You have two two-liter bottles,
one (marked W) containing one liter
of pure water and one (marked A)
containing one liter of pure alcohol.
The two things you can do are to
pour liquid from either bottle down
the drain or pour liquid from one
bottle into the other, but you must
always keep at least one liter of liquid
in bottle A.
What is the minimum
alcohol concentration that can be
achieved in bottle A? Assume that
there is no volume change on mixing
water and alcohol.
Yes, Kenny M got it right. Final concentration n =1/e
This is, in fact, one of the classical examples of the "naturalness"
of the exponential function, arising when the rate of change of a quantity is proportional to the quantity. (e.g, Bernoulli found e while contemplating compound interest.)
If you run the dilution by putting in water at the same rate you are taking out mixed solution (say k=1 liter per min), then the change in concentration n is proportional to the current concentration:
dn/dt = -k n
dn/n = - k dt
integrate both sides
ln n = -k t + c
n(t) = n0 exp(-kt) + C
with n0 =1, k=1 and C=0, and with the water running out after
time t=1 minute, we get n(1) = 1/e
Note: this calculation was part of the solution to my real life problem:
http://perplexus.info/show.php?pid=11391&op=sol
Edited on February 3, 2025, 6:13 pm
soln
