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//developed in windows 8 operating system 64bit
//platform Scilab 5.4.1
//example 25_1w
clc;clear;
//Given Data
mass_ice=1; //Mass of ice (Unit:kg)
init_temp=-10; //Temperature of the ice (Unit : degree centigrade)
final_temp=100; //Temperature of the steam (Unit : degree centigrade)
sp_heat_ice=2100; //Specific heat capacity of ice (Unit: J/kg-K)
latent_heat_ice=3.36*10^5; //Latent heat of fusion of ice (Unit: J/Kg)
sp_heat_water=4200; //Specific heat capacity of water (Unit: J/kg-K)
latent_heat_water=2.25*10^6; //Latent heat of vapourization of water (Unit: J/Kg)
//Calculation
heat_ice_to_0_degree=mass_ice*sp_heat_ice*(0-(init_temp)); //Heat required to take the ice from -10 degree centigrade to 0 degree centigrade (Unit : Joules)
heat_req_melt=mass_ice*latent_heat_ice; //Heat required to melt the ice at o degree centigrade to water (Unit: Joules)
heat_req_zero_to_100=mass_ice*sp_heat_water*(final_temp-0); //Heat required to take the water from 0 degree centigrade to 100 degree centigrade (Unit : Joules)
heat_req_steam=mass_ice*latent_heat_water; //Heat required to convert 1 kg of water at 100 degree centigrade into steam
total_heat=heat_ice_to_0_degree+heat_req_melt+heat_req_zero_to_100+heat_req_steam; //Total heat required (Unit: Joules)
disp(total_heat,"Total Heat required to convert 1 kg of ice at -10 degree centigrade into steam at 100 degree centigrade is (Unit : Joules)");
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