Re: NANFA-- physic problem for aquarists
Wed, 27 Dec 2000 16:01:08 EST

I knew I should have stayed away from this problem hehe. if we were to
include the calories to change the temperature of water to steam then we
would need Centigrade. however in this case we are lucky.

F = ( 9/5 ) * C + 32

we know 0 C is 32 F so that works. and 100 C is 212 F so that works. so
our equation is correct.

now the question involves a change in F or called delta F we see that a
change in F is proportional to change in C

dF = ( 9/5 ) dC where the d is used to indicate change or delta.

sine the final answer is a change of 3 over a change of 17 then if we used C
instead of F we would have a division of the ( 9/5 ) on top and bottom of
the fraction and thus the 9/5 would cancel out leaving the answer 3/17. so
we have the correct answer for this problem.

also the problem answer makes sense. we start with 17 gallons add 3 gallons
of colder water and the temperture goes down a little.

because we are doing a simple change or delta, we do not need absolute
temperature in this case.

the above is my humble opinion. now I have to forget Phyusics and think
about painting. hehe. my thoughts are that it is too cold out to paint

george the mad scientist.

In a message dated 12/27/00 2:46:04 PM Eastern Standard Time, writes:

<< Subj: Re: NANFA-- physic problem for aquarists
Date: 12/27/00 2:46:04 PM Eastern Standard Time
From: (William Allen jr.)

Wouldn't the problem have to be worked in absolute (K) degrees?
----- Original Message -----
From: <>
To: <>
Sent: Tuesday, December 26, 2000 9:11 PM
Subject: Re: NANFA-- physic problem for aquarists

> now I could be in trouble. hehe. but it seems like if you have x gal at
> and y gal at 55 and added them together we would have ( x+y ) at 72.
> 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. >>

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