Respuesta :
Answer:
A) 0.0325 W
B)21.667 W
C)1300 W
Explanation:
A)
THERMAL CONDUCTIVITY OF AIR IS K 0.026 w/m-k
Thermal resistance is given as
[tex]Rth = \frac{L}{KA}[/tex]
and [tex]\Delta T = Q_1 \times Rth[/tex]
therefore we have
[tex]305 - 295 = q_1 \times \frac{0.8\times 10^{-3}}{0.026\times 10\times 10 \times 10^{-6}}[/tex]
[tex]q_1 = 0.0325 w[/tex]
B) WHEN [tex]L = 1.2 \mu m[/tex]
therefore we have
[tex]305 - 295 = q_2 \times \frac{1.2\times 10^{-6}}{0.026\times 10\times 10 \times 10^{-6}}[/tex]
[tex]q_2 = 21.667 w[/tex]
C) When L is 20 nm
therefore we have[tex]305 - 295 = q_1 \times \frac{20\times 10^{-9}}{0.026\times 10\times 10 \times 10^{-6}}[/tex]
[tex]q_1 = 1300 w[/tex]
