# Heating metals with resistance

Tags:
1. Apr 6, 2017

### JoeSalerno

This question is purely theoretical, so don't worry about safety or doability. Is there a way of calculating how many watts it takes to heat different metals? To be specific, how many amps and bolts it would take to get lead to its melting point. Thanks in advance.

2. Apr 6, 2017

3. Apr 6, 2017

### JoeSalerno

It actually sounds pretty similar to an induction furnace, but this would be without direct contact. It would be from an electrical arc. For simplicity, the material would be about a .75 cm cube of lead.

4. Apr 6, 2017

### Nidum

Lead melts in the same way as ice . So you need to use specific heat and then latent heat of fusion to find out how much heat is actually needed to melt the cube

Factor in then the efficiency of the heating process .

I know that this is a theoretical question but please note anyway that arc melting of lead is not recommended for a whole raft of practical and safety reasons .

Lead is usually just melted in a flame heated crucible .

Small amounts can be melted with a soldering iron for that matter .

5. Apr 6, 2017

### JoeSalerno

So, I found the specific heat (0.128 J/kg Kelvin) and the latent heat (23 kJ/kg). How would I find a value for the efficiency?

6. Apr 7, 2017

### Nidum

You'll probably need to do some experiments .

Exactly how is the block to be heated ?

7. Apr 7, 2017

### JoeSalerno

I was afraid somebody might say that. The cube of lead would be heated via an electrical arc.

8. Apr 7, 2017

### Staff: Mentor

Do you understand the safety issues that Nidum points out?

9. Apr 7, 2017

### JoeSalerno

I've thought of some calculations I could use, if I find the formulas.
1. If I use specific heat as well as latent heat of fusion in the correct formula, I believe that tells me how much energy the system needs.
2. I'd calculate the distance of an electrical arc and the wattage that comes through to the other end.
3. Some sort of formula for knowing how to calculate the watts needed to heat a metal using joule heating.

This would tell me how much power I need to melt the lead, of which I can determine what arc distance and for how long the energy has to flow. If that made any sense, does that sound right?

10. Apr 7, 2017

### JoeSalerno

Yeah, I figured this was pretty dangerous, that's why I want to avoid doing real experiments. In the beginning of the thread I stated that I was going to disregard any feaseability or safety issues because this is purely theoretical. I don't plan on doing any of this irl.

11. Apr 7, 2017

### Rive

The question is more related to thermodinamics than electronics. The 'watt' is just the speed of providing energy (by electronics means): the temperature it can reach purely depends on the heat loss/insulation.
The 'watt' value also has effect on the speed of the heating, but then that's all.

12. Apr 11, 2017

### JoeSalerno

Is there really a way to solve this problem without doing real testing? If not, is there at least a dumbed down, high school electronics level way of solving this? If neither of those are an option, I don't have the money, materials, or the level of ignorance to test this for real.

13. Apr 11, 2017

### Rive

You have to rebuild the question first.

Specify the amount and the type of metal you want to use and ask the amount of necessary heat somewhere in the 'general physics' part (of course there are people here too who can solve this, but: even so, it's definitely not electronics).

Then you can ask here that what kind of heating can provide that heat, at what power, in what time? That part will be electronics (more or less).

As the question is now, it's quite hard to 'solve'. Now it is a bit like 'The Almighty Answer to the Meaning of Life, the Universe, and Everything'.

14. Apr 11, 2017

### CWatters

In #9 point 1 gives you the energy needed to melt the lead. Then if you were to assume no heat losses the power required depends on how fast you want to melt it. Power=energy/time. If the heat losses were say 50% then it would take roughly twice as long or you need twice the power.

Try searching for papers on the efficiency of arc smelting? I don't have access but perhaps something like this one would have data