Well think about it. The wort will probably have a higher heat capacity than water does.
Heat capacity is how much energy is required to raise 1 gram of material, 1 degree in temperature (units are Joules per gram per kelvin).
So if your wort has a Cp of 1.5, and water has a Cp of 1, you will have a situation where for every degree your wort cools, your water will increase 1.5 degrees. That is, provided you have equal masses of wort and water.
Let us just assume for a moment, that you are mixing your cooling water, with your wort (which is a way of approximating an ideal heat exchanger)
In addition, you are trying to cool the wort by 70 degrees, and only heat the water by 18.5 degrees.
Even if they have the same heat capacity, you will need 3.5 times as much water as you have wort.
I am assuming we aren't talking about a Heat exchanger setup, and that you are just dunking your kettle in the drum of water?
Remember that you will also have heat transfer between the surroundings and both the wort, and water. If you're water is in say, a stainless steel sink, or metal drum, you probably have 3 or 4 times the surface area of the water exposed to air, as you do the wort (this is considering both area of the container, and surface of the liquid).
So you will be absorbing more heat into your water from the surroundings, than the wort will be losing to the surroundings. So the amount of water required to cool your wort will increase.
There is a few other factors, but thats it in an approximated manner.
I hope I understood your post correctly.
A slightly neater version of the spreadsheet:
View attachment Mixing_Fluids.xls