Water vapour transmission rate of a packaging film is 4g/m2
/day. A food product is packed in a
rectangular pouch measuring 0.12m x 0.16m. The maximum amount of moisture lost in 90 days is:
Answers
Answer:
6.912 grams
Explanation:
We calculate the area of the packaging film as follows.
Area = Length × width
Area = 0.12 × 0.16 = 0.0192 m²
according to the rate :
Rate = 4g/m²/day
1 m² = 4g
0.0192 m² = ?
0.0192 × 4g = 0.0768 g per day.
In 90 days the amount of moisture lost is given by :
0.0768 × 90 = 6.912
In 90 days the maximum amount of moisture lost is :
6.912 grams
Answer;
The maximum amount of moisture lost in 90 days is 6.912 g.
Explanation;
Water vapor transmission rate = 4 g/m^2/day4g/m
2
/day
Area of the pouch = 0.12 m\times 0.16 m =0.0192 m^20.12m×0.16m=0.0192m
2
Days till the food remained in the packet = 90 days
The maximum amount of moisture lost in 90 days is:
4 g/m^2/day\times 0.0192 m^2\times 90 day=6.912 g4g/m
2
/day×0.0192m
2
×90day=6.912g
The maximum amount of moisture lost in 90 days is 6.912 g.