Solving a Heat Transfer problem by using Finite Difference Method (FDM) in Matlab

Consider a large Uranium Plate of thickness, L=4 cm and thermal conductivity, k=28 W/m.Degree Cel in which Heat is generated uniformly at constant rate of Hg=5x10^6 W/m^3. One side of the plate is maintained at 0 Degree Cel by iced water while the other side is subjeted to convection to an environmet at 30 Degree Cel with a heat transfer coefficient of h=45 W/m^2.Degree Cel. Considering a total of 5 equally spaced nodes in the medium,two at the boundaries and three at the middle, estimate the surface temperatures of the plate at each nodes under steady conditions using the FDM and plot the curve showing "Temperature wrt Thickness of the plates"

close all
clear all
clc

%Inputs%

L_total=.04;                %Total Thickness in Meter
n=5;                        %Number of nodes
T(1)=0;                     %Initial Temperature (@ L=0) in Celsius
T(n+1)=30;                  %Ambient Temperature in Celsius
Hg=5000000;                 %Heat Generation in W/m3
K=28;                       %Thermal Conductivity in W/m-C
h=45;                       %Convective HT Coeff in W/m2-C
m=1000;                     %Number of iterations

dX=L_total/(n-1);           %Length of the Divisions in meter

for i=1:n                        
    T(i)=0;       %Necessary condition of initialization to avoid random Temp value
    
    for j=0:m     %1000 no. of iterations for accuracy
        
        for i=2:n-1               
            T(i)=(((Hg*dX*dX)/(2*K))+((T(i-1)+T(i+1))/2));    
            %Upto node 4 i.e. L=0.03 m
        end
            T(n)=((((K*T(n-1))/dX)+(h*T(n+1))+(Hg*(dX/2)))*(dX/(K+(dX*h))));   
            %@ node 5 i.e end of L=.04 m
    end   
end

    L=0:.01:.05              %Thickness array @ which Temperatures to be measured 
    
    %Output%
    
    disp(T);                 %Displaying all the Temperatures
    
    %Plotting%
    
    figure(1)
    plot(L,T,'x','color','b','MarkerSize', 5,'linewidth',5)
    xlim([0 .06])
    ylim([0 140])
    xlabel('Thickness of the Plate')
    ylabel('Temperature')
    title('Temp vs Thickness (POINTS)')
    
    figure(2)
    plot(L,T,'color','r','linewidth',2)
    axis([0 0.06 0 140])
    xlabel('Thickness  of the Plate')
    ylabel('Temperature')
    title('Temp vs Thickness (CURVE)')

RESULTS OBTAINED:

L = 0 0.0100 0.0200 0.0300 0.0400 0.0500

T = 0 60.7958 103.7344 128.8159 136.0403 30.0000

PLOTS OBTAINED: