A newly constructed fish pond contains 1000 liters of water. Unfortunately the pond has been contaminated with 5 kg of a toxic chemical during the construction process. The pond's filtering system removes water from the pond at a rate of 200 liters/minute, removes 50% of the chemical, and returns the same volume of (the now somewhat less contaminated) water to the pond. 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/dt =
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 =

A) dm/dt = 5 - 0.5t (kg)

B) dc/dt = 0.005 -0.0005t (kg/liter)

C) dc/dt = 5 - 0.5t (gram/liter)

D) dc/dt = 5 - t/1200 (gram/liter)

Step-by-step explanation:

The initial mass of the chemical is 5kg in the 1000liter tank.

Since 200liters are removed per minute for purification, it means 1/5th of the tank's volume is removed which implies 1kg of chemical is in it.

But 50% (0.5kg) of the chemical is removed before returning the water.

Hence, the mass of chemical at any minute is initial mass - removal rate

A) dm/dt = 5 - 0.5t (kg)

B) Concentration in 1000 liters of water is mass ÷ volume of water = (5 - 0.5t) / 1000

dc/dt = 0.005 - 0.0005t (kg/liter)

C) Concentration in gram/liter = concentration in kg/liter x 1000. Since 1kg = 1000g

dc/dt = 5 - 0.5t (gram/liter)

D) Previous answers have had unit of t in minutes. So, to convert to hours, we simply divide the time component by 60

dc/dt = 5 - t/1200 (gram/liter).