Chapter he Frst aw o hermodynamcs. developng the concept o heat. etendng our concept o work to thermal processes 3. ntroducng the rst law o thermodynamcs. Heat and Internal Energy Internal energy: s the energy assocated wth the mcroscopc components o a system atoms and molecules. It ncludes knetc and potental energy assocated wth the random translatonal, rotatonal, and vbratonal moton o the atoms or molecules that make up the system as well as an ntermolecular potental energy. Heat: s a mechansm by whch energy s transerred between a system and ts envronment because o a temperature derence between them. It s also the amount o energy Q transerred by ths mechansm. Unts o Heat: he calore s the heat necessary to rase the temperature o g water rom 4. o to. o C. he Mechancal Equvalent o Heat Joule s Eperment: he result that 4.84 J o mechancal energy s equvalent to cal o heat energy s known as the mechancal equvalence o heat. cal = 4. 86 J, Calore = calore
Eample: osng weght that hard way A student eats a dnner contanng Calores o energy. He wshes to do an equvalent amount o work n the gymnasum by ltng a -kg object. How many tmes must he rase the object to epend ths much energy? Assume that he rase t a dstance o m each tme. E = 4.86 = 837J, 837 3 = 8., 8. 3 tmes 9.8. Specc Heat and Calormetry a quantty o energy Q s transerred to a mass m o a substance and changng ts temperature by, he heat capacty: 熱容量 Q C s dened as C = he specc heat: 比熱 c s dened as c = Q m Calormetry: 熱量計 place the object nto a vessel contanng water and measure the change o temperature m c ( ) = m c ( ) c m c ( ) = m ( ) Eample: Coolng a Hot Ingot he temperature o a.-kg ngot o metal s rased to o C and the ngot s then dropped nto a lght, nsulated beaker contanng.4 kg o water ntally at o C. I the nal equlbrum temperature o the med system s.4 o C, nd the specc heat o the metal. 4 (.4 )4.86 J c = =.4( ) (.4) g O C
.3 atent Heat 潛熱 latent heat: Q = m latent heat o usson: latent heat o vaporzaton: Substance Meltng Pont atent Heat o v Fuson Bolng Pont atent Heat o Vaporzaton (J/kg) He.9 K 4. K.9 4 H 4. K. K N 63. K 77.3 K. O 8.7 K 89.9 Alcohol ater o C 3.33 o C.6 6 Pb 37.3 o C 7 o C Al 66 o C 4 o C Ag 96.8 o C 93 o C Au 63 o C 66 o C Cu 83 o C 87 o C 34 o C o C mportant data or thermal evaporaton Part A: Q = mc ce Part B: Q = m.4 ork and Heat n hermodynamcs state varables: pressure, volume, temperature, and nternal energy 3
transer varables assocated wth a change n the state o the system the gas s compressed quas-statcally, that s, slowly enough to allow the system to reman n thermal equlbrum at all tmes r r d = F dr = Fdy = PAdy = PdV V F PdV 對氣體做功? = V n PV dagram, he work done on a gas n a quas-statc process that takes the gas rom an ntal state to a nal state s the negatve o the area under the curve on a PV dagram, evaluated between the ntal and nal states. Eample: Comparng processes An deal gas s taken through two processes n whch P = Pa, V = m 3, P =. Pa, and V = m 3. For process, the temperature remans constant. For process, the pressure remans constant and then the volume remans constant. hat s the rato o the work done on the gas n the rst process to the work done n the second process? PV = P V = C = V V C C V = dv = dv = C ln = ln( ) = 3. V V V V V = P ( V V ) = P ( V V ) =. ( ) =.6, = Isothermal Epanson: A gas at temperature epands slowly whle absorbng energy rom a reservor to mantan the constant temperature. Free Epanson: A gas epands rapdly nto an 4
evacuated regon ater a membrane s broken. Energy transer by heat, lke work done, depends on the ntal, nal, and ntermedate states o the system.. he Frst aw o hermodynamcs he rst law o thermodynamcs s a specal case o the law o conservaton o energy. he quantty Q+ s ndependent o the path. nt E = Q + 暫時不談位能 de = dq + d nt cyclc process: a process that starts and ends at the same state E nt =, Q = n a cyclc process, the net work done on the system per cycle equals the area enclosed by the path representng the process on a PV dagram.6 Some Applcatons o the Frst aw o hermodynamcs Energy Conservaton: E nt = Q +, = P V Adabatc Process: Q = E = nt Isobarc Process: ( ) = P V V
Isovolumetrc Process: = E = Q nt Isothermal Process: A process that occurs at a constant temperature s called an sothermal process. Isothermal Epanson o an Ideal Gas: V PV = nr, = PdV V P = nr V V nr = V V dv V = nr ln V V = nr ln V.7 Energy ranser Mechansms. Conducton. Convecton 3. Radaton In all mechansms o heat transer, the rate o coolng o a body s appromately proportonal to the temperature derence between the body and ts surroundng. Rate hermal conducton: Q H C P = = ka = ka ( I Q t t 阻 ) /, 熱流 = ( I ) ( ka) = IR /, R: 熱 For a compound slab contanng several materals o thckness,,... and thermal conductvtes k, k,, the rate o energy transer through the slab at steady state s A( h c ) P = / k ( = IR + IR +... = I R I = R = A ) / k Eample: wo slabs o thckness and and thermal conductvtes k and k are n thermal contact wth each other. he temperature o ther outer suraces are c and h, 6
respectvely, and h > c. Determne the temperature at the nterace and the rate o energy transer by conducton through the slabs n the steady-state condton. c P = ka, h P = ka when a steady state s reached, h ka k k = ka + k + k h c =, c A( P = ) k h c / k + / EX: he nsde (radus a) o a hollow cylnder s mantaned at a temperature a whle the outsde (radus b) s a lower temperature, b. he wall o the cylnder has a thermal conductvty k. Ignorng end eects, show that the rate o energy conducton rom the nner to the outer surace n the radal drecton s dq = I = dt dq dt k / A a b = π k ln b dq dr dt π r = k( ) ( b / a) a a b dq ( ) = k dt π a b ln b / a. Home Insulaton Convecton: It s transport o energy as heat by the transport o the materal medum tsel. Radaton: electromagnetc radaton Stephan-Boltzmann law: 4 P = σae A s the area, σ s a unversal constant called Stephan s constant 7
8 σ =.673X /(m K 4 ) e s the emssvty o the object. Its value s between and. Eample: he temperature o a lghtbulb lament Estmate the order o magntude o the temperature o the lament o a lghtbulb when t s operatng. o model the lament as a cylnder cm long wth a radus o. mm. P 4 = 4 =.7 8 σae.67 ( π )() = 3 K Melt? Blackbody radaton?? λ ma.898 = ( mm) 4 4 ( ) = e A( + )( + )( ) P = eσ A σ 3 dp 3 I ~ P = 4eσ A ( P = = 4eσ A d coolng law = ) Newton s coee he Dewar Flask 8