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I have Kanthal A1 26AWG (3.21 Ohms). I need to heat a small cardboard box (Tissues box) 15 degrees Celsius over the room temperature. And I am using 2 D Cell Batteries of 1.5V.

I coil the Kanthal with a regular pencil, approximately as long as a finger. Then I connect a regular copper wire to one that goes to the battery and the other end to the opposite side of the battery. The problem is that it doesn't really heat. I need to make the coil as small as nail to make it heat to high levels. Any idea?

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  • \$\begingroup\$ You need to explain what exactly this application is for. As it stands, I'm pretty sure you're building a weapon - either that or something that's just plain dangerous. Either way, we need to know a lot more. \$\endgroup\$ – Sean Boddy May 15 '15 at 7:29
  • \$\begingroup\$ What weapon would use a tissues box heated to 15 degree above ambient?? \$\endgroup\$ – RJR May 15 '15 at 7:30
  • \$\begingroup\$ @RJR, some rapid oxidizing agents only need a little kick. Or to have a bottle overpressurized and shattered. It's easy stuff. \$\endgroup\$ – Sean Boddy May 15 '15 at 7:48
  • \$\begingroup\$ @SeanBoddy ok, but then almost any electrical device is a potential weapon. Try putting your laptop charger in a box... \$\endgroup\$ – RJR May 15 '15 at 11:12
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The internal resistance of a D-cell is about 300 milliohms according to Energizer. That, with the 3.21 ohm wire resistance, should be generating about 9/3.5 = 2.6 Watt (P=VI = VV/R). I'm assuming you are using the batteries in series (to get 3V).
That won't be enough for the temperature rise in the box you are looking for although the wire should get warm/hot to the touch.
To get an idea about the amount of power required, have a look here: https://what-if.xkcd.com/35/

Some more details...
The thermal conductivity of cardboard is about 0.21 W/m/K. So it will conduct 0.21 W for every square meter surface area for every degree Kelvin in temperature difference for every meter thickness. To get the inside to a certain temperature, you have to pus as much power in as the box conducts to the outside (I'm ignoring radiation here as the temperature is relatively low).
Let's assume the area of your box is 10x10x20cm, which is 0.002 m2. Assume also it is one millimeter thick.
That means that to obtain a 15 degree Celsius temperature difference with the outside, you have to put in 0.21 * 15 * 0.002 / 0.001 = 0.21 * 30 = 6 W of energy.
As such, you need about 2 times more power and you should halve the resistance (length) of the wire.

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