Question

A tank contains 1000L1000L of brine with 20kg20kg of dissolved salt. Pure water enters the tank at a rate of 15L/min.15L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. Answer the following questions. 1. How much salt is in the tank after tt minutes

Answer 1

there is 15
`e^{ (-t)/(100)` amount of salt in the tank after t minute.

Explanation:

Given the data in the question;

tank contains 1000L of brine with 20kg of dissolved salt, Pure water enters the tank at a rate of 15L/min.

How much salt is in the tank after t minutes.

Now,

let x represent the amount of salt after t minutes,

so

x(0) = 15

dx/dt = rate in - rate out

Now, since the water entering is pure, the rate in is 0

rate out will be; x/1000 × 10 kg

∴

dx/dt = -x/100

`(1)/(x)`dx = -
`(1)/(100)`dt

∫
`(1)/(x)`dx = -
`(1)/(100)`∫1dt + c

lnx = -
`(t)/(100)` + c

Now since x(0) = 15

ln15 = c

lnx = -
`(t)/(100)` + ln15

x =
`e^{ (-t)/(100) + ln15`

x = 15
`e^{ (-t)/(100)`

Therefore, there is 15
`e^{ (-t)/(100)` amount of salt in the tank after t minute.