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//molar heat of vaporisation and overall energy required
clear;
clc;
printf("\t Example 11.8\n");
m=346;//mass of H2O in g
//from 0 to 100 C
s=4.184;//specific heat of H2O, J/g C
deltaT=100-0;//change in Temp, C
q1=m*s*deltaT/1000;//heating H2O, kJ
//for evaporation at 100 C
deltaH=40.79;//heat of vaporisation in kJ
H2O=18.02;//mol mass of H2O, g
q2=m*deltaH/H2O;//heat of vaporising water, kJ
//for steam from 100 to 182 C
deltaT=182-100;//change in temp of steam, kJ
s=1.99;//specific heat of steam, J/g C
q3=m*s*deltaT/1000;//heating steam, kJ
q=q1+q2+q3;//overall energy required, kJ
printf("\t the overall energy required is : %4.0f kJ\n",q);
//End
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