Question

A tank contains 3000 L of pure water. Brine that contains 15 g of salt per liter of water is pumped into the tank at a rate of 25 L/min. The concentration of salt after t minutes (in grams per liter) is C(t) = 15t 120 + t . As t → [infinity], what does the concentration approach? g/L

Answer 1

Answer:

15 g/L

Step-by-step explanation:

Since Brine is added to the tank containing 15g/L at a rate of 25 L/min, the rate at which the quantity of salt in the tank increases is 15 g/L × 25 L/min = 375 g/min. So, in t minutes, the mass of salt in the tank, m is 375 g/min × t min = 375t

Also, the volume of the tank increases by 25 L/min, in t minutes, its volume increases by 25 L/min × t min = 25t. So, the new volume, V is 3000 + 25t.

Since we know, concentration C(t) = m/V = 375t/(3000 + 25t)

dividing through by 25, we have

C(t) = 375t/25/(3000/25 + 25t/25)

C(t) = 15t/(120 + t)

To find the limit as t ⇒ ∞, we divide C(t) by t

C(t) = 15t/t/(120/t + 1)

C(t) = 15/(120/t + 1)

putting t = ∞, we have

C(∞) = 15/(120/∞ + 1)

C(∞) = 15/(0 + 1)

C(∞) = 15/1

C(∞) = 15 g/L

So, as t  t → [infinity], the concentration approaches 15 g/L