Answer: Energy from the core travels by radiation through the radiative
zone, then by convection through the convection zone.
Explanation:
First, balance the reaction:
_ KClO₃ ==> _ KCl + _ O₂
As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :
2 KClO₃ ==> 2 KCl + 3 O₂
Since we start with a known quantity of O₂, let's divide each coefficient by 3.
2/3 KClO₃ ==> 2/3 KCl + O₂
Next, look up the molar masses of each element involved:
• K: 39.0983 g/mol
• Cl: 35.453 g/mol
• O: 15.999 g/mol
Convert 10 g of O₂ to moles:
(10 g) / (31.998 g/mol) ≈ 0.31252 mol
The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need
(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃
KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of
(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g
of KClO₃.
After one meter, 3.4% of the light is gone ... either soaked up in the fiber
material or escaped from it. So only (100 - 3.4) = 96.6% of the light
remains, to go on to the next meter.
After the second meter, 96.6% of what entered it emerges from it, and
that's 96.6% of 96.6% of the original signal that entered the beginning
of the fiber.
==> After 2 meters, the intensity has dwindled to (0.966)² of its original level.
It's that exponent of ' 2 ' that corresponds to the number of meters that the light
has traveled through.
==> After 'x' meters of fiber, the remaininglight intensity is (0.966) ^x-power
of its original value.
If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of
cable, you'll have
<em>(1,500) · (0.966)^x</em>
lumens of light remaining.
=========================================
The genius engineers in the fiber design industry would not handle it this way.
When they look up the 'attenuation' of the cable in the fiber manufacturer's
catalog, it would say "15dB per 100 meters".
What does that mean ? Break it down: 15dB in 100 meters is <u>0.15dB per meter</u>.
Now, watch this:
Up at the top, the problem told us that the loss in 1 meter is 3.4% . We applied
super high mathematics to that and calculated that 96.6% remains, or 0.966.
Look at this ==> 10 log(0.966) = <em><u>-0.15</u> </em> <== loss per meter, in dB .
Armed with this information, the engineer ... calculating the loss in 'x' meters of
fiber cable, doesn't have to mess with raising numbers to powers. All he has to
do is say ...
-- 0.15 dB loss per meter
-- 'x' meters of cable
-- 0.15x dB of loss.
If 'x' happens to be, say, 72 meters, then the loss is (72) (0.15) = 10.8 dB .
and 10 ^ (-10.8/10) = 10 ^ -1.08 = 0.083 = <em>8.3%</em> <== <u>That's</u> how much light
he'll have left after 72 meters, and all he had to do was a simple multiplication.
Sorry. Didn't mean to ramble on. But I do stuff like this every day.
