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clear;
clc;
disp('Example 1.6');
// Given values
m_cop = 2; // mass of copper vessel, [kg]
m_wat = 6; // mass of water, [kg]
c_wat = 4.19; // specific heat capacity of water, [kJ/kg K]
t1 = 20; // initial temperature, [C]
t2 = 90; // final temperature, [C]
// From the table of average specific heat capacities
c_cop = .390; // specific heat capacity of copper,[kJ/kg k]
// solution
Q_cop = m_cop*c_cop*(t2-t1); // heat required by copper vessel, [kJ]
Q_wat = m_wat*c_wat*(t2-t1); // heat required by water, [kJ]
// since there is no heat loss,so total heat transfer is sum of both
Q_total = Q_cop+Q_wat ; // [kJ]
mprintf(' \n Required heat transfer to accomplish the change = %f kJ\n',Q_total);
//End
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