Answer:
a) 
b) 10.643 kg
Step-by-step explanation:
Solution:-
- We will first denote the amount of salt in the solution as x ( t ) at any time t.
- We are given that the Pure water enters the tank ( contains zero salt ).
- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min
- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.
- The ODE is mathematically expressed as:
( salt flow in ) - ( salt flow out )
- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0
- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).
- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.
- So any time ( t ) the concentration of salt in the 5,000 L is:
- The amount of salt leaving the tank per unit time can be determined from:
salt flow-out = conc * V( flow-out )
salt flow-out = 
salt flow-out = 
- The ODE becomes:
- Separate the variables and integrate both sides:
- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:
- The solution to the ODE becomes:
- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:
- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg
