Answer:
[tex]K_B=2.49\ W/m.K[/tex]
Explanation:
We know that
Thermal resistance due to conductivity
[tex]R=\dfrac{L}{KA}[/tex]
Thermal resistance due to heat transfer coefficient
[tex]R'=\dfrac{L}{KA}[/tex]
We know that heat transfer Q
Q=ΔT/Rth
By equating heat
[tex]\dfrac{600-22}{\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{L_C}{AK_C}}=\dfrac{800-22}{\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{L_C}{AK_C}+\dfrac{1}{Ah}}[/tex]
[tex]\dfrac{600-22}{\dfrac{0.35}{30}+\dfrac{0.25}{K_B}+\dfrac{0.25}{65}}=\dfrac{800-22}{\dfrac{0.35}{30}+\dfrac{0.25}{K_B}+\dfrac{0.25}{65}+\dfrac{1}{25}}[/tex]
Now by solving we get
[tex]K_B=2.49\ W/m.K[/tex]