# Heating up a wire with current

1. Apr 25, 2014

### 5carola5

I have a question about an object heating up when electricity is applied. When putting electricity (say 10 to 15 A) trough a wire the wire will heat up because of it's resistance for electricity, right?

But does anyone know how to do calculations on this? Like how to calculate how warm the wire would get in say 15 min?

I searched trough all my textbooks and the internet (I am not a good internet searcher, it should be there I guess) but I found nothing on it. Hopefully someone here can help ^^. I know some equations about the warming up proces. Like R=rho * l / A and Q=h*A*deltaT. But nothing with a time variable in it.

Hope someone knows it =)

2. Apr 25, 2014

### Simon Bridge

This is basically correct.

Yes.
Lynchpin: power dissipated by the resistor turns into heat.

3. Apr 25, 2014

### 5carola5

But in how much heat? Do you know some kind of formula (or have some idea to get it) to calculate the amound of heat (or temperature) that the wire would get in an amound of time. I would like to be able to calculate say how much my wire would heat up in 15 min or half an hour or so.

4. Apr 25, 2014

### Matterwave

If you know the resistance of the wire, then the heat dissipated by the wire is simply the power: $dQ/dt=I^2R$, expressing this in terms of resistivity, we have: $dQ/dt=I^2\rho l/A$. Assuming a cylindrical wire, per unit volume, the rate that heat is deposited into the wire is:

$$\frac{1}{V}\frac{dQ}{dt}=I^2\frac{\rho}{A^2}$$

So let's say we have a typical copper wire of thickness (radius) 1 mm, resistivity 17 nano-ohms*meter and we run 10 amps into it. Then the heat deposited, per unit time, per unit volume is:

$$\frac{1}{V}\frac{dQ}{dt}=(10A)^2\frac{17*10^{-9}n\Omega\cdot m}{(\pi*(1mm)^2)^2}\approx 170,000\frac{watts}{m^3}$$

The molar heat capacity (at STP) of copper is ~24J/(mol*K), and the density is ~9g/cm^3=9000kg/m^3. The molar mass of copper is ~64g/mol, meaning the molar density is ~140,000mol/m^3, and the heat capacity per unit volume of copper is ~3,400,000J/(m^3*K).

SO! As you can see, running 10 amps across a typical household wire, would heat the wire up at a rate of about (170,000W/m^3)/(3,400,000J/(m^3*K))~.05 kelvin per second.

Running 20 amps would increase the rate of heating by a factor of 4 to ~.2 kelvin per second. Of course, after a while, the wire would get really hot so there must be a similar amount of cooling of the wire happening at the same time (losing heat to the environment). If cooling is negligible, then you'd get about 45 degrees Celsius of heating in 15 minutes of running the copper wire at 10 amps, and 180 (!) degrees Celsius of heating in 15 minutes of running the copper wire at 20 amps. If the wire can't cool down fast enough, it might get hot enough to start a fire, and so we don't want to run too many amps through a copper wire. Additionally, I suspect at higher amps, non-linear effects to the resistivity would come into play, as the wire gets hotter, there should be more resistance, leading to more heating. As I specialize in Astrophysics and not electrical engineering, I hope someone double checks my math for me lol.

5. Apr 25, 2014

### Simon Bridge

I believe I just told you ... you need the equation for power.

matterwave has just spoonfed you the result if you assume no heat escapes and the current is a constant.
$I^2Rt=mc\Delta T$ ... there's your time variable.

In real life, heat loss can be significant, and usually leads to a stable constant temperature.
The exact equations you need depends on the exact situation you have.

Note: you will learn more, and be able to make better use of what you learn, by working through things yourself.

Last edited: Apr 25, 2014
6. Apr 25, 2014

### 5carola5

@Matterwave: I was forgetting the heatcapacity. Not the heat capacity of copper but just that it was there at all. I was just looking at it like I have this and this and I know I can get to this but how again? Thanks so much! I can move on now and see what I get with the formula's you gave me. Hopefully they will still be alright with high currents since I want to calculate the heat for like 100 A tot 400 A =)

7. Apr 25, 2014

### Matterwave

If you run 100A-400A on a wire of 1mm radius, you'll burn that wire out pretty quickly...those standard wires are only rated for maximum ~20A. If you're looking at wires on like a telegraph line, then maybe? But I would guess that at such high amperage, significant modifications to the equations I provided are required due to heat loss (since there will be quite a lot of heat deposited into the wires now), variable resistivity of materials, and effects like that.

8. Apr 26, 2014

### Simon Bridge

I would expect domestic 1mil copper wire to melt under 400A or so yep.

Insert a standard safety warning here: do be careful.

Please make sure you understand the risks. High currents can be quite damaging.

I don't think further comment will be useful without the details of the project in question.

9. Apr 26, 2014

### Mordred

As Simon mentioned you need to check the regulations. There is an amp capacity chart in the NEC

National electrical code. Wire comes in a variety of types, twisted wire, solid and with different shielding types. Do NOT rely on your own ampacity calculations. Use the manufacture ratings and NEC regulations.

here is an ampacity chart, you will also need to consider cable length, longer runs need a higher gauge due to line loss

http://www.usawire-cable.com/pdfs/NEC AMPACITIES.pdf

if this is a school project, check with your instructor. If your not certified DO NOT play with electricity

https://archive.org/details/nfpa.nec.2014 [Broken]

Last edited by a moderator: May 6, 2017
10. Apr 28, 2014

### 5carola5

wow, relax, I have got someone helping me with this (someone who know his stuff) and I am not gonna use a 1mm wire (duh).
I just wanted to know the formula's as a start =). Thanks a lot for them =)

11. Apr 28, 2014

### 1ledzepplin1

Get a length of resistance wire if you have any plans of building something with this information. They are built for sorts of applications like a blow dryer and for electronic cigarette/hookah atomizers.

12. Apr 29, 2014

### Mordred

ok good, we have to to be cautious when advising posters on electrical applications for obvious safety reasons