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
