As a fishkeeper, I would like an efficient way to mix say $25$ litres of water by using $X$ amount of tap water at $10-15$ degrees $C$ and kettle water of $100 C$. At the moment I'm guessing and tinkering will excel. In my head I've figured out that I need the ratio of $n$ parts of $A$ to $n$ parts of $B$. So I may have $3:2$ ratio so that's $5$ parts in total. So I need to divide desired litres($25$) by $5$ to get the multiplier of $5$. So my proportion is $15:10$.
What I'm struggling with is how do I plug in the desired temp. I'm sure I'm missing something very simple here, I feel like I've been hypnotized to count to ten but told there's no such number $7$. My spreadsheet takes the input volumes and Temps and gives me the output litres and temperature and I have to tweak each input (goal seeking) to get my desired output. I hope this isn't too simple for anyone to help me or to tell me off.
You need both of the following to hold true:
1) desired temp = [(amount of tap water)* 10 + (amount of kettle water) * 100]/25
AND
2) 25= (amount of tap water) + (amount of kettle water).
Solving for amount of kettle water in the second equation you find:
25-(amount of tap water) = (amount of kettle water)
Substituting this into initial equation:
desired temp = (amount of tap water)* 10/25 + (25-(amount of tap water)) * 100/25
**
Rearranging for a more desirable solution:
(amount of tap water)= [5(desired temp) - 500 ]/ (-18)
** Example: If desired temp was 60C then you would need 11.11L of tap water.