Using mass-balance solve following problems (include proof table)

Note: unless specified otherwise, assume that skim milk, condensed skim milk and skim milk powder have 0% fat, and there is 9% milk snf in skim or in the skim fraction of liquid milk/cream and butter products.

a) Desired : 800 kg of ice cream mix @ 11% fat, 10.5% msnf, 11% sucrose, 5% corn syrup solids, 0.25% stabilizer. Available : Butter @ 82% fat, skim milk, condensed skim milk @ 32% msnf, sucrose, corn syrup solids, stabilizer.

b)Desired : 3500 kg of ice cream mix @ 10% fat, 10.5% msnf, 13% sucrose, 3% corn syrup solids, 0.2% stabilizer. Available : Cream @ 34% fat; skim milk powder @ 97% msnf; sucrose, corn syrup solids, stabilizer, water.

c)Desired : 1800 kg of ice cream mix @ 12% fat, 10% msnf, 15% sucrose, 0.3% stabilizer. Available : Cream @ 41% fat; milk @ 3.8% fat; condensed skim @ 34% msnf; sucrose; stabilizer.

d)Desired : 400 kg of ice cream mix @ 14% fat, 10% msnf, 12% sucrose, 3% corn syrup solids, 0.2% stabilizer Available : Cream @ 36% fat; milk @ 3.9% fat; condensed skim milk @ 32% msnf; liquid sucrose @ 65% solids; liquid corn syrup @ 75% solids; stabilizer

Example:

EXAMPLE PROBLEM 1 Desired : 100 kg mix @ 13% fat, 11% MSNF, 15% sucrose, 0.5% stabilizer, 0.15% emulsier On hand: Cream 9 40% fat, 5.4% msnf,' skimmilk @ 9% msnf,' skimmilk powder @ 91% msnf; sugar; stabilizer; emulsier. Solution: {Note: only one source of fat, sugar, stabilizer, and emulsifier, but two sources of solids-natlat} Cream 100 kg mix it 13 l_(g fat x 100 l_cg cream = 32.5 kg cream 100 kg mix 40 kg fat Sucrose 100 kg mix it 15 kg sucrose = 15 kg sucrose 100 kg mix Stabilizer 100 kg mix it 0.5 stabilizer = 0.5 kg stabilizer 100 kg mix Emulsifier 100 kg mix it 0.15 kg emulsier = 0.15 kg 100 kg min Skim milk and Skim powder, Note: two sources of MSNF Now, let x = skim powder, y = skim milk MASS BALANCE (All the components add up to 100 kg) x+ y = 100 - (32.5 + 15 + 0.5 + 0.15) (1) MSNF BALANCE (Equal to 11% of the mix and coming from the skim milk, the skim powder, and the cream) 0.97 x + 0.09 y = .11(100) - (.054 x 32.5) (2) x+ y = 51.85 so y = 51.85 - x from (1) 97 x + .09 y = 9.245 from (2) 97 x + .09 (51.85 - x) = 9.245 substituting 97 x - .09 x + 4.67 = 9.245 .88 x = 4.58 x = 5.20 kg skim powder y = 46.65 kg skim milkPROOF: Ingredient Total wt. Wt. of Fat Wt. of SNF Wt. of Total Solids (kg) (kg) (kg) (kg) Cream 32.50 13.00 1.75 14.75 Skim milk 46.65 4.20 4.20 Skim milk 5.20 5.04 5.04 powder Sucrose 15.00 15.00 Stabilizer 0.50 0.50 Emulsifier 0.15 0.15 Totals 100.00 13.00 11.00 39.64 Note: 13% fat + 11% snf + 15% sucrose + 0.50% stab. + 0.15% emul. = 39.65% TS