75 X + 55 Y = ( X + Y ) 72
thus Y = ( 3/17 ) X
so if there is 17 gallons at 75 then the answer is 3 gallons to fill a 20
gallon tank. any more than 3 gallons is too much.
I hate to say this but I am a Physicist. U of Illinois Champaign Urbana.
hehe.
but now I do something useful, I paint houses. hehe. I thought Physics was
going to tell me how the world worked. what was happening. but I found out
they don't know. they are trying to figure it out too. all they were doing
was taking each problem and dividing it in two. hehe. I became discouraged.
Now I know what the world is all about. it is a place of insane madness.
hehe. the mind is insane. everything is insane. with that comes happiness.
hehe. as long as I keep reminding myself that I KNOW NOTHING.
there is one problem with my solution. it is suppose to be worked out in
Centigrade not Farenheit. but from the answer it is obvious that the the
conversion units drop out at the end. but a formal solution would require
the conversion to Centigrade before working it out.
If I am wrong they might want to take my degree in Physics back. hehe.
george arndt
presently aquarist former physicist
In a message dated 12/26/00 8:59:32 PM Eastern Standard Time,
billandbonnie_at_peoplepc.com writes:
<< Subj: NANFA-- physic problem for aquarists
Date: 12/26/00 8:59:32 PM Eastern Standard Time
From: billandbonnie_at_peoplepc.com (William Allen jr.)
Sender: owner-nanfa_at_aquaria.net
Reply-to: nanfa_at_aquaria.net
To: nanfa_at_aquaria.net
OK, all you fishkeepers with a passing interest in physics (or physicists
who keep fish), this ones for you -
Given an aquarium holding X amount of water at 75 degrees F, what is the
maximum amount of water (in terms of X) at 55 degrees F that you could add
and lower the temperature of the total less than 3 degrees F?
For the purposes of the exercise, the air temperature surrounding the
aquarium is also 75 degrees F, and discount any possible radiation or
absorption from rocks, the glass, decorations, etc.
Bill Allen >>
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