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authorpriyanka2015-06-24 15:03:17 +0530
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+//example 5.9
+
+clear;
+clc;
+
+//Given:
+T1=276;//Initial temperature[K]
+Tf=278.7;//Freezing point temperature[K]
+Tb=353.3;//Boiling point temperature[K]
+T2=373;//Final temperature[K]
+Hf=9870;//Standard enthalpy of fusion[J/mol]
+Hv=30800;//Standard enthalpy of vaporization[J/mol]
+Cp=136.1;//Specific heat capacity of benzene[J/K/mol]
+mol.wt=78;//molecular weight of benzene[g/mol]
+mass=200;//weight of solid benzene[g]
+disp(" Cp doesnot change within this temp limit");
+
+//To find the total entropy change of solid
+n=mass/mol.wt;//no. of moles
+S1=n*Cp*log(Tf/T1);//entropy change in heating [J/K]
+S2=n*Hf/Tf;//entropy change in melting[J/K]
+S3=n*Cp*log(Tb/Tf);//entropy change in heating[J/K]
+S4=n*Hv/Tb;//entropy change in vaporization[J/K]
+S5=n*Cp*log(T2/Tb);//entropy change in heating[J/K]
+S=S1+S2+S3+S4+S5;//total entropy change in heating from 276 to 373K
+
+printf("Total entropy change in heating 200g benzene from 3 to 100`C is %f J/K",S);
+ \ No newline at end of file