Roedy Green <see_website@[EMAIL PROTECTED]
> wrote in
news:t4oa54h6a9ojeli3h0sm4a5kh593r9isbu@[EMAIL PROTECTED]
> On Sun, 08 Jun 2008 14:18:30 +0200, Daniele Futtorovic
><da.futt.news@[EMAIL PROTECTED]
> wrote, quoted or indirectly quoted
> someone who said :
>
>>ot high enough to heat anything,
>
> That is a very unscientific statement.
>
> Please show your physics before you patronisingly spit at me for even
> considering the problem. Obviously, the size of the object heated is
> of considerable im****tance. A shorted dry cell gets plenty hot.
>
> To pop a single kernel of corn would take considerably less than the
> amount of energy needed to convert an equal volume of water to steam.
>
> I don't think you can't categorically dismiss the question without a
> little back of the envelope math. In any case, I get really really
> annoyed with folks who try to squash curiosity in others. There is
> nothing wrong with asking questions even when the answers (including
> the wrong answers) are "obvious" to some.
>
> I hope you don't have any kids.
>
getting a BackOfEnvelope estimate is always a good place to start. But it
would help to have better numbers to base this on
> As a ballpark, I will guess you need to va****ise a milligram of water
> to pop a kernel.
>
Why this number?
> The latent heat of va****isation is 2260 joules per gram.
>
> So I would need roughly 2 joules to pop one kernel.
>
Assuming all the energy went into a particular milligram of water in the
the kernal. The average kernel weighs 140 milligrams and has a 13-14%
moisture content. So about 18 milligrams of water. This also ignores the
fat content. Also much of the energy would transfer from the warmed water
to the solids of the kernel. It would probably be better to estimate
temperature increase and increased va**** pressure.
> A joule is a watt-second.
>
> You can get 3 watt cell phone boosters, but most cell phones are about
> 1 watt according to
> http://electronics.howstuffworks.com/cell-phone-radiation1.htm
>
Actually, most are a good deal below that. They moderate the amount of
power based on signal strength. And with most output numbers quoted this
way, you may need to convert to average power from peak power. I'm not
sure and cannot bother to check.
> If you put three of them together you would have available 3 watts.
>
> Presuming all the power went into the heating the kernel, I should be
> able to pop a kernel within a second.
>
But the output is roughly omni-directional so only a small fraction of
that energy would reach the kernel. Or your head. So perhaps 50% would be
a better number
> So I think an experiment is in order to see if this is a hoax, after
> you check my arithmetic of course.
>
>
Or let someone else do it for you.
http://www.snopes.com/science/cookegg.asp
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