Background I am working on a project that involves heating air inside a small box, then measuring the temperature over time. The purpose of the box is to be a basic test chamber for PID experiments.

Initial Test I made a 10 cm * 10 cm * 10 cm box from 3 mm plywood. I then placed a small ~1 W heater inside with a temperature probe and observed the temperature rise... very slowly.

Proposed New Prototype I am going to construct a new prototype once I decide how to calculate the approximate time to warm the air inside the box to a set-point. Some specifications and assumptions are laid out below:

  • The ~1 W heater consists of 4 parallel 100 ohm resistors and a 5 V power supply (there is also a 330 ohm resistor and an LED for visual indication). I don't want to change this heater design, as the intention is that this is easy to replicate and uses readily accessible 5 V power sources and electronic components.
  • Currently no fan has been used but I have a small 20 mm, 5 V fan on order, so will integrate this into the design at some point.
  • The box will be used at room temperature between 21–23 °C
  • Ideally a temperature rise up to 30 °C (or more) would be possible over a 5-10 minute time interval
  • The new proposed box size is 5 cm * 5 cm * 5 cm, 8 times smaller than the previous prototype. This is open to change depending on the previous 2 requirements of maximum temperature rise and time taken to achieve that change.

I am not after exact timings - I am aware that losses through the box will have an effect. But if this effect is minimal, then a simplified approximate solution is preferable as some design parameters may change slightly. I am looking for any guidance and calculations that will save me time instead of having to make different boxes and learn by trial and error.

  • $\begingroup$ I would definitely use a fan to mix the air, otherwise you can have very large temperature gradients. Where is the temperature probe located? Also, wood is not a great thermal insulator, I would use plastic instead (polyprop, polycarb, or similar). $\endgroup$
    – am304
    Commented May 7, 2015 at 9:30
  • $\begingroup$ Hi Am304, I will probably have 2 sides made from acrylic to act as a viewing window, but small losses are not an issue as the PID algorithum will show fluctuations to cope with this. The temperature probe will probably be near the top, offset to the side slightly, so it's not directly above the heating element. A fan will definitely help, just didn't have this to hand at the moment. $\endgroup$
    – Ant
    Commented May 7, 2015 at 9:38
  • $\begingroup$ You really should calculate heat losses at say 30C. AND A fan is close to essential BUT the fan risks adding considerable energy with a 1 Watt heater. My raised wet finger waved in the air (as it were) assessment says that 1 Watt is very small compared to losses liable to be encountered. | If a fan is used motor should be outside with rotor driven on a shaft. almost better - if fan was entirely internal then you need no heater - power the fan at the level required. Input 1 Watt electrical say (eg 5V 200 mA ) and it will end up as heat one way or other. Windows->What's to see? $\endgroup$ Commented May 7, 2015 at 14:27
  • $\begingroup$ I am dealing with a similar scenerio where I have to heat up air very slowly at around 1 degree celsius at a time inside a small container. It woould really help me if you could give me more detail on your heater. can i use a heating element alone to place inside the shut container? $\endgroup$
    – user2903
    Commented Sep 3, 2015 at 17:57
  • $\begingroup$ @sara, for the heater i simply used 4 * 100 Ohm resistors in parallel, this equates to 25 ohms. I used a 5V power supply sourced from a USB port. So using Ohms Law (V/R) 5V / 25 Ohms = 200mA, Watts can be calculated from this by simply multiplying V * A therefore 5V * 200mA = 1W $\endgroup$
    – Ant
    Commented Sep 4, 2015 at 11:44

1 Answer 1


A simple approach would be to just consider the heat added.

The specific heat equation gives the temperature change due to heating as

$$Q=mc\Delta T$$ where $Q$ is heat added, $m$ is mass. $c$ is the specific heat of the material in question (~1.005kJ/kgK for air at room temperature) and $\Delta T$ is the temperature change.

A 5x5x5 cm box will have an air mass of ~0.15 g (air density ~1.2 kg/m$^3$).

For a 1W (1 J/s) heater this gives a temperature change of $$\Delta T = \frac{Q}{mc}= \frac{1}{0.00015*1005}= 6.6 K/s$$

I will suggest this is much higher than you observed with the larger box (even accounting the increased volume) which suggests there is a factor not being considered.

Two factors leap to mind.

1) Heat losses from the box. Unless your box is well insulated it will lose heat via radiation and conduction. However for temperatures near room temp. I would expect these loses to be low.

2) Heating the box. Inevitably if you heat the air in the box this will transfer to the box itself. The box has much larger weight so will require more energy to heat (probably 100x or more). Intuitively I would think the transfer of heat from the air to the box would be slow compared to the heating of the air. But I am not an expert in this so may be wrong.

More likely I think the problem may be that most of the heat is going into the box initially if the heater is in contact with it, due to much much better thermal conductivity of wood than air. This is what I would look at to improve the rate of heating.

  • 2
    $\begingroup$ With a 10cm x 10cm x 10cm box, it goes down to ~1 K/s. $\endgroup$
    – am304
    Commented May 7, 2015 at 9:29
  • $\begingroup$ Hi Nivag, thanks you very much for the response and calculation. I need to try and equate this to actual time now, but can use your calculation on the previous 10*10*10 example I agree with the heat loss being minimal through the box as it's wood. The resistors are on top of a small PCB so are insualted to some degree from the box, but I will elevate them slightly as you suggest, therefore limiting conduction losses. I will wait to see what other answers come from the community and pending repsonses will accept this as an answer. $\endgroup$
    – Ant
    Commented May 7, 2015 at 9:31
  • $\begingroup$ Hi Am304. thanks for your comments, saves me calculating the heat transfer for a 10*10*10 box. So it could be said the 5cm box is 8 times smaller than the 10cm but box, but will heat 6.6 times faster? $\endgroup$
    – Ant
    Commented May 7, 2015 at 9:34
  • 1
    $\begingroup$ @Ant Yes, that's correct. The temperature change is inversely proportional to the mass, which is in turn proportional to the volume. $\endgroup$
    – am304
    Commented May 7, 2015 at 9:39
  • 1
    $\begingroup$ @Ant The 8 times smaller box will heat 8 times faster, but 6.6/8=~1. $\endgroup$
    – nivag
    Commented May 7, 2015 at 9:54

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