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Compare the rate of heat conduction through a 13.0 -cm- thick wall that has an area of 10.0 ${m}^{2}$ and a thermal conductivity twice that of glass wool with the rate of heat conduction through a window that is 0.750 ${cm}$ thick and that has an area of 2.00 ${m}^{2}$ , assuming the same temperature difference across each.

$35 : 1$

Physics 101 Mechanics

Chapter 14

Heat and Heat Transfer Methods

Thermal Properties of Matter

Simon Fraser University

University of Sheffield

University of Winnipeg

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for this question, we are trying to compare the heat conduction rate through a wall versus that of a window. So in order to do that, we want to find the ratio of the two can heat conduction rates. So let's collect our information first from the problem. So I'm gonna use the number one for the wall and the number two for the window. So the thickness of the window is 13 centimeters. That's your 130.13 meters. These are the wall. Rather, the thickness of the window is 0.75 centimeters, which is your 0.75 meters. The area of the wall this tens grew meters. In the area of the window is two square meters. Be thermal. Conductivity of the wall, we are told, is two times the thermal conductivity of glass wool. So if you look that up, this is two times your 20.0 for two Jules per seconds times meters, temps degree Celsius, which is just your 0.8 four and ah, it's a glass window. So the K for the window, if you look up that of glass, is your 0.84 Jules per seconds times, meters times, degrees Celsius. Other thing we need is the equation of the rate of heat conduction, which is Q equals K times, eh? Times t two minutes to you want over the thickness D. So since we're gonna just look at the ratio, we're also told that the temperature changes the same for the wall in the window. So we're going to look at rate if he connection of the wall over the rate of he connection of the window. So on top, we have K one a one T to minus 21 over D one divided by all the same thing for the windows. Okay, too. A to the same temperature change over D too, so we can cancel the temperature change and rearrange our thicknesses on the bottom. So this is Kei Wan a one D too divided by K to a two D one. So this is okay. One to your point is your 84 We know this is gonna be unit lists because all of the, uh, types of values are the same. Size is just a ratio is I'm not gonna bother to write that. Units here says your points. You're 84 times Area one, which is 10 square meters times D too, which is your 0.75 divided by K two to your 20.84 Time's, A two, which is to and D one, which is your 10.13 So you put that in your calculator, you'll get that it's 0.288 This is not really a useful fraction for us. What we can see right off the bat is that the wall rate, if he conduction, is much lower than that of the window. So let's flip this so that it's Q window over Q Wall. So that's one over your 10.288 and that is 34.7, So the window is 34.7 times faster, he conduction than the wall.

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