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soldi70 [24.7K]
3 years ago
5

A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe

r minute and the well-stirred solution runs out at the same rate. How long will it be until there are 23 pounds of salt in the tank? Answer: the amount of time until 23 pounds of salt remain in the tank is minutes.
Mathematics
1 answer:
adoni [48]3 years ago
5 0

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

\frac{dC}{dt}=Ci*Qi-Co*Qo

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

\frac{dC}{dt}=-Co*Qo

Rearranging the equation, it becomes

\frac{dC}{C}=-Qo*dt

Integrating both sides

\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\

The final equation for the concentration of salt at any given time is

C=exp^{-3*t-0.693}

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}=  -\frac{-0.266}{3}=0.088

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