Question

已知冰和水的摩尔热容为：Cp,m(冰)=37.6J.mol^(-1).K^(-1)；Cp,m(水)=75.36J.mol

Answer 1

上述过程可分为三步：

[1] 263.2K的冰变为273.2K,ΔS=nCp,m*ln[T2/T1]=1*37.6*ln[273.2/263.2]=1.402J/K

[2]273.2度时冰熔化为水 ΔS=ΔHm/T==[-60206]/273.2==-220.37J/K

[3]273.2度的水升温到298.2K ΔS=nCp,m*ln[T3/T2]=1*75.36*ln[298.2/273.2]=6.599J/K

整个过程体系熵变==1.402+6.599-220.37==-212.369J/K

ΔH1==nCp,m[T2-T1]==37.6*10==376J/mol

ΔH2=-60206J/mol

ΔH3==nCp,m[T3-T2]==75.36*25=1884J/mol

ΔH=376+1884-60206=-57946J/mol

ΔS环=-ΔH/T==-57946/300==-193.15J/K

ΔS+ΔS环==-405.519J/K