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clear;
clc;
disp('Example 1.8');
// Given values
m = 4; // mass of the liquid, [kg]
t1 = 15; // initial temperature, [C]
t2 = 100; // final temperature, [C]
Q = 714; // [kJ],required heat to accomplish this change
// solution
// using heat equation
// Q=m*c*(t2-t1)
// calculation of c
c=Q/(m*(t2-t1)); // heat capacity, [kJ/kg K]
mprintf('\n The specific heat capacity of the liquid is c = %f kJ/kg K\n',c);
//End
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