# calculating the size of a heating resistor

I'd like to use a ceramic heating resistor to keep a 4-5 cu ft insulated box above freezing on cold nights. A 15w heater belt is too much so I was thinking about using either a 5w or 10w heating resistor. Seemed like something I could eventually design a circuit to control the heat with a DS18B20 sensor and an Arduino.

Does the following look correct or am I missing something?

The resistor I looked at, a Riedon UAL-5, is rated for 1500 VAC. I assume I can run it at 120 VAC which is the RMS value for house AC. I=P/V so 5w/120v = 0.0417 amps. R = V/I so 120v/.04a = 3000 ohms and that would be the size of the resistor I need to generate 5w?

Does the current need to be rectified?

PDF of a part I was considering which will mount nicely on a heat sink:

https://riedon.com/media/pdf/UAL.pdf

Thanks, Dan

EDIT1: I've done some empirical tests with a 15w heater belt that I borrowed off of a friend and I know the insulation is good enough that 5-10w will be enough. It's the calculation for the resistor and question on rectifying that I was sure of. Not to mention that I can always move up in resistor wattage.

I appreciate the answers about area heating and insulation but I purposely didn't ask about that. I guess I should have said ignore the insulation and area heating as I can deal with that. What I really wanted was to understand the calculation and then I would buy the size of resistor I thought was appropriate. Now that I think about it I'll probably go 10 or 2x5 on separate switches.

I was going to use a grounded electrical box ventilation hold to contain the resistor and heatsink along with shrink tubing on the connections. I am aware of the cautions around using 120v.

I appreciate all the answers. Thanks Barleyman. Barley?

• Without knowing the thermal resistance of the insulating box, you are guessing. If ΔT('C)/W matches your intuition, you may be lucky. Otherwise you should measure the temp rise ΔT, for a fixed W then regulate or compensate depending on ΔT. – Tony Stewart Sunnyskyguy EE75 Feb 22 '18 at 3:03
• Yes, that would be the right value to get 5W. No, it doesn't need to be rectified. – user253751 Feb 22 '18 at 3:16
• As a general rule 1.5 watts of heat raises the local ambient temperature about 10 to 15 degrees C. That would cover about 4 cubic inch's, depending on insulation. – Sparky256 Feb 22 '18 at 3:46
• what about a heater for lizards ... get one at a pet shop ... or a coffee cup heater – jsotola Feb 22 '18 at 7:15
• If you know that 15W is enough, why not just put together something to regulate the temperature? Thermistor and a comparator and something to turn on the heater when it gets too cold in the box. That way, you can also handle nights when it gets colder than expected. – JRE Feb 22 '18 at 12:12

Your calculations are correct and no, for generating heat there's no need to rectify anything. Short form of power is $P = \frac{u^2}R$ so in other words $\frac{120^2}{3000} = 4.8W$ which is what you wanted.