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They said wake me up at 6 AM (lemmy.world)

submitted 10 months ago by Sgt_choke_n_stroke@lemmy.world to c/memes@lemmy.ml

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[-] Enkers@sh.itjust.works 54 points 10 months ago* (last edited 10 months ago)

I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don't think they'd get hot enough to burn you, though. (Don't take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.

Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.

Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)

[-] Shard@lemmy.world 12 points 10 months ago

How long did you keep the nail on the battery for?

A 9V battery can be used as a foam cutter.

Styrofoam and most art foams melt at about 200°C

https://youtu.be/4Hj9PJstexk?si=_NEMZZU4Yu_CSN0a

[-] Enkers@sh.itjust.works 9 points 10 months ago* (last edited 10 months ago)

About 10-20s, I left it on until it didn't seem to be getting much hotter. I also didn't want the battery to overheat and fail catastrophically. I think because the "wire" is such a large gauge, there's not enough current for it to get seriously hot. In a foam cutter, you're passing all that current through a much smaller cross-sectional area.

Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter's case, and there's more total resistance, so more heat dissipation as well.

This is because:

A = π r²

R = ρ × L / A

So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.