Re: Bad News for 'Moon Hoax' Buffs

From: bz (bz+sp_at_ch100-5.chem.lsu.edu)
Date: 03/20/05 

Date: Sun, 20 Mar 2005 20:59:18 +0000 (UTC)

"Brad Guth" <ieisbradguth@yahoo.com> wrote in news:1111345828.404401.72700
@g14g2000cwa.googlegroups.com:

> "bz; Stefan's Law P = sigma A T^4
> Power radiated = stefans constant (5.67e-8 W m^-2 K^-4) * Area * Temp^4
>
>
> "bz; Take a rock with 1 m^2 area, at 123 deg C, you will find that the
> rock
> radiates 1.4 kW.
>
> I'm not per say doubting what the average moon accomplishes on a global
> scale. I'm talking about a few specific HOT ROCKS.

How are they going to get to be specially hot rocks? If they absorbe more
heat from the sun (are closer to true black bodies than the surrounding
rocks), then they ALSO radiate heat better than the surrounding rocks. Yes,
they may be a little hotter quicker than some other rocks around them, but
they are not going to be hundreds of degrees hotter.

>
> I can somewhat understand what happens to a uniform surface area of
> black anodised *** of aluminum foil, however doesn't a factor of mass
> and thereby time matter?

It matters, but not the way you are thinking.

If your rock has the same specific heat as oh... lets say iron, and another
nearby rock has the same specifc heat as styrafoam, Then starting at lunar
dawn, the foam rock would heat up quickly and reach radiative equilibrium
with space and the sun fast. The iron rock would heat up slower, but when
it reached equilibrium it would be at about the same temperature as the
other rock. Sure, one may be a bit hotter than the other, but the main
effect of differences in blackness and differences in thermal capacity will
be that one gets hot FASTER than the other.

When sun sets, the foam rock will cool off much faster than the iron one.

>
> Aren't there other rocks nearby and/or touching one another?

Probably. Thermal conductivity comes into play then. Heating and cooling by
conduction. Since there are no gases, convection will not play any part.
But this is not going to make the rocks in the sun hotter, it will allow
them to transfer their heat to other, nearby rocks that are in the shade.

This explains why the average lunar temperature is LOWER than 123 degC

>
> Isn't the bulk of solar influx being converted into secondary/recoil IR
> photons?

If a rock isn't an ideal black body then some of the photons will bounce.

The rock is ALWAYS radiating photons. The amount depends on its
temperature. After sunrise, as the rock reaches equilibrium under the new
conditions, it will be radiating 1.4 kw/m2 into space and absorbing 1.4
kw/m2 from the sun.

>
> Are you then stipulating that 1.4 kw/m2 imposes the exact same transfer
> rate of heat and thereby stored energy per second as it does per hour,
> per day or per week regardless of whatever mass, and regardless of
> whatever insulation factors?

Watts are Joules per second. The number of Joules absorbed will depend on
the time.

The number of Watts is independent of time because time is already included
in its definition. So, your question is meaningless.

>
> Isn't that m2 that's associated with roughly 3 t/m3 as being nicely
> heated by the sun and by all that's surrounding as situated in a near
> vacuum, thus also receptive to secondary/recoil photons, but otherwise
> remaining as somewhat well insulated?

Although the moon has some residual heat at its center, if the sun were to
go out, the surface of the moon, to quite some depth, would quickly
approach absolute zero in temperature. (actually the big bang background
temperature of the universe which is a bit above absolute zero, IIRC about
4k)

This means that the energy available to warm up something, a rock, a space
suit[neglecting the 100W the man puts out], etc., MUST ALL COME FROM THE
SUN. If the solar flux provides 1.4 W / m2 at the distance we(earth/moon)
are from the sun, then that is ALL the energy that is available to heat
that rock.

Of course, if you build a reflector and focus solar energy, you can get
much higher temperatures. But you are not talking about that.

>
> Clearly this raw heat, plus whatever's obtained from secondary sources
> that upon the surface of the moon can be derived from as far away as 10
> km, thereby secondary applied thermal energy factor is continuously
> applied from every visible other nearly black basalt rock that's to be
> seen.

The moon can't radiate any more heat than it receives. And it radiates the
heat in all directions. It doesn't focus the radiation on something.

At best, if the moon were a totally reflective sphere, if you were in orbit
around it, you would have sunlight directly on your ship on one side and
reflected sunlight from the moon on the other. But the image of the sun as
reflected by the moon would be 'much further away' and the energy reduced
corresponding to that extra distance.

Try looking at the reflection of a light as seen in on the surface of a
ball bearing. You will see what I mean about the image being 'much further
away'. Shine a laser on the surface of a ball bearing, the beam diverges.
The sunlight re-radiated by the moon diverges.

>
> At least I'd still think a dark basalt rock that's been heated for days
> on end by the sun at roughly 1.4 kw/m2 is going to be radiating
> something,

Yes, it will be radiating 1.4 kW/m2, which it does very well when it is 123
degrees C.

If it were hotter, it would need to radiate more energy than it receives
and that would require a revision of the laws of physics.

> as obviously at night is where that same basalt rock
> eventually gets seriously cold, in part due to factors of conduction
> into the lunar surface.
>
> Per tonne of whatever substance (say a m3 block of basalt, thus
> 3+t/m3), how long does it take (without sun or within moon nighttime)
> to extract 100°C worth of thermal energy to the point of the core of
> that 3+tonne rock becoming 0°C?

Iron has a specific heat of 0.45 Joule/(gm degC), so you can calculate how
many joules are required to heat a 1 ton iron rock. Iron has a density of
about 7.87 gm/ml. So you can figure out the volume of a rock weighing a
ton. Assume a spherical cow... assume a sphere, you can calculate the
surface area. Take into account that light doesn't fall directly on all of
that surface, or pound your iron out into a *** that is 1 cm thick and
square, and figure out how long it takes it to receive enough joules to
raise its temperature to 123 deg C. At that point, it will be radiating as
much energy as it is receiving and it will NOT get any hotter.

>
> "bz; I support you when you're right, but when you're wrong I support
> you by
> correcting you. Your ideas about heat and vacuum are wrong.
>
> Am I still wrong about the factors of mass, time and of
> secondary/recoil photons?

See discussion above. In brief, yes, you are wrong.

>
> Such as if applying 1.4 MW/cm3, would the entire m3 block of basalt
> manage to get to the same thermal balance in the same time frame?

Now you need to also look at thermal conductivity. 0.803 watt/(cm K) for
Iron.

If you have ever picked up one end of an iron rod that is being heated, you
know that there is some time lag but for an object resting on a body with a
'28 day' day, it is unimportant.

If the surface heats up faster because the energy migrates inward slowly,
then the surface reaches equilibrium temperature faster.

>
> What if each side of a m3 were given 233 w/m2 X 6, thus the same energy
> value as 1.4 kw/m2, would that basalt block arrive at the same thermal
> balance of 123°C in the same amount of time?

close enough for government work.

>
> Isn't thermal energy transfer with regard to mass, such as per a m3 of
> basalt, the sum of all sources?

Yes. And solar energy is the only [significant] source in the case under
discussion.

>
> Other available topics by; Brad Guth / GASA-IEIS
> http://guthvenus.tripod.com/gv-topics.htm
> 

-- 
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an 
infinite set.
bz+sp@ch100-5.chem.lsu.edu   remove ch100-5 to avoid spam trap
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