Question

When sugar is dissolved in water, the amount A that re- mains undissolved after t minutes satisfies the differential equation dA=dt D!kA (k>0). If 25% of the sugar dis- solves after 1 min, how long does it take for half of the sugar to dissolve?

Answer 1

Answer:

Sugar will dissolve to half of it's amount in 2.41 minutes.

Step-by-step explanation:

The given question is incomplete; here is the complete question.

When the sugar is dissolved in water, the amount A that remains undissolved after t minutes satisfies the differential equation

$\frac{dA}{dt}=-kA, (k>0)$

If 25% of the sugar dissolves after 1 min, how long does it take for half of the sugar to dissolve.

The given differential equation is $\frac{dA}{dt}=-kA$

In other form, dA = -kA.dt

We further integrate the equation,

$\int\limits {dA}\,=\int{-kA}\, dt$

$\int\limits{\frac{dA}{A} }\,=-k\int\, dt$

lnA = -kt + c

Or $A=e^{c-kt}$

A = $e^{c}.e^{-kt}$ -----(1)

Here $e^{c}$ is a constant.

For t = 0,

$A=e^{c}$

Let $e^{c}=A_{0}$

Therefore, equation (1) will become

A = $A_{0}e^{-kt}$

If sugar dissolves 25% in 1 minutes then undissolved sugar will be

100 - 25 = 75%

Now from the equation

$0.75A_{0}=A_{0}e^{-k\times 1}$

$e^{-k}=0.75$

By taking natural log on both the sides

$ln(e^{-k})=ln(0.75)$

k = 0.2877

Now we have to calculate the time to dissolve half of the sugar, that means half the sugar will be undissolved.

Form the equation,

$0.5A_{0}=A_{0}e^{-0.2877t}$

$0.5=e^{-0.2877t}$

ln(0.5) = $ln(e^{-0.2877t})$

0.6931 = 0.2877t

t = 2.41 minutes