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Answer:
t₁ = 378.668 K
Explanation:
From the given information:
The heat dissipated (q) from the electrical device = [tex]10^4 W/m^2[/tex]
The contact resistance between device and aluminium;
(R_{const} ) = 0.5× 10⁻⁴ m² K/W
Conductivity of aluminum (k) = 238 \ W/mK
The thickness of aluminum plate (L) = 0.004 m
Convection coefficient (h) = 100 \ W/m^2 K
Surrounding temperature (t_2) = 278 \ k
According to Fourier's law of heat conduction.
[tex]q = \dfrac { t_1- t_2 }{R_{const} + \dfrac{L}{k} + \dfrac{1}{h} }[/tex]
[tex]10^4 = \dfrac{t_1 - 278}{0.5 \times 10^{-4} + \dfrac{0.004}{238} + \dfrac{1}{100} }[/tex]
[tex]10^4 *( 0.5 \times 10^{-4} + \dfrac{0.004}{238} + \dfrac{1}{100}) = t_1 - 278[/tex]
[tex]100.668 = t_1 -278[/tex]
t₁ = 378.668 K