大学物理(热学) 1)Review 2 Examples 3) Order and Disorder 2006-1227 2021/9/5
2021/9/5 1 大学物理（热学） 1) Review 2) Examples 3) Order and Disorder 2006-12-27
Clausius' deduction (1865) FIGURE 20-5 The Carnot cycle Heat In a cyo the cycle for the Carnot engine Isothermal begins at point a on this Pl ue20-1 schemata sus曰e expansion diagram (1)The gas is first panded isothermally, with addition of heat @), along path ab at temperature TH. (2)Next d→ he gas expands adia Adiabatic Adiabatic from b to c-no heat is exchanged compression lexpansion but the temperature drops to TL (3)The gas is then compressed at int temperature Ti, path c to d d 0=0 and heat e. flows out. (4)Finally he gas is compressed adiabatically path da, back to its original state. No Carnot engine actually exists, but as a theoretical engine it played e= W/QH c→d Isothermal important role in the development of thermodynamics 1-Q/QH compression = H Q /T HLL 0 1-TL/TH dorer ∑(Q/T)=0 0
e= W/ QH = 1-QL /QH =1-TL /TH Clausius’ deduction (1865) QL / TL=QH /TH, QH/ TH-QL /TL =0 (Q/T)=0 = 0 T dQrev
dor rev T FIGURE 20-12 The integral, gas, of the entropy for a reversible cycle is zero. Hence the difference in entropy between states a and b b S ∫adS, is the same for path I as for path II
= ? T dQrev
do, rev 0 T T FIGURE 20-11 Any reversible cycle can be approximated as a series of carnot cycles (The dashed lines represent isotherms.) (Q1T1+Q22)+(-Q2T2+Q33)+…+(-Q6T6+Q7T7)=0
= 0 T dQrev T1 T2 T3 T4 T5 T6 T7 (Q1 / T1+Q2 /T2 )+(-Q2 /T2 +Q3 /T3 )+…+ (-Q6 /T6 +Q7 /T7 )=0
)2+=0—)学=(n)男 ds FIGURE 20-12 The integral, gas, of the entropy for a reversible cycle is zero. Hence the difference in entropy between states a and b b S ∫adS, is the same for path I as for path II
T dQ dS = = b a T dQ b a T dQ ( ) + ( ) = 0 (I) (II) a b T dQ b a T dQ I II
Irreversible: e=W/ QH=1-Q/QHQH HIH do, rel ∑(QT)<0 0
Irreversible: e= W/ QH = 1-QL /QH QH /TH (Q/T)<0 0 T dQrev
()学+(m)∫孥 I B0A do R&R FIGURE 20-12 The integral, ds, of the entropy for a reversible cycle is zero. Hence the difference in entropy between states a and b, SB-SA2 r dQ b- Sa= jads, is the same for path I as for path Il For isolated system:△s≥0
− B A B A T dQ S S ( ) + ( ) 0 a b T dQ b a T dQ I II b a T dQ b a T dQ (I) (II) − B A B A T dQ S S R&IR For isolated system: S 0
Clausius Statement The energy of the universe is a constant. The entropy of the universe approaches a maximum
“The energy of the universe is a constant. The entropy of the universe approaches a maximum. ” Clausius’ Statement
Example 20-6 Entropy change in melting. A 1.00-kg piece of ice at 0oC melts very slowly to water at 0oC. Assume the ice in contact with a heat reservoir whose temperature is only infinitesimally greater than 0oC Determine the entropy change of (a) the ice cube and(b)the heat reservoir
Entropy change in melting. A 1.00-kg piece of ice at 0 oC melts very slowly to water at 0 oC. Assume the ice in contact with a heat reservoir whose temperature is only infinitesimally greater than 0 oC. Determine the entropy change of (a) the ice cube and (b) the heat reservoir. Example 20-6
EXAMPLE 20-6 Entropy change in melting. A 1.00-kg piece of ice at o'C melts very slowly to water at 0C. Assume the ice is in contact with a heat reser. voir whose temperature is only infinitesimally greater than 0 C. Determine the entropy change of (a)the ice cube and(b) the heat reservoir SoLUTION(a) The process is carried out at a constant temperature T=273K and is done reversibly, so we can use Eq 20-12 do 1 △S;。= TT do Since the heat needed to melt the ice is 0= mL, where the heat of fusion L=797 kcal/kg=3.33 X 10 J/kg, we have mL(1.00 kg)(797 kcal/kg) △Sice=T 273K 0.292 kcal/K or 1220J/K (b)Heat @- mL is removed from the heat reservoir, so(since T=273 K and is constant △Sres =-0.292kcal/K Note that the total entropy change, ASice ASres, is zero
