 23-15ivfvifvvvvnRTvdvnRTdvvnRTPdVWfiln=(23-18)vnRTP=(d).Work done in thermal isolationThermal isolation is also called“adiabatic”process.Q=0.T can be changed.(23-19)const.PV=const.pv=const.pv=iPfPivfvPVFig 23-16If we know ,and the initialiiV,PconstantVPPVi=i,we have）8.11.1(=We can now find the adiabatic work:=VVPPii 1)()(11