/*! 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 16 A certain double-pane window con... [FREE SOLUTION] | 91Ó°ÊÓ

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A certain double-pane window consists of two glass sheets, each \(80 \mathrm{~cm} \times 80 \mathrm{~cm} \times 0.30 \mathrm{~cm}\), separated by a \(0.30-\mathrm{cm}\) stagnant air space. The indoor surface temperature is \(20^{\circ} \mathrm{C}\), while the outdoor surface temperature is exactly \(0{ }^{\circ} \mathrm{C}\). How much heat passes through the window each second? \(k_{T}=0.84 \mathrm{~W} / \mathrm{k} \cdot \mathrm{m}\) for glass and about \(0.080\) \(\mathrm{W} / \mathrm{k} \cdot \mathrm{m}\) for air.

Short Answer

Expert verified
Approximately 284.09 watts of heat pass through the window each second.

Step by step solution

01

Understand the Problem

We need to find the heat transfer through the double-pane window. The window has two glass sheets separated by air. Each glass sheet measures 80 cm x 80 cm x 0.30 cm, and the air gap is also 0.30 cm. The heat transfer rate through each medium (glass and air) can be calculated using thermal conductivity.
02

Calculate the Area of the Window

The area of the window is the same as the area of one glass sheet. Given that each sheet is 80 cm x 80 cm, convert this to meters: \[ A = 0.80 ext{ m} imes 0.80 ext{ m} = 0.64 ext{ m}^2 \]
03

Calculate the Thermal Resistance for Glass

For the glass sheet:\[ R = \frac{d}{kA} \]where \(d = 0.0030 ext{ m} (converted from 0.30 cm)\), \(k = 0.84 ext{ W} / ext{K} ext{m}\), and \(A = 0.64 ext{ m}^2\).Calculate:\[ R_{glass} = \frac{0.0030}{0.84 \times 0.64} \approx 0.0059 ext{ K/W} \]
04

Calculate the Thermal Resistance for Air

For the air space:\[ R = \frac{d}{kA} \]where \(d = 0.0030 ext{ m}\), \(k = 0.080 ext{ W} / ext{K} ext{m}\).Calculate:\[ R_{air} = \frac{0.0030}{0.080 \times 0.64} \approx 0.0586 ext{ K/W} \]
05

Calculate Total Thermal Resistance

The total thermal resistance is the sum of the resistance of both glass panels and the air gap. Since there are two sheets of glass, the total resistance is:\[ R_{total} = 2R_{glass} + R_{air} = 2(0.0059) + 0.0586 \approx 0.0704 ext{ K/W} \]
06

Calculate Heat Transfer Rate

Use the formula:\[ Q = \frac{\Delta T}{R_{total}} \]where \(\Delta T = 20^{\circ}C - 0^{\circ}C = 20^{\circ}C\).Calculate the heat transfer rate:\[ Q = \frac{20}{0.0704} \approx 284.09 ext{ W} \]
07

Conclusion

The amount of heat passing through the window each second is approximately 284.09 watts.

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Key Concepts

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

Thermal Resistance
Thermal resistance is a measure of how well a material resists the flow of heat. It acts like an obstacle that slows down the transfer of heat between one side of the material and the other.
Imagine wearing multiple layers of clothing in the winter to stay warm. Each layer of clothing provides more resistance to the outflow of body heat to the cold air outside. This is similar to the concept of thermal resistance in materials.

Thermal resistance (\(R\)) is calculated using the formula:\[ R = \frac{d}{kA} \]where
  • \(d\) is the thickness of the material in meters,
  • \(k\) is the thermal conductivity of the material,
  • and \(A\) is the area through which heat is being transferred.
A higher thermal resistance means better insulation. In the case of the double-pane window, both the glass panes and the air gap have their own thermal resistances, which together determine how much heat is transferred from the inside to the outside.
Thermal Conductivity
Thermal conductivity is a property of a material that describes how well it can conduct heat. It’s represented by \(k\), with units of watts per meter per Kelvin (\(\mathrm{W/mK}\)).
Materials like metals have high thermal conductivity, which means heat moves easily through them. In contrast, materials like wood or air have low thermal conductivity, so they are better insulators.

When you’re considering heat transfer through a window, knowing the thermal conductivity of each material involved is crucial. For the given problem:
  • Glass has a thermal conductivity of \(0.84\, \mathrm{W/mK}\), a relatively high value compared to air.
  • Air’s thermal conductivity is \(0.080\, \mathrm{W/mK}\), making it a better insulator.
Understanding the difference in thermal conductivity between materials can help predict how well the window will insulate. The lower the thermal conductivity, the more the material resists heat flow, and thus, the better it insulates.
Double-Pane Window
A double-pane window is designed to provide better insulation than a single-pane window. It consists of two sheets of glass with a layer of air (or sometimes gas) trapped between them. This setup results in reduced heat transfer due to the thermal resistance offered by the air gap and the two sheets of glass.

The effectiveness of a double-pane system depends on:
  • The thickness of the air gap and glass sheets
  • The thermal resistance of each component
  • The thermal conductivity of glass and air
The stagnant air space is key because air is a relatively poor thermal conductor. It slows down the rate at which heat is conducted from the inside to the outside of the window.
This is why double-pane windows are often used in energy-efficient buildings to minimize heat loss in winter and keep out heat in summer. Overall, using a double-pane window helps maintain a more stable indoor climate while reducing energy costs.
Heat Transfer Rate
The heat transfer rate, often denoted as \(Q\), tells us how much heat energy flows from one side of the window to the other per second. It is typically measured in watts (\(\mathrm{W}\)).

In the context of the double-pane window, the heat transfer rate can be calculated using the equation: \[ Q = \frac{\Delta T}{R_{total}} \]where:
  • \(\Delta T\) is the temperature difference between the inside and outside surfaces,
  • and \(R_{total}\) is the total thermal resistance of the window.
A higher heat transfer rate means that more heat is escaping through the window, which can lead to higher energy costs for heating or cooling a building.
In our problem, the heat transfer rate is calculated to be approximately \(284.09\, \mathrm{W}\), meaning this amount of heat passes through the window every second. By reducing \(\Delta T\) or increasing \(R_{total}\), one could reduce the heat loss through the window.

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Most popular questions from this chapter

Suppose the area of a window on a passenger jet is doubled in size and all else is kept constant. Compared with the original window, what will happen to the rate at which heat passes out through the new window from the cabin?

A beverage cooler is in the shape of a cube, \(42 \mathrm{~cm}\) on each inside edge. Its 3.0-cm-thick walls are made of plastic \(\left(k_{T}=0.050\right.\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) ). When the outside temperature is \(20^{\circ} \mathrm{C}\), how much ice will melt inside the cooler each hour? We have to determine the amount of heat conducted into the box. The cubical box has six sides, each with an area of about \((0.42\) \(\mathrm{m})^{2}\). From \(\Delta Q / \Delta t=k_{T} A \Delta T / L\), we have, with the ice inside at \(0{ }^{\circ} \mathrm{C}\), $$ \frac{\Delta Q}{\Delta t}=(0.050 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})(0.42 \mathrm{~m})^{2}(6)\left(\frac{20^{\circ} \mathrm{C}}{0.030 \mathrm{~m}}\right)=35.3 \mathrm{~J} / \mathrm{s}=8.43 \mathrm{cal} / \mathrm{s} $$ In one hour, \(\Delta Q=(60)^{2}(8.43)=30350\) cal. To melt \(1.0 \mathrm{~g}\) of ice requires 80 cal, so the mass of ice melted in one hour is $$ m=\frac{30350 \mathrm{cal}}{80 \mathrm{cal} / \mathrm{g}}=0.38 \mathrm{~kg} $$

A spherical body of \(2.0 \mathrm{~cm}\) diameter is maintained at \(600{ }^{\circ} \mathrm{C}\). Assuming that it radiates as if it were a blackbody, at what rate (in watts) is energy radiated from the sphere? $$ \begin{array}{l} A=\text { Surface area }=4 \pi r^{2}=4 \pi(0.01 \mathrm{~m})^{2}=1.26 \times 10^{-3} \mathrm{~m}^{2} \\ \mathrm{P}=A \sigma T^{4}=\left(1.26 \times 10^{-3} \mathrm{~m}^{2}\right)\left(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)(873 \mathrm{~K})^{4}=41 \mathrm{~W} \end{array} $$

An incandescent lamp filament has an area of \(50 \mathrm{~mm}^{2}\) and operates at a temperature of \(2127^{\circ} \mathrm{C}\). Assume that all the energy furnished to the bulb is radiated from it. If the filament's emissivity is \(0.83\), how much power must be furnished to the bulb when it is operating?

A copper tube (length, \(3.0 \mathrm{~m}\); inner diameter, \(1.500 \mathrm{~cm}\); outer diameter, \(1.700 \mathrm{~cm}\) ) extends across a 3.0-m-long vat of rapidly circulating water maintained at \(20^{\circ} \mathrm{C}\). Live steam at \(100^{\circ} \mathrm{C}\) passes through the tube. ( \(a\) ) What is the heat flow rate from the steam into the vat? ( \(b\) ) How much steam is condensed each minute? For copper, \(k_{L}=1.0 \mathrm{cal} / \mathrm{s} \cdot \mathrm{cm} \cdot{ }^{\circ} \mathrm{C}\) To determine the rate at which heat flows through the tube wall, approximate it as a flat sheet. Because the thickness of the tube is much smaller than its radius, the inner surface area of the tube, $$ 2 \pi r_{i} L=2 \pi(0.750 \mathrm{~cm})(300 \mathrm{~cm})=1410 \mathrm{~cm}^{2} $$ nearly equals its outer surface area, $$ 2 \pi r_{0} L=2 \pi(0.850 \mathrm{~cm})(300 \mathrm{~cm})=1600 \mathrm{~cm}^{2} $$ As an approximation, consider the tube to be a plate of thickness \(0.100 \mathrm{~cm}\) and area given by $$ A=\frac{1}{2}\left(1410 \mathrm{~cm}^{2}+1600 \mathrm{~cm}^{2}\right)=1500 \mathrm{~cm}^{2} $$ (a) \(\frac{\Delta Q}{\Delta t}=k_{T} A \frac{\Delta T}{L}=\left(1.0 \frac{\mathrm{cal}}{\mathrm{s} \cdot \mathrm{cm} \cdot{ }^{\circ} \mathrm{C}}\right) \frac{\left(1500 \mathrm{~cm}^{2}\right)\left(80^{\circ} \mathrm{C}\right)}{(0.100 \mathrm{~cm})}=1.2 \times 10^{6} \mathrm{cal} \mathrm{s} / \mathrm{s}\) (b) In one minute, the heat conducted from the tube is $$ \Delta Q=\left(1.2 \times 10^{6} \mathrm{cal} / \mathrm{s}\right)(60 \mathrm{~s})=72 \times 10^{6} \mathrm{cal} $$ It takes 540 cal to condense \(1.0 \mathrm{~g}\) of steam at \(100{ }^{\circ} \mathrm{C}\). Therefore, Steam condensed per min \(=\frac{72 \times 10^{6} \mathrm{cal}}{540 \mathrm{cal} / \mathrm{g}}=13.3 \times 10^{4} \mathrm{~g}=1.3 \times 10^{2} \mathrm{~kg}\) In practice, various factors would greatly reduce this theoretical value.

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