- #1

Lord Anoobis

- 131

- 22

## Homework Statement

A vat with 2000L of beer contains 4% alcohol (by volume). Beer with 6% alcohol is pumped into the vat at a rate of 20L/min and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour?

## Homework Equations

## The Attempt at a Solution

At ##t = 0## the amount of alcohol is 80L, let the amount at time ##t## be ##y##.

##\frac{dy}{dt} = rate in - rate out##

##rate in = (20L/min)(0.06) = 1.2L/min##

##rate out = \frac{y}{2000}(20L/min) = \frac{y}{100}L/min##

Which leads to:

##\frac{dy}{dt} = \frac{1200 - y}{t}##

##\int\frac{dy}{1200 - y} = \int\frac{dt}{100}##

##-\ln|1200 - y| = \frac{t}{100} + k##

##1200 - y = Ae^{-t/100}##, yielding

##y(t) = 1120e^{-t/100}##

Substituting ##t = 1## gives ##y \approx 91.14## and converting to a percentage gives about ##4.6%## while the actual answer is ##4.9##. The general solution works fine for ##t = 0## which implies that the value of ##A## is correct. Where lies the error here?