## Salt Tank Differential Equation

Ok so this is the question

A tank contains 100 kg of salt and 2000 L of water. A solution of a concentration 0.025 kg of salt per liter enters a tank at the rate 5 L/min. The solution is mixed and drains from the tank at the same rate.

A.What is the concentration of our solution in the tank initially

B.Find the amount of salt in the tank after 4.5 hours

C.Find the concentration of salt in the solution in the tank as time approaches infinity.

I found A to be .05 and C to be .025. But I'm not sure how to get B... I know I have to set up an equation but I'm not very good at these word problems and could use some help setting up the equation and answering B so any help would be great!

Thanks!

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 Rate of change of salt in tank = amount of salt in/min - amount of salt out/min For the amount of salt in, just plug in the values you are given $$Concentration * flowrate = \frac{.025kg}{L}\left(\frac{5L}{min}\right)$$ For salt out, the concentration will depend on t. More explicitly, since concentration is kg/L, and the number of liters will never change (since the rate of flow in and the rate of flow out are the same), the amount of salt will change with time.
 Thread Tools

 Similar Threads for: Salt Tank Differential Equation Thread Forum Replies Calculus & Beyond Homework 1 Calculus & Beyond Homework 7 Calculus & Beyond Homework 3 Calculus & Beyond Homework 4 Differential Equations 5