Question

Let's assume raspberries are 10 wt% protein solids and the remainder water. When making jam, raspberries are crushed and mixed with sugar, in a 45:55 berry to sugar ratio, by mass. Afterward, the mixture is heated, boiling off water until the remaining mixture is 0.4 weight fraction water, resulting in the final product, jam. How much water, in kilograms, is boiled off per kilogram of raspberries processed

Answer 1

Answer:

The mass of water boiled off is $0.0 \overline{185}$ kg

Explanation:

The given percentage by weight of protein solids in raspberries = 10 weight%  

The ratio of sugar to raspberries in ja-m = 45:55

The mass of the mixture after boiling = 0.4 weight fraction water

Let 's' represent the mass of sugar in the mixture, and let 'r' represent the mass of raspberry

The mass of raspberry, r = 1 kg

The percentage by weight of water in raspberry = 90 weight %

The mass of water in 1 kg of raspberry =  90/100 × 1 kg = 0.9 kg

The ratio of the mass of sugar to the mass of raspberry in jam = r/s = 45/55

∴ s = 1 kg × 55/45 = 11/9 kg

The mass of the mixture before boiling = 1 kg + 11/9 kg = 20/9 kg

The weight fraction of water in the remaining mixture after boiling = 0.4 weight fraction

Let 'w' represent the mass of water boiled off, we have;

(0.9 - w)/(20/9 - w) = 0.4

(0.9 - w) = 0.4 × (20/9 - w)

0.9 - w = 8/9 - 0.4·w

9/10 - 8/9 = w - 0.4·w = 0.6·w = (6/10)·w

(81 - 80)/(90) = (6/10)·w

1/90 = (6/10)·w

w = ((10/6) × 1/90) = 1/54

w = 1/54  

The mass of water boiled off, w = (1/54) kg = $0.0 \overline{185}$ kg