Answer:
The answer is below
Explanation:
Given that:
x = 5 cm = 0.05 m, width (w) = 5 cm = 0.05 m, thickness (t) = 1 mm = 0.001 m, k = thermal conductivity = 200 W/mK, h = heat transfer coefficient = 22 W/m².K, Tb = base temperature = 40°C, T∞ = 20°C
[tex]m=\sqrt{ \frac{hp}{kA} }=\sqrt {\frac{22*(2*0.05+2*0.001)}{200*(0.05*0.001)}} =15\ m^{-1}\\\\\frac{T-T_{\infty}}{T_b-T_{\infty}} =e^{-mx}\\\\\frac{T-20}{40-20} =e^{-15*0.05}\\\\T-20=9.45\\\\T=29.45^oC\\\\The\ rate\ of\ heat\ loss\ is:\\\\\dot{Q}=\sqrt{hpkA}(T_b-T_{\infty})\\ \\\dot{Q}=\sqrt{22*(2*0.05+2*0.001)*200*0.05*0.001}*(40-20)\\\\\dot{Q}=3\ W[/tex]
