Mixing Water Masses: Calculating Proportions for Resultant Mass R

[groovayness]

Proportions help needed!

Three water masses (I, II, III) are mixed together to form a resultant water mass (R).
I: T=5 deg C, S=35.50 ppt
II: T=2 deg C, S=34.50 ppt
III: T= 1 deg C, S=35.25 ppt
R: T= 3 deg C, S=35.00 ppt
What were the relative proportions of I, II, and III that were mixed together to form R?

Someone please help, I don't even know where to start...

 

[tiny-tim]

Hi groovayness!

In questions like this, start by saying:

"Let the mass of I be a, of II be b, and of III be c".

(So the mass of R will be a + b + c.)

Hint: if a combination of a at 5º, b at 2º, and c at 1º, is at 3º … what formula does that give you connecting a b and c?

 

[groovayness]

ok so would the equation be 5a+2b+1c=3(a+b+c)? but then what am i solving for?
And then where do i factor in the salt concentration (s)?

 

[tiny-tim]

Yes, that's right!

Now put the a's b's and c's together, to make 2a = b+ 2c.

And then do the same thing for the salt … then you'll have another equation for a b and c, from which you can find the ration b/c.

 

[groovayness]

ok i think i got it.

i got a:b:c is 3.5 : 5 : 1

is that correct?

 

[tiny-tim]

ok i think i got it.

i got a:b:c is 3.5 : 5 : 1

is that correct?

Noo … it should be 5 : 6 : 2.

Show us your working.

 

[aquirk]

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[tiny-tim]

new thread

see aquirk's https://www.physicsforums.com/showthread.php?t=253094"