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
outside.
george the mad scientist.
In a message dated 12/27/00 2:46:04 PM Eastern Standard Time,
billandbonnie_at_peoplepc.com writes:
<< Subj: Re: NANFA-- physic problem for aquarists
Date: 12/27/00 2:46:04 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
Wouldn't the problem have to be worked in absolute (K) degrees?
----- Original Message -----
From: <Arndtg_at_aol.com>
To: <nanfa_at_aquaria.net>
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
75
> 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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