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authorpriyanka2015-06-24 15:03:17 +0530
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+clear;
+clc;
+
+//Example 1.1(Heating of a copper ball)
+
+//(a)
+//density of the copper ball
+rho= 8950;//[kg/m^3]
+//Diameter of the copper ball
+D=0.1;//[m]
+//mass of the ball
+m=rho*(%pi/6)*(D^3);//[kg]
+//Specific Heat of copper
+Cp=0.395;//[kJ/Kg/m^3]
+//Initial Temperature
+T1=100;//[degree C]
+//Final Temperature
+T2=150;//[degree C]
+// The amount of heat transferred to the copper ball is simply the change in it's internal energy and is given by
+// Energy transfer to the system=Energy increase of the system
+Q=(m*Cp*(T2-T1));
+disp("kJ",Q,"Heat needs to be transferred to the copper ball to heat it from 100 to 150 degree celsius is ")
+//b
+//Time interval for which the ball is heated
+dT=1800;//[seconds]
+Qavg=(Q/dT)*1000;//[W]
+disp("W",Qavg,"Average Heat Transfer by the iron ball is ")
+
+//(c)
+//Heat Flux
+qavg=(Qavg/(%pi*(D^2)));//[W/m^2]
+disp("W/m^2",qavg,"Average flux is")