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Judging from its unit \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\), can we define thermal conductivity of a material as the rate of heat transfer through the material per unit thickness per unit temperature difference? Explain.

Short Answer

Expert verified
Answer: Yes, based on Fourier's Law and the relationship between thermal conductivity, heat transfer rate, material thickness, and temperature gradient, the definition of thermal conductivity as the rate of heat transfer through the material per unit thickness per unit temperature difference is accurate, as it matches the units (\(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\)).

Step by step solution

01

Understand Thermal Conductivity

Thermal conductivity, denoted by the symbol \(k\), is a property of a material that indicates its ability to conduct heat. It is expressed in \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\). A material with a higher thermal conductivity can transfer heat more efficiently than a material with a lower thermal conductivity.
02

Relate Thermal Conductivity to Heat Transfer

The rate of heat transfer through a material can be described by Fourier's Law of heat conduction, which states that the heat transfer rate, \(q\), is proportional to the product of the thermal conductivity (\(k\)), the cross-sectional area of the material (\(A\)), the temperature gradient (temperature difference per unit thickness, \(\Delta T / d\)): $$q = kA\frac{\Delta T}{d}$$ Where \(d\) represents the thickness of the material.
03

Evaluate the Temperature Gradient

In Fourier's Law, the temperature gradient (\(\Delta T / d)\) represents the change in temperature with respect to the thickness of the material. It is expressed in units of \(\mathrm{K} / \mathrm{m}\). The temperature gradient plays a crucial role in determining the heat transfer rate.
04

Confirm the Definition

Given the relationship between the rate of heat transfer, thermal conductivity, cross-sectional area, and temperature gradient as per Fourier's Law, we can rewrite the equation and isolate thermal conductivity (\(k\)): $$k = \frac{q}{A}\frac{d}{\Delta T}$$ Since \(k\) is expressed in the same units as \(\frac{q}{A}\frac{d}{\Delta T}\), which are \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\), we can confirm that the definition of thermal conductivity as the rate of heat transfer through the material per unit thickness per unit temperature difference is accurate based on its units.

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Key Concepts

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

Heat Transfer
Heat transfer is the process through which thermal energy moves from one substance to another. It can occur in three primary ways: conduction, convection, and radiation. In the context of this exercise, we are focusing on conduction. This is when heat moves directly through a substance or between substances that are in direct contact. Conduction happens because particles in hotter regions vibrate more energetically and transfer energy to neighboring cooler particles.

Several factors affect how efficiently heat is transferred:
  • Material properties: Some materials, like metals, conduct heat very well, while others, like wood, do not.
  • Cross-sectional area: The larger the area, the more heat that can be transferred.
  • Temperature difference: A greater difference increases the rate of transfer.
  • Distance: The thickness of the material affects how easily heat can travel through it.
Understanding heat transfer is crucial in designing systems for heating, cooling, and insulating materials.
Fourier's Law
Fourier's Law is fundamental in describing how heat conduction occurs through a material. Named after the French physicist Jean-Baptiste Joseph Fourier, this law states that the heat transfer rate through a material is proportional to the negative gradient of temperatures and the material's cross-sectional area. This law can be mathematically expressed as:\[ q = -kA \frac{dT}{dx} \]Where:
  • \(q\): Heat transfer rate, measured in watts (\(\mathrm{W}\)).
  • \(k\): Thermal conductivity, characteristic of the material.
  • \(A\): Cross-sectional area perpendicular to heat flow.
  • \(dT/dx\): Temperature gradient along the material's direction.
The law highlights how thermal energy flow is affected by material properties and temperature changes. It's essential in engineering fields where temperature control is vital, such as in designing buildings, engines, and electronic devices.
Temperature Gradient
The temperature gradient is a crucial component in the study of heat conduction and plays a pivotal role in Fourier's Law. It describes how temperature changes along the length of a material. The gradient is defined as the difference in temperature per unit distance, typically expressed in \(\mathrm{K}/\mathrm{m}\).

The significance of the temperature gradient lies in its direct influence on the rate of heat transfer. A steeper gradient—where temperature changes more rapidly over a shorter distance—means heat flows faster. This happens because heat naturally moves from hotter to cooler areas to equalize temperatures.

Factors affecting the temperature gradient include:
  • Material properties: Materials with high thermal conductivity have less steep gradients compared to insulating materials.
  • Environmental conditions: The surrounding temperature can affect how steep the gradient is.
  • Initial temperature distribution: How heat is initially spread inside the material also impacts the gradient.
Being able to manipulate and understand temperature gradients is key for efficient thermal management, especially in specialized applications like electronics cooling and climate control systems.

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Most popular questions from this chapter

Using the parametric table and plot features of \(\mathrm{EES}\), determine the squares of the number from 1 to 100 in increments of 10 in tabular form, and plot the results.

Four power transistors, each dissipating \(12 \mathrm{~W}\), are mounted on a thin vertical aluminum plate \(22 \mathrm{~cm} \times 22 \mathrm{~cm}\) in size. The heat generated by the transistors is to be dissipated by both surfaces of the plate to the surrounding air at \(25^{\circ} \mathrm{C}\), which is blown over the plate by a fan. The entire plate can be assumed to be nearly isothermal, and the exposed surface area of the transistor can be taken to be equal to its base area. If the average convection heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the aluminum plate. Disregard any radiation effects.

An electronic package in the shape of a sphere with an outer diameter of \(100 \mathrm{~mm}\) is placed in a large laboratory room. The surface emissivity of the package can assume three different values \((0.2,0.25\), and \(0.3)\). The walls of the room are maintained at a constant temperature of \(77 \mathrm{~K}\). The electronics in this package can only operate in the surface temperature range of \(40^{\circ} \mathrm{C} \leq T_{s} \leq 85^{\circ} \mathrm{C}\). Determine the range of power dissipation \((\dot{W})\) for the electronic package over this temperature range for the three surface emissivity values \((\varepsilon)\). Plot the results in terms of \(\dot{W}(\mathrm{~W})\) vs. \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) for the three different values of emissivity over a surface temperature range of 40 to \(85^{\circ} \mathrm{C}\) with temperature increments of \(5^{\circ} \mathrm{C}\) (total of 10 data points for each \(\varepsilon\) value). Provide a computer generated graph for the display of your results and tabulate the data used for the graph. Comment on the results obtained.

Consider two houses that are identical, except that the walls are built using bricks in one house, and wood in the other. If the walls of the brick house are twice as thick, which house do you think will be more energy efficient?

A cylindrical resistor element on a circuit board dissipates \(1.2 \mathrm{~W}\) of power. The resistor is \(2 \mathrm{~cm}\) long, and has a diameter of \(0.4 \mathrm{~cm}\). Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a 24-hour period, \((b)\) the heat flux, and \((c)\) the fraction of heat dissipated from the top and bottom surfaces.

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