Answer:
1.1⁰C
Explanation:
Width W = 5mm = 0.005
Thickness t = 1 mm = 0.001
K = thermal conductivity = 150W/m.K
P = q = heat transfer rate = 4W
We are to find the steady state temperature between the back and the front surface
We have to make these assumptions:
1. There is steady state conduction
2. The heat flow is of one dimension
3. The thermal conductivity is constant
4. The heat dissipation is uniform
We have:
∆t = t*P/k*W²
= (0.001m x 4W)/150x(0.005)²
= 0.004/0.00375
= 1.06667
This is approximately,
1.1⁰C
Thank you!
The awnser is D cause it is
Answer:
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Good luck friend :)
