Answer to Question #305390 in Differential Equations for Daemus

Question

Under a certain condition , cane sugar in water is converted into dextrose at a rate proportional to the amount that is unconverted at any time . If 75 kg at t =; 0, 8 kg are converted during the first 30 minutes , find the amount converted in 2 hours .

Answer 1

Let $m_0 = 75kg$ be the initial mass of the shugar, $m$ - is the amount of converted sugar. Thus, the differential equation of the process will be:

$$\dfrac{dm}{dt} = r(m_0-m)$$

where $r$ is the coefficient of proportionality. Solving the euqiation, obtain:

$$m = m_0(1 - e^{-rt})$$

Since $m = 8kg$ at $t = 0.5hrs$, obtain the coeffitient $r$:

$$r = -\dfrac{1}{t}\ln(1-\dfrac{m}{m_0} ) = -\dfrac{1}{0.5hrs}\ln(1-\dfrac{8kg}{75kg} ) \approx 0.23hrs^{-1}$$

At $t = 2hrs$ obtain:

$$m = 75kg\cdot (1-e^{-0.23hrs^{-1}\cdot 2hrs})\approx 48kg$$

Answer. 48kg.