Answer:
a)[tex]q=14.5\ W/m^2[/tex]
b)Q= 58 W
Explanation:
Given that
Thermal conductivity ,K = 0.029 W/m.k
The temperature difference ,ΔT= 10°C
The thickness ,L = 20 mm
We know that
[tex]Q=\dfrac{KA}{L}\times \Delta T[/tex]
Now by putting the values
[tex]Q=\dfrac{0.029\times 4}{0.02}\times 10\ W[/tex]
Q= 58 W
The heat flux through the sheet is given as
[tex]q=\dfrac{Q}{A}\ W/m^2[/tex]
[tex]q=\dfrac{58}{2\times 2}\ W/m^2[/tex]
[tex]q=14.5\ W/m^2[/tex]
a)[tex]q=14.5\ W/m^2[/tex]
b)Q= 58 W
