The discussion revolves around calculating the power required to melt lead, focusing on theoretical approaches rather than practical applications. Participants explore various methods of heating, including electrical arcs and induction melting, while considering the specific heat and latent heat of fusion for lead.
Participants generally agree on the theoretical nature of the question and the need for calculations involving specific heat and latent heat. However, there are multiple competing views regarding the best method to approach the problem, the significance of safety concerns, and the role of thermodynamics versus electronics in the discussion.
Participants note limitations in their ability to conduct real experiments and express uncertainty about the efficiency of different heating methods. The discussion also highlights the need for clear definitions and specifications regarding the type and amount of metal involved in the calculations.
This discussion may be useful for individuals interested in theoretical physics, thermodynamics, and electrical engineering, particularly those exploring the heating and melting processes of metals.
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.berkeman said:Sounds like you are asking about Induction Furnaces and Induction Melting:
https://en.wikipedia.org/wiki/Induction_furnace
What type of metal, and how thick?
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?Nidum said: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 .
I was afraid somebody might say that. The cube of lead would be heated via an electrical arc.Nidum said:You'll probably need to do some experiments .
Exactly how is the block to be heated ?
JoeSalerno said:The cube of lead would be heated via an electrical arc.
Do you understand the safety issues that Nidum points out?Nidum said:please note anyway that arc melting of lead is not recommended for a whole raft of practical and safety reasons .
I've thought of some calculations I could use, if I find the formulas.Nidum said:You'll probably need to do some experiments .
Exactly how is the block to be heated ?
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.berkeman said:Do you understand the safety issues that Nidum points out?
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.Rive said: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.
You have to rebuild the question first.JoeSalerno said:... is there at least a dumbed down, high school electronics level way of solving this?