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The heat flux through a wood slab \(50 \mathrm{~mm}\) thick, whose inner and outer surface temperatures are 40 and \(20^{\circ} \mathrm{C}\), respectively, has been determined to be \(40 \mathrm{~W} / \mathrm{m}^{2}\). What is the thermal conductivity of the wood?

Short Answer

Expert verified
The thermal conductivity of the wood is \(0.1 \, \text{W/m} \cdot ^\circ \text{C}\).

Step by step solution

01

Identify Given Data

We are given the following information:- Thickness of the wood slab, \( L = 50 \, \text{mm} = 0.05 \, \text{m} \) (converted from mm to meters)- Inner surface temperature, \( T_1 = 40^{\circ} \text{C} \)- Outer surface temperature, \( T_2 = 20^{\circ} \text{C} \)- Heat flux, \( q = 40 \, \text{W/m}^2 \)
02

Write the Heat Flux Formula

The formula to calculate heat flux through a flat surface is given by Fourier's law of heat conduction, which is: \[ q = k \cdot \frac{(T_1 - T_2)}{L} \]Where:- \( q \) is the heat flux,- \( k \) is the thermal conductivity,- \( T_1 - T_2 \) is the temperature difference between surfaces,- \( L \) is the thickness of the material.
03

Calculate the Temperature Difference

Compute the temperature difference between the inner and outer surfaces of the wood slab:\[ T_1 - T_2 = 40^{\circ} \text{C} - 20^{\circ} \text{C} = 20^{\circ} \text{C} \]
04

Rearrange the Formula to Solve for Thermal Conductivity

Since we need to find the thermal conductivity \( k \), rearrange Fourier’s law formula to solve for \( k \):\[ k = q \cdot \frac{L}{(T_1 - T_2)} \]
05

Substitute Known Values and Solve

Insert the known values into the rearranged formula to calculate \( k \):\[ k = 40 \, \text{W/m}^2 \cdot \frac{0.05 \, \text{m}}{20 \, ^\circ \text{C}} \]Calculate the result:\[ k = 0.1 \, \text{W/m} \, ^\circ \text{C} \]

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Key Concepts

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

Thermal Conductivity
Understanding thermal conductivity is essential when analyzing heat transfer within materials. It measures how well a material can conduct heat. This property is crucial, as different materials will either retain heat or allow it to pass through more freely.
To visualize this, consider a winter day when you touch a metal and a wooden object. Even if both are at the same temperature, metal feels colder. This difference is due to metal's higher thermal conductivity compared to wood. Metals rapidly transfer heat away from your hand, giving a sensation of cold. Wood, with lower thermal conductivity, transfers heat less efficiently, feeling warmer by comparison.
In mathematical terms, thermal conductivity, denoted by the symbol \( k \), is determined through the formula derived from Fourier's Law. This formula provides us with a concrete method to quantify how well a specific material transfers heat under a known temperature gradient and thickness. Thus, knowing thermal conductivity is essential for designing and evaluating materials in various applications.
Fourier's Law
Fourier's Law is fundamental for understanding heat conduction. It describes how heat flows through a material, governed by the temperature gradient and the thermal conductivity of the material.
The law is expressed through the formula:
  • \( q = k \times \frac{(T_1 - T_2)}{L} \)
  • Where \( q \) is the heat flux, \( k \) is thermal conductivity, \( T_1 - T_2 \) is the temperature difference, and \( L \) is the material's thickness.

This equation tells us that heat will flow more easily if there is a large temperature difference or if the material has high thermal conductivity. Conversely, thicker materials or small temperature differences result in less heat transfer.
By rearranging the formula, we can also determine unknown quantities such as thermal conductivity when the other variables are known. Fourier's Law thus provides a framework for making predictions about heat transfer in a variety of situations, from engine components to building materials.
Heat Flux
Heat flux is the rate of heat energy transfer through a given surface per unit area. It is an essential parameter in thermal analysis, allowing us to quantify how much heat passes through a surface.
The unit of heat flux is \( ext{W/m}^2 \), signifying watts (energy per time) per square meter (area). When evaluating materials, knowing the heat flux can help determine whether a material can effectively manage thermal loads.
For example, in the exercise we examined, the heat flux value was given as \(40 ext{ W/m}^2 \). This indicates a moderate level of heat transfer through the wood slab, which helped us determine the slab's thermal conductivity. By understanding heat flux, we can assess the thermal efficiency and suitability of materials for different thermal management applications.

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Most popular questions from this chapter

A freezer compartment is covered with a \(2-\mathrm{mm}\)-thick layer of frost at the time it malfunctions. If the compartment is in ambient air at \(20^{\circ} \mathrm{C}\) and a coefficient of \(h=2\) \(\mathrm{W} / \mathrm{m}^{2}\) - \(\mathrm{K}\) characterizes heat transfer by natural convection from the exposed surface of the layer, estimate the time required to completely melt the frost. The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{k} / / \mathrm{kg}\).

The clectrical-substitation radiometer shown schematically determines the optical (radiant) power of a beam by measuring the electrical power required to heat the receiver to the same temperature. With a beam, such as a luser of optical power \(P_{\text {epv }}\) incident on the receiver, its temperature, \(T_{p}\) increases above that of the chamber walls held al a uniform temperature, \(T_{\text {we }}=77 \mathrm{~K}\), With the optical beam blocked, the heater on the hackside of the receiver is encrgized and the electrical power, \(P_{\text {eine: }}\) required to reach the same value of \(T_{\text {, is measured. The }}\) purpose of your analysis is to determine the relationship tetween the clectrical and optical power, considering heat transfer processes experienced by the receiver. Consider a radiomeler with a \(15-\mathrm{mm}\)-diameter receiver having a blackened surface with an emissivity of \(0.95\) and an absorptivity of \(0.98\) for the optical beam. When operating in the optical mode, conduction heat losses from the backside of the receiver are negligible. In the electricul mode, the loss amounts to \(5 \%\) of the electrical power. What is the optical power of a beam when the indicated electrical power is \(20.64 \mathrm{~mW}\) ? What is the corresponding recciver temperature?

During its manufacture, plate glass at \(600^{\circ} \mathrm{C}\) is cooled by passing air over its surface such that the convection heat transfer coefficient is \(h=5 \mathrm{~W} / \mathrm{ma}^{2}+\mathrm{K}\). To prevent cracking, it is known that the temperature gradient must not exceed \(15^{\circ} \mathrm{C} / \mathrm{mm}\) at any point in the glass during the cooling process. If the thermal conductivity of the glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its surface emissivity is \(0.8\), what is the lowest temperature of the air that can initially be used for the cooling? Assume that the temperature of the air equals that of the surroundings.

The concrete slab of a basement is \(11 \mathrm{~m}\) long. \(8 \mathrm{~m}\) wide, and \(0.20 \mathrm{~m}\) thick. During the winter, temperatures are nominally \(17^{\circ} \mathrm{C}\) and \(10^{\circ} \mathrm{C}\) at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what is the rate of heat loss through the slab? If the basemient is heated by a gas furnace operating at an cfficiency of \(\eta_{r}=0.90\) and natural gas is priced at \(C_{r}=50.01 \mathrm{MJ}\), what is the daily cost of the heat loss?

The roof of a car in a parking lot absorbs a solar radiant flux of \(800 \mathrm{~W} / \mathrm{m}^{2}\), while the underside is perfectly insulated. The convection coefficient between the roof and the ambient air is \(12 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). (a) Neglecting radiation exchange with the surruundings, calculate the temperiture of the roof under seady. state conditions if the ambient air temperature is \(20^{\circ} \mathrm{C}\). (b) For the sume ambient air temperature, calculate the temperature of the roof if its surface emissivity is \(0.8\). (c) The convection coefficient depends on airfiow conditions over the roof, increasing with increasing air specd. Compute and plot the roof temperature as a function of \(h\) for \(2 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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