the Sirerigtlis and Values of Spirituous Liquors. 333 



If the gallon of proof spirit at 60", therefore, be taken as 

 the standard, this liquor loses more than two per cent, in 

 its real value per gallon by the elevation of its ten)perature 

 from 33° to 75". 



§ 35. Problem V. — The per-centage and temperature of 

 a liquor leing given, to find the concentration per cent., or 

 that diminution in measure ivhich luould take place on re- 

 ducing 100 parts of it to proof, or proof to 100 parts of 

 such liquor at the same temperature. 



Practical PiUle. 



Firstly, Look into Column I of the tahle adapted to the 

 eiven temperature for the liquor answering to '6157 in Co- 

 lumn Fill (this leing proof in all the tables) ; divide the 

 correspo?ident mnnher in Column VI by that in Column V; 

 multiply the quotient ftvhich at the temperature of 60" is 

 '027768J by the given per-centa^e, and the product ivill 

 be the concentration of the equivalent quantity of proof, or 

 that contained in 100 parts of the given liqnor, if it were 

 inade np from M. Gilpin's alcohol, according to the suppo- 

 sition on ivhich his tables are founded. 



Secondly, Look for the given liquor in the same table 

 Chaving first found its correspondent number in Column VIII 

 ly Problem HI), and divide the ntimber in Column VI 

 ("taking all the three figures as integers) by that in Co- 

 lumn V, and the difference between the former product and, 

 this quotient will be the concentration required. 

 Example. 



What is the concentration per cent, on reducing rum of 

 145 (or 45 O.P.) to proof at 77°-^ 

 Solution. 



Firstly, Against '6157 in Column VIII (which, as be- 

 fore mentioned, indicates proof), in the table for 77°? the 

 correspondent numbers in Column VI and V are 4.21 and 

 160-3 7 ; the quotient of the former by the latter is '02625 

 and -02625 X 145 is = 3-8065. 



Secondly, Against -8928 in Column VIII of the same table, 

 the correspondent numbers in Column VI and V arc 1-26 



126 . 



and 110*9 respectively, and -^-^ is = 1-14. 



And 3-80C5 — ri4 is = 2-667, or 2f. 



The concentration, therefore, under th(;se circumstances, 

 would be 2|- per cent, on the quantity of the over-proof re- 

 duced, and it would therefore require 47-J per cent, of water 

 to reduce it. 



§36. 



