(i) Number of nodes and elements

(ii) Element matrix equation
{[hcA000]+kAhe[1−1−11]}{θ1θ2}={Q1Q2}
For element 1
hc = 30 W/m2; A = 1m2; K = 25; he = 0.3
∴KAhe=25×10.3=83.33
{[30000]+[83.33−83.33−83.3383.33]}{θ1θ2}={Q1Q2}
[113.33−83.33−83.3383.33]{θ1θ2}={Q1Q2}
For element 2
hc = 0; A = 1m2; K = 30; he = 0.2
∴KAhe=30×10.2=150
[150−150−150150]{θ2θ3}={Q2Q3}
For element 3
hc = 30 W/m2; A = 1m2; K = 70; he = 0.15
∴KAhe=70×10.15=466.67
{[00030]+[466.67−466.67−466.67466.67]}{θ3θ4}={Q3Q4}
[466.67−466.67−466.67466.67]{θ3θ4}={Q3Q4}
(iii) Global matrix equation
[113.33−83.3300−83.33233.33−15000−150616.67−466.6700−466.67496.67]{θ1θ2θ3θ4}={Q1Q2Q3Q4}
Imposing boundary condition,
θ1=800∘Cθ4=20∘Cθ2=θ3=0(Forbalancing)
[113.33−83.3300−83.33233.33−15000−150616.67−466.6700−466.67496.67]{800θ2θ320}={Q100Q4}
Now,
(113.33 x 800) - 83.33 θ2 = θ1 ...(1)
(-83.33 x 800) + 233.33 θ2 - 150 θ3 = 0 ...(2)
(-150 θ2 + 616.67 θ3 - 466.67 x 20) = 0 ...(3)
(-466.67 θ3 + 496.67 x 20) = θ4 ...(4)
By solving equation (2) and (3), we get