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clear ;
clc;
// Example 4.2
printf('Example 4.2\n\n');
// Page no. 93
// Solution
// Given
// Heat capacity = 139.1 + (1.56*10^-1)Tc J/(g mol* degree C), T is in degree C
// First convert Tc (Temperature in degree celsius) to TR (in degree R) to get c + dTR, where
c = 139.1 + (1.56*10^-1)*(-460-32)/1.8 ;
d = (1.56*10^-1)/1.8;
//Now convert c +dTR to (Btu/lb mol*degree R) to get answer of form a + bTR,where
a = c*(454/(1055*1.8)) ;
b = d*(454/(1055*1.8)) ;
printf('The required answer is %.2f + (%.2e)T Btu/(lb mol*degree R) , where T is in degree R . \n',a,b);
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