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+// Example 3_7
+clc;funcprot(0);
+// Given data
+T_1=240;// °F
+T_2=80;// °F
+p_1=150;// psia
+p_2=14.7;// psia
+c_p=0.240;// Btu/lbm · R
+c_v=0.172;// Btu/lbm · R
+
+// Solution
+// (a)
+deltau=c_v*((T_2+459.67)-(T_1+459.67));// Btu/lbm
+deltah=c_p*(T_2-T_1);// Btu/lbm
+printf('\n(a)The change in specific internal energy,u_2-u_1=%2.1f Btu/lbm \n   The change in specific enthalpy,h_2-h_1=%2.1f kJ/kg',deltau,deltah);
+// (b)
+// Values for u and h for variable specific heat air can be found in Table C.16.
+T_1=T_1+459.67;// R
+h_1=167.56;// Btu/lbm
+u_1=119.58;// Btu/lbm
+T_2=T_2+459.67;// R
+h_2=129.06;// Btu/lbm
+u_2=92.04;// Btu/lbm
+deltau=u_2-u_1;// Btu/lbm
+deltah=h_2-h_1;// Btu/lbm
+printf('\n(b)The change in specific internal energy,u_2-u_1=%2.1f Btu/lbm \n   The change in specific enthalpy,h_2-h_1=%2.1f kJ/kg',deltau,deltah);