A newly designed square semi conductor chip (k = 175 W/m∙K) has a width w = 9 mm on a side and a thickness t = 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 92 W are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between back and front surfaces?

__Hint: __Does this sound complicated to you? If you do your research, you will find that most FE Exam questions can be solved by using only one formula! You can pass the exam by just becoming familiar with the basic equations in each topic area. It's that simple. This is what our book is all about. Forget theory and over complications used by our professors. It's time to transition to real life. All of the equations are found in the FE Handbook. The secret to passing is Problem Recognition (literally).

Doesn't this problem sound like conduction? This is a basic heat transfer problem. What famous equation uses conduction? The essence of our book describes the importance of knowing the layout of the FE Handbook (since the FE Exam is an open book exam!). If you haven't already, download the FE Handbook from NCEES.org. You have to create an account first. Now, use the search function (Control -F) on the pdf to look up "conduction".

What equation do you see? Yes, it's that easy!

Solution: In the FE Handbook - you will find the equation:

The question is asking for the Temperature Difference, so let's rearrange the equation to find delta T.

The rest is just plug and chug. Be careful though, you need to convert all the mm to m.

ΔT=kw2t⋅P= 0.003m⋅ 92W

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175 W/m∙K (.009 m)^2

= 19.47 K