Radiation & Equilibrium Temperature Video Lecture | Heat Transfer - Mechanical Engineering

welcome to a lecture online and here our
next example is going to illustrate
something really interesting just I
think the interesting part about
thermodynamics and heat transfer is that
hot objects tend to transfer more heat
out than they will receive and cold
objects will tend to receive more heat
in than they will transfer out and
eventually things will then reach an
equilibrium so a hot sunny day let's say
that the sun is shining down on an
asphalt road and let's assume that we
take a look at one square meter of that
road eventually that road will reach
what we would call an equilibrium
temperature if more energy is received
and the road radiates out then the
temple will continue to increase and
eventually as the temperature increases
the radiation out will increase until
the radiation out equals the radiation
in and if and at that point the
temperature Road will have reached its
maximum or its equilibrium point so how
do we find that well let's assume that
the energy received per unit time is
seven liter watts per square meter so
what that means is that each square
meter surface on the road will receive
700 joules of energy every single second
if the radiation out from the road is
less than that the temperature Road will
keep going up if it's more the
temperature will start going down and if
it equals the temperature will then be
constable stay the same so if we use
this equation DQ DT is equal to the
emissivity of an asphalt road times
Sigma times the surface area times
temperature to the fourth power and if
we then assume that the radiation out
will equal the radiation and we can
substitute this in for the DQ DT and
figure out what the temperature then is
and we'll have to solve that for the
temperature all right so then we write t
to the fourth power is equal to DQ DT
divided by e Sigma n a so e Sigma and a
and then if we take the fourth root of
both sides we then say that e is equal
to the fourth root of 4 there should be
a 4 right here of DQ DT
/ e Sigma na
so now let's plug in those numbers so
we're going to use some hundred watts
per square meter so that's a square root
of 700 watts per square meter we then
divide this by actually we shouldn't say
per square meter we're just gonna say
700 watts otherwise we may have a little
bit of a problem here okay
700 watts then we have Sigma E which is
about 0.9 for asphalt Sigma that's five
point six seven times ten to the minus
eight
that's what's per meter squared times
Kelvin to the fourth power
and finally the area here that would be
one meter squared and then of course
this meter squared will cancel out with
that meter squared this watts will
cancel off without watts and one over
Kelvin that fourth common denominator
will give me Kelvin to the fourth power
in the numerator take the cube root of
that don't forget to I'm not to Cuba the
fourth root of that and that will give
us the temperature at which the Rho will
no longer increase all right so we take
seven hundred divided by point nine
divided by five point six seven
e to the eighth minus and then we take
the fourth root of that and the
temperature is 342 Kelvin all right now
how hot is 342 Kelvin well if we convert
that to centigrade degrees we have to
subtract to nine 273 from that so minus
273 that will give a 69 so this is equal
to 69 degrees centigrade
wow that's pretty hot and if you're not
that familiar with centigrade degrees
then let's convert the foreign high
degrees so what we do there is we divide
by 5 multiplied times 9 and at 32 so
divided by 5 multiplied times 9 and not
32 degrees of that and it gives us a
hundred and fifty-seven that's about a
hundred and fifty seven degrees
Fahrenheit so that's why on a very hot
summer day you do not want to walk there
put on an asphalt road because you will
burn your feet and it turns out when you
go to the beach on a hot day and the Sun
absorbs all that heat and it radiates
heat back out but again the temperature
of that sand will reach a certain
equilibrium and when the incoming energy
equals the outgoing energy and you'll
find out that the numbers will not be
all that different from beachside sand
just a little bit less because it's a
lighter in color so the emissivity will
probably not be as high but what you'll
find is that that you will also have the
beach sand near in those kinds of
temperatures in the game you don't want
to walk across beach sand when it's
really hot outside because you will also
burn your feet but that's how you
calculate the equilibrant temperature of
an object that radiates heat at the same
time that it's receiving Heat