In a tank of l00 liters of brine containing dissolved salt. Pure water is alloved to run into the tank at the rate of 3 li/min . Brine runs out of the tank at rate of 2 liters minute. The instantaneous concentration in the tank is kept uniform by stirting. How much salt is in the tank at the end of 1 Hr?

Name: In a tank of l00 liters of brine containing dissolved salt. Pure water is alloved to run into the tank at the rate of 3 li/min . Brine runs out of the tank at rate of 2 liters minute. The instantaneous concentration in the tank is kept uniform by stirting. How much salt is in the tank at the end of 1 Hr?
Uploaded: 2022-02-23T17:22:57+00:00
Description: In a tank of l00 liters of brine containing dissolved salt. Pure water is alloved to run into the tank at the rate of 3 li/min . Brine runs out of the tank at rate of 2 liters minute. The instantaneous concentration in the tank is kept uniform by stirting. How much salt is in the tank at the end of 1 Hr?

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Orlando Paz · Accepted Answer

Let Q= amount of salt in the mixture  100+(3−2)t=100+t= volume of mixture at any time t  q100+t= concentration of salt  dqdt= rate of gain-rate of loss  dqdt=0−2(q100+t)  ∫dqdt=−2∫dt100+t+c  ln⁡q=−2ln⁡(100+t)+c  ln⁡q=−22ln⁡(100+t)2+c  When t=0:q=50kg  ln⁡50=−ln⁡(100+1)2+c  c=13.12236  When t=60min,Q=?  ln⁡q=−ln⁡(100+60)2+13.12236  ln⁡q=2.972  q=19.53kg

veiga34 · Accepted Answer

Suppose that the amount of salt in the mixture be Q. The volume of mixture at time t will be as follows: 100+(3−2)t Recall the fact that the concentration of salt will be given as follows: Concentration of salt=Amount of salt in mixtureVolume of mixture at any time t =Q100+(3−2)t =Q100+t Also, the rate of gain minus rate of loss will be dQdt, where Q is the amount of salt in the mixture. Here rate of gain is 0 and rate of loss is 2(Q100+t). That is, dQdt=0−2(Q100+t) dQQ=−2(dt100+t) ∫dQQ=−2∫(dt100+t) ln⁡Q=−2ln⁡(100+t)+C ln⁡Q=−2ln⁡(100+t)2+C Initially the amount of salt in the mixture is 55 kg. That is, Q=55,t=0. Substitute Q=55,t=0 ∈ ln⁡Q=−ln⁡(100+t)2+C: ln⁡55=−ln⁡(100+0)2+C C=13.218 By the end one 1 hours, t will be 60 minutes. Substitute t=60 ∈ ln⁡Q=−ln⁡(100+t)2+13.218: ln⁡Q=−ln⁡(100+60)2+13.218 ln⁡Q=3.06765 eln⁡Q=e3.06765 Q=21.4913 Thus, the amount of salt in tank after one hour is 21.4913 Kg.

Vasquez · Accepted Answer

100+(3−2)t=100+t= volume of mixture at any time t q100+t=dqdt=dqdt=0−2(q100+t)∫dqdt=−2∫dt100+t+cln⁡q=−2ln⁡(100+t)+cln⁡q=−22ln⁡(100+t)2+cWhen t=0:q=50kgln⁡50=−ln⁡(100+1)2+cc=13.12236When t=60min,Q=?ln⁡q=−ln⁡(100+60)2+13.12236ln⁡q=2.972q=19.53kg

In a tank of l00 liters of brine containing dissolved salt.

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