WEBVTT
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This is College Physics Answers
with Shaun Dychko.
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We are going to compare the rate of
heat transfer through a wall,
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which I have labeled with subscripts 1,
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to the rate of heat transfer
through a window,
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which has all of its details labeled
with a subscript 2.
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So the wall is 13.00 centimeters thick
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and since we are comparing two things...
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I know we are going to be dividing and so
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if we have units that are the same—
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centimeters for the thickness of the wall
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and centimeters for the thickness of
the window—
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we probably don't need to
change those units.
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We'll check to make sure when
we get further along
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but if we are gonna divide centimeters by
centimeters that will be fine;
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there's no need to convert this into meters.
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So the area of the wall is 10.0 square meters
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and the thermal conductivity of the wall is
2 times that of glass wool
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because it's probably full of fiber glass
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so fiber glass is another
name for glass wool
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and so it's 0.084 joules per second per
meter per Celsius degree.
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The thickness of the window... we assume
it's just a single pane window
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because it has such a small thickness of
0.750 centimeters
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and we need to figure out what
the thermal conductivity of it is
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and we'll have to use that of glass
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which is 0.84 joules per second per
meter per Celsius degree
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and the area of the window is
2.00 meters squared.
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So the rate of heat conduction
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is the thermal conductivity multiplied by
the area of something
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times the temperature difference between
one side and the other
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divided by the distance between
these two sides.
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And so for the wall that is
*Q* over *t* number 1,
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we have the thermal conductivity of the wall
times the area of the wall
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times this temperature difference
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which we don't need to label with
subscripts 1 or 2
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because it's the same temperature difference
on the inside of the wall
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and the outside of the wall as it is
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in the inside of the window and
the outside of the window
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so they are both in contact
with the same air
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inside the house and the same air
outside the house.
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But the thickness of the wall is distinct
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and so it has a subscript 1 there for
the thickness of the wall.
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Okay!
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So this is getting divided by the same
expression for the window
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but I find it confusing to divide
a fraction by a fraction
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and so instead I am multiplying by
the reciprocal of the denominator.
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So this is this rate of heat conduction
across the window
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but I flipped it over since we are
dividing by it.
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So this is the thickness of the window
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divided by the thermal conductivity of
the window
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times its area times the same
temperature difference.
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So this bracket *T a* minus *T b*
can be canceled—
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it's a common factor on
the top and bottom—
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and so we are left with this expression:
*k 1A 1d 2* over *k 2A 2d 1*.
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So we have 0.084—thermal
conductivity for the wall—
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times the wall area of 10.0 square meters
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times the window thickness of
0.750 centimeters
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and here we can see that we are
dividing 0.750 centimeters by
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13.0 centimeters and since the centimeter
units are the same—
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this is all good—we don't need to do
unit conversions there.
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And so we are dividing by the thermal
conductivity of the window—0.84—
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times its area of 2.00 square meters
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and we get 0.029.
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So this says the rate of heat transfer
through the wall
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is only 2.9 percent, if you convert
that into a percent,
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that of the window
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and that's quite dramatic
because the wall is
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5 times larger area than the window
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and yet it has only 2.9 percent of
the rate of heat loss.