A CPU chip with a footprint of 2.5 cm by 2 cm is mounted on a circuit board. The chip rejects heat to the environment at 28oC by convection and radiation. The outer casing of the chip has an emissivity of 0.88 and the heat transfer coefficient is 40 W/m2·K. If the chip temperature is 82 oC, calculate the heat transfer rate at the outer surface of the chip. Neglect the thickness of the chip and any conduction into the circuit board

The heat transfer rate is 0.9504 W

Explanation:

Heat transfer rate (Q) = ekA∆T

e is emissivity of the chip = 0.88

k is heat transfer coefficient = 40 W/m^2.K

A is surface area of chip = 2.5 cm × 2 cm = 5 cm^2 = 5/10000 = 5×10^-4 m^2

∆T is change in temperature = 355 - 301 = 54 K

Q = 0.88 × 40 × 5×10^-4 × 54 = 0.9504 W

answer:

d. attempt to find the owner, then leave your information if you cannot find them.

explanation:

d. represents both (b) and (c).

more logical as well.

b and c are to complicated.

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