Giancoli's Physics: Principles with Applications, 7th Edition

14

Heat

Change chapter14-1: Heat as Energy Transfer

14-3 and 14-4: Specific Heat; Calorimetry

14-5: Latent Heat

14-6 to 14-8: Conduction, Convection, Radiation

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.

Problem 45

Q

A

$340 \textrm{ Btu/h}$

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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The rate of heat conduction across any material is the thermal conductivity of the material times its cross sectional area times the difference in temperatures divided by the thickness of the material. Now, when we have, so, you could say a compound material here where it consists of two different substances touching each other. Then if it's in a steady state where this temperature 1 has been maintained for a long time and this temperature 2 has been maintained for a long time, that means that the rate of heat transfer across the first material equals the rate of heat transfer across the second material, and that equals the total rate of heat transfer across the combination as well. Because if that was not true then it would mean the temperature is changing somewhere in here. But that contradicts the idea that things have been, you know, things have been maintained with these temperatures here for a long time. So, after a while, there's not going to be any temperature changes within the wall. Eventually is gonna reach some steady state where the rate of heat transfer across all the sections are the same. So that is to say that *Q* over *t*, the overall rate of heat transfer across the entire wall is equal to *Q1* over *t*, the rate of heat transfer across the first material, and that equals *Q2* over *t*, rate of heat transfer across the second material. So, we'll call the temperature inside the room *t1* and outside the room *t2*, and then this temperature here in between them is going to be *t3*. And... And so we can write the rate of heat transfer in terms of this first section here which has the insulation. So, that's the thermal conductivity of insulation times its cross sectional area times *t3* minus *t1* divided by the thickness of this insulation. And we can solve or we can rearrange this to solve for *t3* minus *t1*, and multiply both sides by *l1* and divide by *k1 A*. And you get *Q* over *t* times *l1* over *k1 A* is *t3* minus *t1*. And the reason that's useful is because when we express this same *Q* over *t* in terms of the brick, we can solve it for the temperature difference and we get *t2* minus *t3*. And then we can add these equations together and we'll end up having the *t3's* cancel each other. So, this is the thermal conductivity of brick, and it's *t2* minus *t3* and divided by the thickness of the brick, *l2*, and when we add the two left sides, we get this. Not much we can do with that, except just rewrite them with an addition between them. And on the right hand side, though, we have *t3* minus *t1* plus *t2* minus *t3*, and the *t3's* cancel away because they have opposite signs. Now, recall that that *R* value is thickness divided by thermal conductivity. So, we can replace this *l1* over *k1* with *R1*, the *R* value of insulation. And then likewise, *l2* over *k2* can be replaced with the *R* value of brick. And we can factor out the *Q* over *t*, and so that means we have *Q* over *t* times *R1* over *A* plus *r2* over *A* is the difference in temperatures *t2* minus *t1*. And solve this for *Q* over *t* and by multiplying both sides by *A* over *R1* plus *R2*. And that cancels on the left, leaving us with *Q* over *t*. And on the right hand side we have *t2* minus *t1* times *A* over *R1* plus *R2*. And the area is 195 square feet times 35 Fahrenheit degrees, temperature difference, between *t1* and *t2* inside and outside of the room, divided by 19 is the *R* value of the insulation. And brick will take to be an *R* value of 1, it tells us a range of possibilities here for our value of brick. It says it could be anywhere between 0.6 and 1. And that's for 3.5 inch thickness. And we have a 4 inch thickness of brick. So, let's take the upper possibility of 1 for the *R* value. And that gives about 341.25 or 340 BTUs per hour of heat transfer across the wall.

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