/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 106 One experimental method of measu... [FREE SOLUTION] | 91Ó°ÊÓ

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One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 \(\mathrm{W}\) is required to keep the interior surface of the box 65.0 \(\mathrm{C}^{\circ}\) (about 120 \(\mathrm{F}^{\circ}\) ) above the temperature of the outer surface. The total area of the box is \(2.18 \mathrm{m}^{2},\) and the wall thickness is 3.90 \(\mathrm{cm} .\) Find the thermal conductivity of the material in SI units.

Short Answer

Expert verified
The thermal conductivity is approximately 0.0495 W/m · K.

Step by step solution

01

Understanding the Problem

To find the thermal conductivity of the material, we need to use the formula for heat conduction. The power input maintains a temperature difference across a known thickness of the material, which allows us to find thermal conductivity.
02

Identify Given Values

From the problem, we know: - Power input ( P ) = 180 W - Temperature difference ( ΔT ) = 65.0 °C - Total area ( A ) = 2.18 m² - Wall thickness ( L ) = 3.90 cm = 0.039 m.
03

Apply the Thermal Conductivity Formula

The formula to calculate thermal conductivity (k) is given by:\[P = k \frac{A \Delta T}{L}\]Rearranging the formula to solve for k, we have:\[k = \frac{P \cdot L}{A \cdot \Delta T}\]
04

Substitute Values into the Formula

Substitute the given values into the rearranged formula:\[k = \frac{180 \mathrm{W} \cdot 0.039 \mathrm{m}}{2.18 \mathrm{m}^2 \cdot 65.0}\]\[k = \frac{7.02}{141.7}\]
05

Perform the Calculation

By calculating the above expression:\[k = 0.0495\, \mathrm{W/m \cdot K}\]Thus, the thermal conductivity of the material is approximately 0.0495 W/m · K.

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Key Concepts

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

heat conduction
Heat conduction is a mode of heat transfer where thermal energy moves through a material without the overall movement of the material itself. Think of it like the way warmth moves through a metal spoon when you stir hot soup. This happens because of the transfer of kinetic energy among particles in the material. The rate at which heat is conducted depends on several factors:
  • The temperature difference across the material—which acts as the driving force for the transfer.
  • The area through which the heat is conducted, as a larger area allows more room for energy transfer.
  • The thickness of the material, where a thicker material resists heat flow.
In mathematical terms, heat conduction can be described by Fourier's law, which in simple cases, is expressed as: \[ q = -k A \frac{dT}{dx} \]where:
  • \( q \) is the heat transfer rate,
  • \( k \) is the thermal conductivity,
  • \( A \) is the cross-sectional area,
  • \( dT/dx \) is the temperature gradient—the change in temperature over distance.
This formula helps us calculate how efficiently heat is transferred within a material.
temperature difference
Temperature difference is a key concept in understanding heat transfer processes like conduction. It represents the measure of how much the temperature varies between two points.
In the context of heat conduction, the temperature difference (\( \Delta T \) in formulas) is what causes heat to move from one area with higher temperature to another with lower temperature, seeking equilibrium.
This principle is similar to how a ball naturally rolls downhill due to gravity. In thermal systems, it's the temperature difference that causes thermal energy to 'flow' or transfer between regions.
Maintaining a significant temperature difference across a material's thickness enhances the rate of heat conduction, as it increases the 'push' of heat across the material.
In the exercise, the interior of the box is kept 65.0 \( ^\circ C \) warmer than the outside, acting as the driving force to measure the material's insulating properties.
SI units
The International System of Units (SI) is a globally accepted metric system for scientific measurements. It standardizes measurements so they are understood universally, crucial for fields such as physics and engineering.
In the context of the problem, several SI units are important:
  • Power (\( P \)) is measured in watts (\( W \)). It's equivalent to one joule per second, representing the rate of energy transfer.
  • Area (\( A \)) is measured in square meters (\( m^2 \)), a basic unit for measuring surface area.
  • Length or thickness (\( L \)) is measured in meters (\( m \)). Although the thickness was initially given in centimeters, it was converted to meters, as SI units require.
  • Temperature (\( \Delta T \)) is expressed in degrees Celsius (\( ^\circ C \)) but is often converted to Kelvin when used in formulas for enhanced precision in scientific contexts.
By using these units, calculations for thermal conductivity yield consistent and understandable results, facilitating communication of scientific ideas.
insulating material
Insulating materials are substances that resist the flow of thermal energy, making them ideal for reducing heat loss or gain in a system.
Their effectiveness is quantified by a property called thermal conductivity (denoted \( k \)), which measures how well a material conducts heat.
Lower thermal conductivity values indicate better insulating capability.Insulating materials perform a crucial role in many engineering applications, such as:
  • Building construction, where they maintain comfortable interior temperatures by acting as barriers to unwanted heat gain or loss.
  • Electrical devices and appliances, where they prevent overheating and improve energy efficiency.
In the exercise, the insulating material forming the box's walls is evaluated for its thermal conductivity to ensure effective heat retention inside the box, maintaining a temperature 65.0 \( ^\circ C \) warmer than the outside environment with a power input of 180 watts. This process illustrates how insulating materials optimize energy usage by minimizing unnecessary heat flux.

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Most popular questions from this chapter

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