Answer:
The question has some other details missing, but here are the details ;
Write a differential equation for the time (measured in minutes) evolution of:
a) The total mass (in kilograms) of the chemical in the pond dm/dt =
b) The concentration (in kg/liter) of the chemical in the pond: dc/d t=
c) The concentration (in grams/liter) of the chemical in the pond: dc/dt =
d) The concentration (in grams/liter) of the chemical in the pond, but with time measured in hours: dc/dt =
Step-by-step explanation:
The detailed steps from first principle and the solution of each differential equation is as shown in the attached file.
starting from Concentration = mass/volume ; mass = volume x concentration, and then find the differential of both sides to generate the solved differential equation.
Suppose that its chemical compound's mass anywhere at moment t is m kg. The equation for total mass is as follows:
[tex]\to \frac{dm}{dt} = \text{inflow rate} - \text{outflow rate}[/tex]
Throughout this case, the rate of chemical intake is 0 and the rate of chemical outflow is [tex]200 \ \frac{lit}{min}[/tex], which removes [tex]70\%[/tex] of the chemical.
Therefore, the calculation of the outflow rate:
[tex]\to 200 \ \frac{lit}{min} \times \frac{m}{1000} \frac{kg}{lit} \times 70\% \\\\\to \frac{2m}{10} \frac{kg}{min} \times \frac{70}{100}\\\\\to \frac{14m}{100} \frac{kg}{min} \\\\\to \frac{7m}{50} \frac{kg}{min} \\\\[/tex]
So, calculating the differential equation:
[tex]\to \frac{dm}{dt}= 0- \frac{7m}{50}= - \frac{7m}{50} \ \frac{kg}{min}[/tex]
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