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Chapter 1: Problem 34

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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Most popular questions from this chapter

A thin metal plate is insulated on the back and exposed to solar radiation on the front surface. The exposed surface of the plate has an absorptivity of \(0.7\) for solar radiation. If solar radiation is incident on the plate at a rate of \(550 \mathrm{~W} / \mathrm{m}^{2}\) and the surrounding air temperature is \(10^{\circ} \mathrm{C}\), determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Take the convection heat transfer coefficient to be \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and disregard any heat loss by radiation.

An ice skating rink is located in a building where the air is at \(T_{\text {air }}=20^{\circ} \mathrm{C}\) and the walls are at \(T_{w}=25^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\) and its density is \(920 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?

Solve this system of three equations with three unknowns using EES: $$ \begin{array}{r} x^{2} y-z=1.5 \\ x-3 y^{0.5}+x z=-2 \\ x+y-z=4.2 \end{array} $$

Consider a sealed 20-cm-high electronic box whose base dimensions are \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\) placed in a vacuum chamber. The emissivity of the outer surface of the box is \(0.95\). If the electronic components in the box dissipate a total of \(120 \mathrm{~W}\) of power and the outer surface temperature of the box is not to exceed \(55^{\circ} \mathrm{C}\), determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible.

Consider a flat-plate solar collector placed on the roof of a house. The temperatures at the inner and outer surfaces of the glass cover are measured to be \(33^{\circ} \mathrm{C}\) and \(31^{\circ} \mathrm{C}\), respectively. The glass cover has a surface area of \(2.5 \mathrm{~m}^{2}\), a thickness of \(0.6 \mathrm{~cm}\), and a thermal conductivity of \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is lost from the outer surface of the cover by convection and radiation with a convection heat transfer coefficient of \(10 \mathrm{~W} /\) \(\mathrm{m}^{2} \cdot \mathrm{K}\) and an ambient temperature of \(15^{\circ} \mathrm{C}\). Determine the fraction of heat lost from the glass cover by radiation.
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