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clear;
clc;
disp('Example 1.5');
// Given values
m = 5; // mass, [kg]
t1 = 15; // inital temperature, [C]
t2 = 100; // final temperature, [C]
c = 450; // specific heat capacity, [J/kg K]
// solution
// using heat transfer equation,[1]
Q = m*c*(t2-t1); // [J]
mprintf('\n The heat required is = %f kJ\n',Q*10^-3);
//End
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