"Rule" bilge pump questions

[Nutmegmooner]

I have recently acquired a Rule 500 G.P.H. bilge pump for nada because the automatic feature doesn't work. Happily the manual override does so I'm thinkin of using it as a circulating pump with my kitchen sink as a reservoir and frozen tupperware ice blocks as needed. I know ice is discouraged, but I think that is when used with a traditional copper coil and flake bucket. Am I correct on this? I'm using a Liebig and think it will be okay. Correct me if I'm wrong.
Also I am going to put a valved by pass on the feed side, 500 G.P.H. just seems crazy...
Lastly I am planning on using a 12 volt "wall wart" as a power supply. How long do you suppose (or know) one of those little transformers will last? I have bunches of them.
Thanks for considering these queries and helping me out with your collective experience!

[ranger_ric]

6.7' of head in a $20 pump that you got for free...
I couldnt find how big of a power supply you would need. I saw one comment about a 12V 3Amp power supply. Pretty powerful pump.
I think you will find that your sink is a pretty small reservoir. I think you will need a LOT of ice to try to keep the temp down in that small of a reserve.
Perhaps you can commandeer a bigger trash can or use the bathtub.
Good Luck

[Hound Dog]

Yes, check the amperage on your power supply to see if it will handle the pump. Then also, the 6.7' lift. Don't know how tall your rig is. As Ranger Ric said though, I would look into a bigger water reservoir. Is your water really that expensive? A large trash can would serve you much better. No matter what the entire process is all a transfer of energy. There is still a cost to transfer the extra cold into the water to make ice to transfer it to the still and condense your vapor. You think, but my freezer is already running. But it runs harder using more electrical energy to convert water to ice. All the energy transfers have a cost.

[Swag]

A wall wart will probably be insufficient to power that size pump. A car battery would be better. Look at the wattage rating on the motor, and divide it by 12. That will give you the current needed.