WEBVTT
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This is College Physics Answers
with Shaun Dychko.
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Suppose a 1.00 megaton bomb explodes
a few kilometers above the ground,
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we convert this energy into joules by
multiplying by
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4.2 times 10 to the 12 joules
for every kiloton
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and we convert the megatons
into kilotons
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by multiplying by a 1000 there
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and that's 4.2 times 10 to the 15 joules.
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25 percent of this energy is
turned into radiant heat
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and so we multiply that by 0.25 to get
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1.05 times 10 to the 15 joules of
heat energy
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and a distance of 10.0 kilometers away is
10.0 times 10 to the 3 meters
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and the question is
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what will be the calories per
square centimeter
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at this distance of 10.0 kilometers
away from the explosion?
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So this is a type of flux so
I have this symbol here
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equals the energy divided by
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the amount of area over which
that energy is spread.
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So the area is the area of a circle
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with this radius of 10.0 kilometers
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and that is 4*πr squared* is
the surface area of a circle
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so we substitute that in place of area here.
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So that's 1.05 times 10 to the 15 joules
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multiplied by 1 calorie for every 4.186 joules
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because the question asks us to
give our answer in
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calories per square centimeter
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and we divide that by 4*π*
times this radius,
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which we also convert into centimeters
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and to square that
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and we get 20.0 calories per
square centimeter.
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In part (b), we are asked]
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supposing that the energy is
absorbed into
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a depth of 1.00 centimeter of skin tissue
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by how much will it increase
the temperature of that skin?
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So we need to look up
the specific heat of skin
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and we look at table [14.1]
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and the human body has
a specific heat of about
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0.83 kilocalories per kilogram
per Celsius degree
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and we also need to know the density of
the human body,
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which we look up in table [11.1]
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and we will assume it's the same
as that of water
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which is 1.000 times 10 to the 3
kilograms per cubic meter
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and yeah, the depth is 1.00 centimeters
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so we have written down all the stuff
that we need to know here.
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So here's a formula from chapter 14,
which says the heat absorbed
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equals the mass times the specific
heat of the substance
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times its change in temperature
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and ultimately we want to find
what is this *ΔT*?
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Now we don't know the energy absorbed
but we do know the energy per area.
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So if we divide this left side by area
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we can also divide the right-hand side
by area
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and the equation is still true,
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we are just dividing both sides
by the same thing
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and so let's find an expression then
for mass per area
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because this *Q* per area,
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this energy per area is
the answer to part (a)
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this is *Q* per *A*—
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that's the left side—
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so now the task is then
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to figure out what is this mass per
area of the human body?
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So the density is mass divided by volume
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and volume is the depth
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multiplied by the area for a rectangle
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so if you have a rectangle here with
an area of *A* on the top
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and it has some depth here—
we'll call this *d*—
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then the volume of this is the area
multiplied by the depth *d*
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so we substitute *dA* in place of *V*.
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Now we can solve this for
*m* over *A* then
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by multiplying both sides by *d*
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so *m* over *A* then is density
times the depth.
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So we substitute this in
place of *m* over *A*
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and we do that here and
now we'll solve for *ΔT*
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so we divide both sides by density times
depth times specific heat.
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And so the change in temperature then
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is the energy absorbed per area
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divided by the density times
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the depth through over which it's absorbed
times specific heat.
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So that's 19.96 calories per
square centimeter
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and we are going to convert this into
kilocalories per square meter.
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We need to have square meters
in order to match with this
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density, which has units of
kilograms per cubic meter;
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this depth also has to be converted into
meters for the same reason
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and we want kilocalories here because our
specific heat has units of kilocalories.
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So we have 19.96 calories per square meter
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times 1.00 kilocalorie for
every 1000 calories
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times 100 centimeters for every meter
and we do that twice
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and we have meters squared on the top then
or per meters squared I should say
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and then divide that by 1.00 times 10 to
the 3 kilograms per cubic meter
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times 1.00 centimeter
converted into meters
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times 0.83 kilocalories per kilogram
per Celsius degree
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and we get a temperature change then
of 24 Celsius degrees.