WEBVTT
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This is College Physics Answers
with Shaun Dychko.
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So fasten your seat belt, we have
a long solution coming up
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and you are brave and you can actually
congratulate yourself for
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doing a challenging problem like this.
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Alright and I will try and make it as
easy as I can and here we go!
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So we have this temperature
inside the house of *T 1*
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and that is 15.0 degrees Celsius and
the temperature outside, *T 4*,
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is minus 10.0 degrees Celsius.
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This is a double pane window so
there's a layer of glass here
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that has a thickness of 0.800 centimeters
which we convert into meters
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and there's an air gap in between
the two panes of glass
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with a thickness of 1.00 centimeter
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and then an identical pane of
glass here as well.
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So this temperature *T 2* labels
the temperature just
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touching this inside
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interior portion of this
first pane of glass.
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I am picturing the heat going
this direction here
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so this is the first pane of glass
that the heat hits
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and then it will go across this air gap
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and until it reaches some reduced
temperature, *T 3* here,
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just before this pane of glass
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and then it will go across
this final pane of glass.
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And when this system is in equilibrium
which is to say that
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when the temperature's are
not changing at all
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that means that the rate at which heat is
entering a certain position
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is going to equal the rate at which
the heat leaves that position
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otherwise temperatures would change.
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So I mean of course there's gonna be
a gradient of temperature—
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the temperature is changing
with position clearly,
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it's higher here and then
going lower here—
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but the temperature at a particular spot
is gonna stay constant
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so this will be constantly some temperature
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much less than 15.0 and probably
close to negative 10.0, right here
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and when you have that equilibrium state
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then you can say that the rate at which
heat is going across
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this first pane of glass—
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we'll call that *Q*/*t* subscript *p* for
pane of glass—
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is going to equal the rate at which
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the heat is traveling across this air gap
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so *Q* per *t A* for air gap
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and then again there's gonna be
a rate of heat transfer
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across the second pane of glass
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but it's gonna equal the rate of heat transfer
across the first pane of glass
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because the question tells us
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that the temperature difference across
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the second pane of glass
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the difference between *T 3* and
*T 4* is the same as
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the difference between *T 1* and *T 2*
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and the pane's are also otherwise identical;
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same thickness and the same material so
the same thermal conductivity and so on.
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Okay!
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So with all that in mind, let's figure out
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what the rate of heat transfer will be.
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So if we can find the rate of heat transfer
for any of these three portions
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that will be the answer because
they are all equal to each other
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and therefore they equal the general
sort of *Q* versus *t*
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or *Q* over *t* for any overall.
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Okay!
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So here's the rate of heat transfer across
the first pane of glass
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the thermal conductivity times the glass
area times the temperature difference
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between *T 1*—inside the room—
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and *T 2*—just on the other side of
this first pane of glass—
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divided by the thickness of the pane
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and that equals the thermal
conductivity of the air
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multiplied by the same area
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there's no need for subscript here because
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the area of this air is gonna be
the same as the glass
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and that is the area that's facing,
you know, sort of
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eyeballs that are looking this way.
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Okay!
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And so multiplying by the temperature
difference between
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the two sides of the air gap
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so that's *T 2* minus *T 3* divided by
the thickness of the air gap.
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Now we don't know *T 2* nor
do we know *T 3*
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and so we have two unknowns.
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Now if we could solve for one of them
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then we could answer this question because
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if we can solve for *T 2* for example,
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we could then use this formula to get
our answer if knew what *T 2* was.
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Or likewise,
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well, I didn't write it down but we could
also have *k PA* times
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*T 3* minus *T 4*—
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this would be the rate of heat transfer
across the second pane of glass—
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divided by *d P*.
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If we know what *T 3* was
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then we could use this to figure out
*Q* over *t*.
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So there's three different equations that
will give us the same answer
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and our strategy is to figure out
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one of the unknown temperatures
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and I am gonna try and find *T 3*
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and then we are gonna end up
using this equation here.
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So in order to reduce the amount of writing
so just for our convenience,
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we are gonna replace this *k P* over *d P*
with a single thing called *x P*
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and likewise *k a* over *d a* is gonna be
replaced by a single thing *x a*.
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So rewriting it with this new
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convenience factor you might say,
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we have *x P* times *T 1* minus *T 2*
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equals *x a* times *T 2* minus *T 3* where
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*k a* over *d a* is replaced with *x a*
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and *k P* over *d P* was
replaced with *x P*
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and the areas being the same, they canceled
and we are left with this.
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And then we'll multiply into the brackets
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and then I collect the *T 3* term
on one side
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because remember my goal is to figure out
what is *T 3*.
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So yes, this term got moved to the left hand
side making it positive
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and then this term went to the right hand
side making it positive as well
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and I factored out the *T 2* so *T 2* is
multiplied by *x a* plus *x P*
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and then this term got subtracted
from both sides
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and it becomes minus *x PT 1*.
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Okay so that's equation 1, we'll call it.
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Now equation 2 is this idea that
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the temperature difference across
the panes of glass are the same
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so *T 1* minus *T 2* equals
*T 3* minus *T 4*
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and we can use this to solve for *T 2*
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and then we'll make a substitution in
equation 1 with *T 2*
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and then we'll be able to solve for *T 3*.
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So if we move this to the right hand side
that it makes it positive
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and this goes to the left making it negative
and this goes to the left making it positive
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and then switching the sides around, we have
*T 2* equals *T 1* minus T 3 plus *T 4*.
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So *T 2* in equation 1 then gets replaced
by all of this—
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*T 1* minus *T 3* plus *T 4*
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so this is a substitution of equation 2
into equation 1.
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And then we multiply these
two brackets together
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so the *T 1* gets multiplied by
both factors
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which makes a positive *T 1x a* and
then plus *x PT 1*
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and then the *T 3* gets multiplied into
both terms in the bracket
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and so that's a minus *x aT 3* and
then minus *x PT 3*
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and then *T 4* gets multiplied
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by both terms and that becomes plus
*x aT 4* and plus *x PT 4*
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and then copy down this minus *x PT 1*
and copy this over here as well.
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So let's collect all the *T 3* terms
on the left side
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so we add *x aT 3* to both sides and
add *x PT 3* to both sides
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and then the *x aT 3* plus *x aT 3*
is 2*x aT 3*
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and then the *T 3* gets factored out
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what becomes two terms in total
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so that's gonna be *T 3* times
*2x a* plus *x P*;
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that's on the left.
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Then on the right hand side,
we have *x aT 1* copied there;
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this *x PT 1* cancels with this one and
they make zero
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and then we have dealt with
these two already
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and then this becomes
*T 4* (*x a* plus *x P*).
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And then we divide both sides by
this 2*x a* plus *x P*
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and now we have an expression for *T 3*.
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All the things on the right hand side
are things that we know:
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we have *T 1* and *T 4* and we can figure out
what these factors *x a* and *x P* are.
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So *x a* is the thermal conductivity of air
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divided by the thickness of the air—
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that's what we set up here—
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and so we plug in numbers.
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0.023 divided by 1.00 times 10 to
the minus 2 makes 2.3
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and *x P* is the thermal conductivity of
the pane of glass
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divided by the glass thickness
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and that's 0.84 divided by 0.800 times
10 to the minus 2 which is 105.
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So temperature 3 then is
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2.3 times 15.0 degrees Celsius—
which is temperature 1—
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plus negative 10.0 degrees Celsius—
which is temperature 4—
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times 2.3 plus 105
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divided by 2 times 2.3 plus 105
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so that's negative 9.47536 degrees Celsius.
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Now before we continue, let's make sure
that number makes sense.
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We expect some number
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that is significantly less than
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the room temperature of positive 15.0
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because we are quite a long ways
away from the room here
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the heat has gone across the glass and
has gone across the whole air gap
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and the only place left to go is
just across this pane of glass
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and then it's at minus 10.0
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so it should be closer to minus 10.0
than it is to positive 15.0
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and it certainly is
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so negative 9.47536 degrees Celsius
seems reasonable.
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I was a bit surprised that it was
so close to 10.0 but
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I think that is a testament to how
poorly insulating glass is.
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Okay!
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So the overall rate of heat transfer is
the heat transfer across any portion
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including the last pane of glass
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and so we have the thermal conductivity of
that pane of glass times its area times
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*T 3*—which we now know—
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minus *T 4* all divided by the thickness of
the pane of glass will be our answer.
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So that's 0.84 times 1.50 meters squared
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times negative 9.47536 degrees Celsius
minus negative 10.0 degrees Celsius
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all divided by the pane of glass thickness
of 0.800 times 10 to the minus 2 meters
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and that is 83 watts.
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So that's the rate of heat transfer across
the overall window is 83 watts.
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Then part (b) asks us to consider
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what the rate of heat transfer would be
across a single pane of glass
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that was 1.60 meters thick?
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So the total thickness of this
double pane window
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is the 1.00 centimeter air gap
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plus the 0.800 centimeters on each side.
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So 1.60 is...
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I think they should have made
a thickness of glass
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that's exactly the same as this
whole double pane window
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but anyway... it's close!
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So we have 0.84—the thermal conductivity of
the glass—times 1.50 square meters—
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this is the same area—
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times 15.0 degrees Celsius minus
negative 10.0 degrees Celsius
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all divided by 1.60 times 10 to the minus
2 meters—thickness—
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and that is 2.0 times 10 to the 3 watts.
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And so that thick single pane window
loses heat at a rate which is
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24 times that of the double pane window
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and this calculation did not
consider convection
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which would be to the detriment of
the double pane window
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and so this difference would be less
in the real world
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but nevertheless still significant.