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+//example 5
+//calculating heat transfer for the given process
+clear
+clc
+Vliq=0.05 //volume of saturated liquid in m^3
+vf=0.001043 //in m^3/kg
+Vvap=4.95 //volume of saturated water vapour in m^3
+vg=1.6940 //in m^3/kg
+m1liq=Vliq/vf //mass of liquid in kg
+m1vap=Vvap/vg //mass of vapors in kg
+u1liq=417.36 //specific internal energy of liquid in kJ/kg
+u1vap=2506.1 //specific internal energy of vapors in kJ/kg
+U1=m1liq*u1liq+m1vap*u1vap //total internal energy in kJ
+m=m1liq+m1vap //total mass in kg
+V=5 //total volume in m^3
+v2=V/m //final specific volume in m^3/kg
+disp('by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 MPa')
+u2=2600.5 //specific internal energy at final state in kJ/kg
+U2=m*u2 //internal energy at final state in kJ
+Q=U2-U1 //heat transfer for the process in kJ
+printf("\n hence,heat transfer for the process is Q = %.0f kJ. \n",Q)
+\ No newline at end of file