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Can someone please help me build the differential equation for this word problem?

  • Thread starter skyturnred
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  • #1
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Homework Statement



A tailing pond at a mine contains a million liters of water and 10 kg of the very toxic substance mercuric chloride (HgCl2), mixed evenly throughout the water in the pond. Water from the mining operation flows into the tailing pond at a rate of 10 liters per second, and each liter of water is contaminated with 1 mg of HgCl2, which mixes instantly into the pond water upon arrival. Unfortunately, the mining company does absolutely nothing to remove the mercury from the pond water, and allows it to exit the pond unfiltered, at the same rate of 10 L/sec. How much HgCl2 is found in a liter of pond water after this process continues for (a) an hour (b) a day (c) If the process were to go on indefinitely, how much HgCl2 would the pond end up containing?

Homework Equations





The Attempt at a Solution



I know how to do differential equations if they are given, but I cannot seem to construct one from the text above. That is the extent of my attempt at a solution, because I don't even know where to start.

Thanks in advance!
 

Answers and Replies

  • #2
tiny-tim
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hi skyturnred! :wink:

the volume of liquid is fixed, so call the weight of HgCl2 "m"

there are two things contributing to dm/dt …

the amount entering, and the amount leaving …

so write out dm/dt for each, and add

what do you get? :smile:
 
  • #3
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OK, that makes much more sense. I was trying to somehow incorporate the volume of the water.

So the amount going in is 10 kg / s. Do dy/dt going in is 10. But I'm confused about the amount going out. The initial amount leaving would be dy/dt=-0.0001, but this value obviously changes over time. So I have to somehow add a time variable in there. But I have no idea how to do this.

Thanks by the way, you have already cleared it up a lot.
 
  • #4
tiny-tim
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So the amount going in is 10 kg / s.
noooo
… I'm confused about the amount going out. The initial amount leaving would be dy/dt=-0.0001, but this value obviously changes over time. So I have to somehow add a time variable in there.
that would be difficult!

no, the "word problem" doesn't give it as a function of time, does it?

it's a function of how much HgCl2 is still in the water! :wink:
 
  • #5
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OK, hate to be hard headed but that just makes it more confusing again. I don't quite understand how this question could possibly be done without using a function of time.

The amount changing initially is + one mg (or 0.000001 kg) per liter, but there are ten liters going in every second.
 
  • #6
tiny-tim
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I don't quite understand how this question could possibly be done without using a function of time.
a differential equation may have a solution which is a function of time, but that doesn't mean the equation itself contains a function of time

eg x'' = -k2x has no function of t, but its solutions are x = Acoskt + Bsinkt :wink:
 

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