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+// Example 4_4
+clc;funcprot(0);
+// Given data
+P=400;// kPa
+V_1=2;// m^3
+T_2=200;// °C
+Q=3500;// The amount of heat added in kJ
+
+// Calculation
+// Using the steam tables
+v_1=0.5342;// m^3/kg
+u_1=2674;// kJ/kg
+m=V_1/v_1;// kg
+// V_2=m*v_2
+// Q-(P*(V_2-V_1))=(u_2-u_1)*m---->(a)
+// This requires the trial and error process.
+// For example,guess
+v_2=1.0;// m^3/kg
+u_2=((Q-(P*((m*v_2)-V_1)))/m)+u_1;// kJ/kg
+// From the steam tables at P=0.4 MPa
+T_2=654;// °C
+// The v_2 gives 
+T_2=600;// °C
+// Guess
+v_2=1.06;// m^3/kg
+u_2=((Q-(P*((m*v_2)-V_1)))/m)+u_1;// kJ/kg
+// The tables are interpolated to give 
+T_2=640;// °C
+// The v_2 gives 
+T_2=647;// °C
+// The final temperature being approximately
+T_2=644;// °C
+printf("\nThe final temperature being approximately,T_2=%3.0f°C",T_2);