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+// Example 3_6
+clc;funcprot(0);
+// Given data
+T_1=20;// °C
+T_2=100;// °C
+p_1=0.100;// MPa
+p_2=1.00;// MPa
+rho=515;// kg/m^3
+c=1.76;// kJ/kg.K.
+
+// Solution
+deltau=c*((T_2+273.15)-(T_1+273.15));// The change in specific internal energy in kJ/kg
+v=1/rho;// The specific volume in m^3/kg
+deltah=deltau+(v*((p_2*10^3)-(p_1*10^3)));// The change in specific enthalpy in kJ/kg
+printf('\nThe change in specific internal energy,u_2-u_1=%3.0f kJ/kg \nThe change in specific enthalpy,h_2-h_1=%3.0f kJ/kg',deltau,deltah);