I'm trying to design a growing box that will maintain a stable temperature for the seeds germination.

I'm planning to use an Arduino with a temperature sensor to control a heating element. I'm thinking of using a resistance wire as the heating element, as it is a simple and cheap solution. Let's say the size of the box will be 1 m3. I'm not sure what the insulation of the box will be, but I'm considering either a plastic box, or an old refrigerator. The target temperature is 30°C (86°F).

I need to choose the right heating element and determine what power supply to use for that heating element.

I've already done some calculations to reaching this goal.

Power needed to heat up the box

First of all, I need to determine how much power is required to heat up the box to the target temperature. Let's assume that the temperature is 20°C and I want to heat it up to 30°C in 30 minutes.

I used an equation from https://www.engineeringtoolbox.com/heat-up-energy-d_1055.html to determine the energy required to heat up a substance.

m = 1.2 kg (the mass of 1 m3 of air)
cp = 1.006 kJ/(kg°C) (specific heat of the substance)
dT = 10°C (required change of temperature [30° - 20° = 10°])
t = 1800 s (heat up time)

\begin{align} &\mathrm{q = \frac{m \cdot c_p \cdot dT}{t}} \\\\ & \mathrm{q= \frac{1.2~kg \cdot 1.006~kJ/(kg°C) \cdot 10°C}{1800~s}=} \\\\ & \mathrm{= \frac{1.2 \cdot 1.006~kJ \cdot 10}{1800~s}=6.7~W} \end{align}

So the required power to heat up the box to the target temperature in 30 minutes is 6.7 watts. This is the first time that I have calculated something like this. Please tell me if I made a mistake. Does this mean that I need a heating element that consumes 6.7 watts? If so, let's continue:

Calculating required voltage to power the heating element

I have three resistance wires. Each one of them is a meter long. It is specified that they have the following resistances: 5.4 ohm/m, 20 ohm/m, and 55.4 ohm/m. Assuming that I need 6.7 watts to heat up the box, and that I choose the 5.4 ohm/m wire, I can calculate the required voltage from the formula: $$\mathrm{P = \frac{V^2}{R}}$$ $$\mathrm{V = \sqrt{6.7~W \cdot 5.4~\Omega} = 6~V}$$

Therefore, the required voltage using this wire would be 6 V, which I can achieve with 4 AA alkaline batteries in series.


I am a beginner in electronics, so I need a bit of help. Am I on the right track, or is it what I wrote above complete nonsense? What would you recommend as a power supply for this task? From what I know, the lower the resistance, the greater the power. But is it okay to use a resistance that low in a circuit? If you have any other recommendation, feel free to share them.

  • 1
    \$\begingroup\$ If you really need 7W or so, I'd get a 10~20W heating element and control it with some temperature sensing. 4AA batteries and 3 different wires with different resistances sounds more like a homework problem than an actual viable solution. \$\endgroup\$
    – Wesley Lee
    Sep 1, 2020 at 22:28
  • \$\begingroup\$ I intend to use only one wire, not all of them. I think it's a nice DIY though. \$\endgroup\$ Sep 1, 2020 at 22:35
  • \$\begingroup\$ it would be safer and last longer to simply heat a 50 or 100w aluminum cased ceramic filled wire wound resistor, which are only a few bucks and will last forever without presenting a burn hazard. You can just drive a 1 or 2 ohm at 5v with PWM to get a desired average current. Look into PID control if you want to be really bad ass. \$\endgroup\$
    – dandavis
    Sep 1, 2020 at 22:35
  • \$\begingroup\$ @dandavis thank you for your answer! I found 50 watt 680 ohm resistor on the internet. I assume I'd need to use a greater power supply than regular alkaline batteries, right? I'll certainly look into PWM and PID. I'm new to this! \$\endgroup\$ Sep 1, 2020 at 22:51
  • 1
    \$\begingroup\$ Batteries are a very poor choice to operate any sort of heater, unless you plan on replacing the batteries twice a day. You would be much better to use some AC line powered heater - a small incandescent light bulb may work well. \$\endgroup\$ Sep 1, 2020 at 23:11

1 Answer 1


First of all, I need to determine how much power is required to heat up the box to the target temperature. Let's assume that the temperature is 20°C and I want to heat it up to 30°C in 30 minutes.

Not the way most of us would go about it. Your long term load is the heat loss through the walls. You have to make sure you have enough power to support that. Once that's done, the actual heating time doesn't really need to be specified, it will take whatever time it takes to get up to temperature. With the timescales associated with a grow box, 30 minutes seems unreasonably fast. We can see how near to that we'll get later when we have estimates for power, and the mass of what's in the box.

Estimate the heat loss of a box with volume 1 m3. Make some assumptions, it's easy to modify the answer later if any are grossly wrong. This will produce an absolute minimum answer, you'll probably want to use 2 to 10 times this figure, for safety.

Surface area of a regular 1 m3 cube A = 6 m2
Thermal conductivity for new PU insulation foam K = 22 mW/K/m
Foam thickness t = 50 mm
Temperature rise of box dT = 10 °C

Heat loss = dT.K.A/t = 26 W

That's the best case. As the foam ages, K will increase. Any wires passing through, or air leaking in or out will increase the loss. Using a cuboid rather than a cube will increase the surface area per volume. Any thin spots, or structure passing through the foam, aka 'cold bridging', will increase the loss. If your ambient drops and you need more than 10 °C rise, or if you use a lesser insulation material like polystyrene or rockwool, or a thinner wall for cost reasons, your power requirement will increase.

You might get away with 50 watts if you use new PU of at least that thickness, well air-sealed, but 100 watts would give you more margin for construction defects.

At this power level, incandescent bulbs, for instance auto bulbs or MR16 spot lights, make an excellent low cost self-indicating heater. Needless to say, you will not be using batteries. For those 12 V heaters, an old ATX PC power supply would be good, having as it does a 5 V output for the electronics as well.

Don't forget you need to raise the temperature of the soil, its moisture content, and the inside of the walls of the box, as well as the air. The temperature may take a long time to become uniform throughout the soil, 30 minutes would be nowhere near enough.

The thermal capacity of soil varies enormously depending on whether it's mostly dry vermiculite, or very wet loam. If we make the assumption of 50 mm soil thickness and 30 % moisture, that's 15 kg of water, even neglecting the soil itself. The thermal capacity of 15 kg water through 10 °C is 15x4200x10 = 630 kJ, which is about 2 hours at 100 watts. This suggests a warmup in less than 24 hours is quite possible, and a 30 minute one would need unreasonably high power.

If this is a grow box, then presumably you will need a light source in there as well? Perhaps switching your light source between LEDs and MR16s would modulate the heat input into the box enough to give you control?

  • \$\begingroup\$ That is a wonderful answer! The wattage of MR16 lamps ranges from 3 W to 7 W. Doesn't it mean that I'd need to use many of them? Also, I don't know how much heat they make. Does the wattage always corresponds with the produced heat? I'm not sure if light bulbs are the right choice as a heating element, because I don't want to have the lights turned on during the night, as the plants need darkness. \$\endgroup\$ Sep 2, 2020 at 8:53
  • \$\begingroup\$ @LadislavOndris I was thinking of the 50 W halogen ones, not the few watt LED ones. But I take the point about the light. You could put them in a metal box with ventilation holes, but a bit of resistance wire is probably less trouble. \$\endgroup\$
    – Neil_UK
    Sep 2, 2020 at 9:46
  • \$\begingroup\$ Thank you for all the suggestions! I'm really glad you mentioned the ATX power supply. It sounds like a great idea! \$\endgroup\$ Sep 2, 2020 at 9:55

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