A tank originally contains 100 gal of fresh water. Then water containing 1/2 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 8 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 min.

Question

Arabella Singleton

Physics

23 June, 18:02

A tank originally contains 100 gal of fresh water. Then water containing 1/2 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 8 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 min.

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Answers (1)

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Zaire Heath · Answer 1

1,98 lb of salt

Explanation:

Gal at the beginning = 100 gal fresh water.

Inside rate1 = 2 gal/min x 10min x 1/2lb/gal = 10 lb at the end of ten minutes.

outside rate1 = 1 lb / 100 gal = 0,001 lb / gal x 10 min = 0,01 lb at the end of ten minutes.

Final quantity of salt at ten minutes = 10 - 0,1 = 9,9 lb.

Inside rate 2 = 8 gal/min x 0 lb = 0lbatthe end of next 10 minutes, + 9,9 lb from the initial state = 9,9 lb.

Outside rate 2 = 9,9 lb / 100 gal = 0,099 lb/gal x8 gal/min x 10min = 7,92 lb.

Final quantity = 9,9 lb - 7,92 lb = 1,98 lb of salt.