The capacity of a car\'s radiator is 8 liters. The radiator is full of a dirty,

Question

The capacity of a car's radiator is 8 liters. The radiator is full of a dirty, two-year-old 50/50 solution of antifreeze and water. The owner, deciding that it is time to replace the old antifreeze with new antifreeze, starts the car, removes the radiator cap and the drain plug, and inserts a garden hose into the radiator. The faucet is turned up just enough so that the radiator remains full, without overflowing, while the solution runs out at an estimated rate of 4 liters per minute. Since the engine is kept idling, the solution in the radiator is continually stirred. Find the minimum time that is needed to ensure that at least 99% of the old antifreeze is removed. The answer is 9.21 minutes. I need to know how to get there. Thanks in advance.

Explanation / Answer

At time 0, there are .5(8) = 4 liters of antifreeze in the 8 liters.
Let the antifreeze be A.
A(0) = 4

Since there are always 8 liters in the radiator, and the solution is flowing out at 4 liters/minute, the rate of change in antifreeze is A*4/8 = A/2
dA/dt = -A/2 t

A = C e -1/2t

At time 0, there are 4 liters of anti-freeze, so C = 4

We are trying to find when there is 1% of 4 liters left, or .04

Then, 4  e -1/2t = .04

e -1/2t = .01

Taking logs of both sides, -1/2t = ln(.01) = -2 ln(10)

t = 4 ln(10)

t 9.21034037197618 minutes, which rounds to 9.21 minutes.

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