Answer:
The correct answer is part 'c': 6.4 W/m-K
Explanation:
We know that the thermal resistance of a wall is given by
[tex]R_{wall}=\frac{L}{kA}[/tex]
where,
L = thickness of wall
k = thermal conductivity of wall
A = Surface area of wall
Applying values we get
[tex]k=\frac{L}{R_{wall}A}[/tex]
[tex]k=\frac{80\times 10^{-2}}{0.0125\times 2.5\times 4}\\\\\therefore k=6.4W/mK[/tex]
