Answer:
Explanation:
The volume of fluid in the tank =350 liters
Initial Amount of Salt, A(0)=50 grams
<u>Amount of Salt in the Tank
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Pure water is then pumped into the tank at a rate of 5 L/min. Therefore:
=(concentration of salt in inflow)(input rate of brine)
=(concentration of salt in outflow)(output rate of brine)
Therefore:
We then solve the resulting differential equation by separation of variables.
Taking the integral of both sides
Recall that when t=0, A(0)=50 grams (our initial condition)
Therefore, the number of grams of salt in the tank at time t is: