a 100 gallon tank initially contains 60 gallons of water holding 20lbs of salt.

Question

a 100 gallon tank initially contains 60 gallons of water holding 20lbs of salt. A water solution containing 2 pounds of salt per gallon runs into the tank at the rate of 4 gallons per minute and the well-mixed solution in the tank runs out at the rate of 2 gallons per minute.
1)how much salt will be in the tank at any time t?
2)how much salt will be in the tank 10 minutes after the spigots are open?
3)how much salt will be in the tank when the tank reaches capacity?

Explanation / Answer

Initially, a 100 gallon tank contains 60 gallons of water that holds 20 lbs of salt. This is an initial concentration of 1/3 lbs/gallon. The input to the tank is a water solution with 2 lbs salt/gallon at the rate of 4 gallons/minute. Thus, the rate of increase of salt is ( 2 lbs/gallon * 4 gallons/minute) = 8 lbs/minute At time t, the well-mixed solution in the tank will contain 60+4t gallons on water holding 20+8t lbs of salt, which is a concentration of (20+8t)/(60+4t). This well-mixed solution is leaving the tank at the rate of 2 gallons/minute. Thus, the amount of solution in the tank at time t is 60+2t, i.e. 4 gallons in and 2 gallons out, wherein the solution leaving the tank has a concentration of (20+8t)/(60+4t). 1) Thus, at time t, the amount of salt in the tank is (60+2t)*[(20+8t)/(60+4t)] 2) After 10 minutes, the amount of salt in the tank is (60+20)*[(20+80)/(60+40)] = 80 lbs 3) The amount of solution in the tank is given by 60+2t, and thus the time taken to reach capacity is 60+2t=100 => t = 20 minutes. After 20 minutes, the amount of salt in the tank is (60+40)*[(20+160)/(60+80)] = 100*[180/140] = 128.57 lbs

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