220 MILK HYGIENE 



milk, as well as of the whey. Feser gives the following 

 example : 



Start with 9 liters of milk of the composition : 3.95 

 per cent., fat; 8.9 per cent., solids not fat; 1.031, specific 

 gravity. 



To this add 1 liter of water of this composition : per 

 cent., fat; per cent., solids not fat; 1.000, specific 

 gravity. 



This gives 10 liters of adulterated milk of the com- 

 position: 3.55 per cent., fat; about 8.0 per cent., solids 

 not fat ; 1.028, specific gravity. 



Further evidence of this adulteration is found in the 

 lower specific gravity of the whey and in the fact that 

 the specific gravity of the dry solids (m) and the fat 

 content (p) of the solids are, practically, normal; in the 

 above example, then : ®* 



ST 1.031 X 12.85 13.24 



ST — (lOOS — 100) 1.031 X 12.85 — (100 X 1-031 — 100) 10.14 

 St 1.028 X 11.65 



1.307 



St — (100s — 100) 1.028 < 11.55 — (100 X 1.028 — 100) 9.073 



P 3 95 



P = -i- X 100 .^ X 100 = 30.738 



f 3 55 



p = -:- X 100 = rf^ X 100 - 30.736 



t 11.00 



If milk samples are available which may be justly com- 

 pared with the milk under suspicion (herd samples 

 taken under fixed conditions ; milk from the same large 

 herd; other milk sent by the same shipper or from the 

 same can in the dealer's possession), then the percent- 

 age of water added can be computed according to Vo- 

 gel 's formula : 



x=-^X 100-^100 



^•' [M and P represent the specific gravity of the total solids and 

 the percentage of fat in the total solids of whole milk, and m and 

 p the same factors in adulterated milk.] 



