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+clear ;
+clc;
+// Example 23.3
+printf('Example 23.3\n\n');
+// Page no. 693
+// Solution
+
+// Given
+// Heat capacity = 2.675*10^4 + (42.27)Tk - 1.425*10^-2Tk^2 J/(kg mol K)
+// First convert heat capacity to Btu/(lb mol*F)  to get c + dT - eT^2, where
+c = (2.675*10^4*.4536)/(1055*1.8) ;
+d = (42.27*.4536)/(1055*1.8) ;
+e = (1.425*10^-2*.4536)/(1055*1.8) ;
+
+//Now convert Tk (Temperature in K) to TF (temperature in F) to get answer of form x + yT - zT^2,where
+x = c + d*460/1.8 - e*((460/1.8)^2) ;
+y = d/1.8;
+z = e/(1.8*1.8) ;
+
+printf('The required answer is %.2e + (%.2e)T - (%.3e) T^2 Btu/(lb mol*F) , where T is in degree F .  \n',x,y,z);
+
+// Note answer in textbook seems wrong by order of 10^-3
+\ No newline at end of file