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A square silicon chip \((k=150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is of width \(w=5 \mathrm{~mm}\) on a side and of thickness \(t=1 \mathrm{~mm}\). 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 4 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?

Short Answer

Expert verified
The temperature difference is approximately 1.07 Kelvin.

Step by step solution

01

Understand the Problem

We need to find the temperature difference between the back and front surfaces of a silicon chip, where heat is dissipated through the front surface, and the other surfaces are insulated. A total of 4 W is dissipated.
02

Collect Given Information

- Thermal conductivity, \( k = 150 \, \text{W/m} \cdot \text{K} \)- Width and length, \( w = 5 \, \text{mm} = 0.005 \, \text{m} \)- Thickness, \( t = 1 \, \text{mm} = 0.001 \, \text{m} \)- Power dissipated, \( P = 4 \, \text{W} \)
03

Determine Area of Front Surface

Calculate the area of the front surface of the square chip: \[ A = w^2 = (0.005 \, \text{m})^2 = 0.000025 \, \text{m}^2 \]
04

Apply Fourier’s Law of Heat Conduction

Fourier's law states that the rate of heat transfer \( Q \) through an area is given by:\[ Q = -kA \frac{\Delta T}{t} \]Where:- \( A \) is the area of the surface,- \( \Delta T \) is the temperature difference between surfaces,- \( t \) is the thickness of the material.
05

Rearrange to Solve for Temperature Difference

Solve for \( \Delta T \):\[ \Delta T = \frac{Qt}{kA} \]
06

Insert Given Values and Solve

Insert the known values into the formula:\[ \Delta T = \frac{(4 \, \text{W})(0.001 \, \text{m})}{(150 \, \text{W/m} \cdot \text{K})(0.000025 \, \text{m}^2)} \]Calculate the result:\[ \Delta T = \frac{0.004}{0.00375} = 1.07 \, \text{K} \]
07

Interpret the Result

The steady-state temperature difference between the back and front surfaces of the silicon chip is approximately 1.07 Kelvin.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Conductivity
Thermal conductivity is a fundamental property that describes how well a material can conduct heat. In simpler terms, it indicates how easily heat can travel through a material.
Materials with high thermal conductivity, like metals, are great for transferring heat, making them ideal for applications requiring quick and efficient heat dispersal. On the other hand, materials with low thermal conductivity, such as wood or foam, are good insulators and impede the flow of heat.
The thermal conductivity of a material is denoted by the symbol \( k \) and measured in watts per meter-Kelvin (\( \text{W/m} \cdot \text{K} \)). For example, in this exercise, silicon has a thermal conductivity of \( 150 \, \text{W/m} \cdot \text{K} \), meaning it's quite effective at conducting heat compared to many other materials.
Understanding the thermal conductivity is crucial when designing systems where temperature control is imperative, such as in electronic components or thermal insulation.
Fourier's Law
Fourier's Law provides a mathematical framework for understanding heat conduction. It states that the rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area through which the heat is flowing. The law can be expressed with the equation: \[ Q = -kA \frac{\Delta T}{t} \]
In this expression,
  • \( Q \) represents the rate of heat transfer (in watts, W),
  • \( k \) is the thermal conductivity of the material,
  • \( A \) is the surface area through which the heat is passing,
  • \( \Delta T \) is the temperature difference between two surfaces, and
  • \( t \) is the thickness of the material through which heat is being conducted.
In situation where heat is dissipated from a solid to a surrounding fluid, understanding this relationship helps engineers and scientists calculate how effectively a material can transfer heat across surfaces. By rearranging this law to solve for different variables, like temperature difference, as done in the original exercise, you can make key design decisions or diagnose thermal performance issues.
Steady-State Temperature Difference
The concept of a steady-state temperature difference involves understanding how temperatures stabilize over time in a thermal system. In a steady-state scenario, the temperature distribution does not change as time progresses. This means the heat entering and exiting a system remains balanced.
In the exercise, for instance, the silicon chip reached a steady-state temperature difference of approximately 1.07 Kelvin between its back and front surfaces. At this point, the heat generated from the chip's electronic circuits is being dissipated to the coolant at a constant rate, leading to a stable temperature profile.
Reaching a steady-state is crucial in applications where continuous and predictable temperature behavior is necessary for the operation of devices or systems. It creates conditions where you can confidently predict the thermal behavior of a system and ensure that temperatures remain within safe operational limits. Understanding steady-state conditions is a fundamental aspect of designing effective thermal management systems.

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Most popular questions from this chapter

The inner and outer surface temperatures of a glass window \(5 \mathrm{~mm}\) thick are 15 and \(5^{\circ} \mathrm{C}\). What is the heat loss through a window that is \(1 \mathrm{~m}\) by \(3 \mathrm{~m}\) on a side? The thermal conductivity of glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Air at \(40^{\circ} \mathrm{C}\) flows over a long, 25-mm-diameter cylinder with an embedded electrical heater. In a series of tests, measurements were made of the power per unit length, \(P^{t}\), required to maintain the cylinder surface temperature at \(300^{\circ} \mathrm{C}\) for different freestream velocities \(V\) of the air. The results are as follows: \begin{tabular}{lccccc} \hline Air velocity, \(V(\mathrm{~m} / \mathrm{s})\) & 1 & 2 & 4 & 8 & 12 \\ Power, \(P^{\prime}(\mathrm{W} / \mathrm{m})\) & 450 & 658 & 983 & 1507 & 1963 \\\ \hline \end{tabular} (a) Determine the convection coefficient for cach velocity, and display your results graphically. (b) Assuming the dependence of the convection coeffcient on the velucity to be of the form \(h=C V^{\prime \prime}\), determine the parameters \(C\) and \(n\) from the results of part (a).

A freezer compartment consists of a cubical cavity that is \(2 \mathrm{~m}\) on a side. Assume the bottom to be perfectly insulated. What is the minimum thickness of styrofoam insulation \((k=0.030 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that muse be applied to the top and side walls to ensure a heat loed of less than \(500 \mathrm{~W}\). when the inner and outer surfaces are \(-10\) and \(35^{\circ} \mathrm{C}\) ?

An inexpensive food and beverage container is fabricuted from 25-mm-thick polystyrene \((k=0.023 \mathrm{~W} / \mathrm{m}-\mathrm{K})\) and has interior dimensions of \(0.8 \mathrm{~m} \times 0.6 \mathrm{~m} \times 0.6 \mathrm{~m}\). Under conditions for which an inner surface temperature of approximately \(2^{*} \mathrm{C}\) is maintained by an ice-water mixture and an outer surface temperature of \(20^{\circ} \mathrm{C}\) is maintained by the ambient, what is the heat flux through the container wall? Assuming negligible heat gain through the \(0.8 \mathrm{~m} \times\) \(0.6 \mathrm{~m}\) base of the cooler, what is the total heat load for the prescribed conditions?

Consider a surface-mount type transistor on a circuit board whose temperature is muintained at \(35^{\circ} \mathrm{C}\). Air at \(207 \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm}\) by \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm}\) by \(0.25 \mathrm{~mm}\) and length 4 nim, conduct heat from the case to the circuit board. The gap between the case and the board is \(0.2 \mathrm{~mm}\). (a) Assuming the case is isothermal and neglecting rudiation, eximate the case temperature uhen \(150 \mathrm{~mW}\) is dissipoted by the thansistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal condoctivitios of the wire leadk, air, and cunductive paste are 25, \(0.0263\), and \(0.12\) W/m - \(\mathrm{K}\), respectively. (b) Using the conductive paste to fill the gap, we wish to determine the extent to which increased heat dissipation may be accommodated, subject to the constraint that the case lemperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a lager convection coefficient \(h\) and/or changing the lead wire material to one of larger thermal conductivity. Independently considering leads fabricated from materials with thermal conductivities of 200 and \(400 \mathrm{~W} / \mathrm{m}+\mathrm{K}\), compute and plot the maximum allowable hean dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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