# If I want to use as little current as possible, then what resistance wire should I use to get the hottest temperature?

I am trying to build a circuit that is able to melt wax (130˚F) and use as few amps as possible since it will need to be powered using a battery. Do I need to be looking for wire that has a high resistance or low resistance to get the most heat per mAh in the circuit?

I'm using about 20 feet of nichrome right now and it takes about 28 volts to get to that temperature with a current draw of about 2.4 amps but I would like to use less current that that if possible and that's why I'm looking into alternative choices of wire with varying resistance.

I love learning so explain your answer in as much detail as you'd like. :)

• You need to say how much wax you want to melt, and how fast. If you're trying to melt 1kg as in Phil's answer you're looking at 3000s not 30s and a very big battery. Or are you trying to carve it with a hot wire? – Brian Drummond Sep 8 '17 at 9:35
• What do you mean by melting wax? Are you heating a vessel or crucible to melt a bulk quantity of wax? Are you heating the wire to make a hot thin cutting blade to slice wax? What kind of battery? A 12V lead battery or a Li watch coin battery? – J... Sep 8 '17 at 12:55
• How much wax are you melting? – user3528438 Sep 8 '17 at 14:01

If you currently require 28 volts and 2.4 amps, that's a power of:

$$28\:\mathrm V \cdot 2.4 \:\mathrm A = 67.2\:\mathrm W$$

67.2 watts. Unless you can somehow use the heat more efficiently, this is how much power you need, period. If you want less current you can do that by increasing the voltage such that the product of voltage and power equals 67.2 W. But a higher voltage battery, all else equal, will have a lower capacity, so you haven't actually increased battery life.

This is because while the battery geometry can be altered to have a higher voltage, unless you make it bigger, or use a different chemistry, the chemical energy in the battery has not changed.

The conversion of electrical energy to heat is nearly 100% efficient, so if you need to increase the battery life you'll have to look for efficiency improvements elsewhere. That means not heating things that aren't wax, such as by adding insulation or making the heating vessel as small as possible or using a more efficient battery.

There's a theoretical bound to the minimum amount of energy required to melt wax. Paraffin wax has a heat of fusion of about 200 joules per gram. A joule is one watt for one second. So if you need to melt 1000 grams of paraffin wax, you require at a minimum:

$$1000\:\mathrm g \cdot {200\:\mathrm J \over \mathrm g} = 200,000\:\mathrm J$$

Additionally the wax must be heated to its melting point. Paraffin wax has a specific heat capacity) in the neighborhood of 2.5 joules per gram kelvin. A change of one kelvin is the same as one degree Celsius. So say the 1000g of wax wax starts at 20°C and must be heated to the melting point of 37°C:

$$1000\:\mathrm g \cdot {2.5\:\mathrm J \over \mathrm {gK}} \cdot (37-20)\mathrm{^\circ C} = 85,000\:\mathrm J$$

This brings the total energy required to 285,000 joules. Given the 67 watt power of your current setup, that means it must run for a minimum of:

$${285,000\:\mathrm J} \cdot {\mathrm s \over 67.2\:\mathrm J} = 4241\:\mathrm s = 71\:\text{minutes}$$

71 minutes for each 1000 grams of paraffin melted.

In practice, additional energy is lost to heating other things like the air, the container, and the battery, so for best efficiency all you need to worry about is heating the wax, and only the wax.

• Umm. .. About 67 Watts – Dwayne Reid Sep 8 '17 at 1:57
• Don't forget the energy needed to heat the wax up to the melting point from what I presume is room temperature. – Arthur Sep 8 '17 at 8:43
• @Christoph apparently the math part of my brain was not working when I wrote this. Thanks for the correction. – Phil Frost Sep 8 '17 at 12:05
• "efficiency improvements elsewhere. That means not heating things that aren't wax," Absolutely. Note that that also includes the battery - it may be possible to use a different battery geometry that results in less heating of the battery (possibly with some circuitry to adjust the output voltage/current to work better with the wires). – psmears Sep 8 '17 at 13:50