A square silicon chip (k = 152 W/m·K) is of width 7 mm on a side and of thickness 3 mm. The chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If 6 W are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between the back and front surfaces?

The steady-state temperature difference is 2.42 K

Explanation:

Rate of heat transfer = kA∆T/t

Rate of heat transfer = 6 W

k is the heat transfer coefficient = 152 W/m.K

A is the area of the square silicon = width^2 = (7/1000)^2 = 4.9×10^-5 m^2

t is the thickness of the silicon = 3 mm = 3/1000 = 0.003 m

6 = 152×4.9×10^-5×∆T/0.003

∆T = 6×0.003/152×4.9×10^-5 = 2.42 K

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answer: 5

explanation:

6-1