332 Relation heiween the Specific Gravities and 

 Solution. 



Here, 129*48 x -006157 is = 'ZSTS. 



Or, 3-7893692 + 2-1122027 (= log. of 129*48) =3 

 1-9015719 = log. of -7972. 



Which number being found in Column VIII of the table 

 tnarked " heat 48°," gives 887 for the correspondent spe- 

 cific gravity in Column II, and which is accordingly that 

 of the hquor. 



2. What is the specific gravity of a liquor at 75°, which 

 is 62-16 (or 37-84 U.P.) at that temperature? 

 Solution. 



In this case 62-16 x -006157 (or the natural number 

 answering to the logarithm 1-5828802, being the sum of 

 1-7935110, the log. of 62-16 and "3-7B93692) is -3827, 

 which being found in Column VIII of the table marked 

 *' HEAT 75°," gives 9531- in Column II for the specific 

 gravity of the liquor at tliat temperature, 



§ 34. Problem IV. — The per-centoge of a liquor at a 

 given temperature being given, to Jind that whicli it pos- 

 sesses at any other temperature. 



Pkactical Rule. 

 Multiply the given per-ceni age of the liquor ly '00'6I57 

 (either by means of natural numbers or by the addition of 

 3-7893692 to its logarilhmj, as in Probicm III. Fi?id the 

 product in Column VIII of the table corresponding to its 

 teynperatiire, and take out the correspondent numbers from 

 Column I. Enter the table adapted to the temperature at 

 which its per-centage is required with these numbers ; take 

 cut the correspondent number from Column VIII aiid mul- 

 tiply it by 162-42, as in Probicm II. Tlie product Is the 

 per-centage at tJtat temperature. 



Example. 

 What is the per-centage of brandy at 75", wliich Is 105 

 (or 3 0.P.) at 33°? 



Solution. 

 105 X -006157 = -6465. 

 And -6465, in Column VIII of the table for 33°, answers 

 to a compound of 100 + 73*8 in Column I. 



This same compound being found in Column I of the 

 table for 75°, gives -633 7 in Column VIII ; and this mul- 

 fiplicd by 162-42 is ;^ 102-93, the required per-ceutage at 

 that temperature. 



