Basically I need heating element which can heat up small volume of air (lets say shoes box size) up to about 60 C degrees. I am going to use 25 V source supply with 2 A current.

What I thought about doing is:

Cp= 1.00 J/gK (air specific heat)
q=1275g/m^3 (density of air)
V=0.05m^3 (volume of air I need to heat up)
T=333.15K - 293.15K = 40 K
Lets assume room temperature is 20 C = 293.15 K

How much energy does it need to heat the 0.05m^3 volume of air:

U = cp * m * T = 1*1275*0.05*40 = 2550 J

If I want it to heat that kind of volume in 2 minutes how much watt do I need?

P=U/t = 2550/120 = 21.25 W

So I know how much watts should heating element produce, but what about ohms? Should I calculate Resistance needed by:

P=I^2 * R   where I=2A, P=21.25W 
R=21.25/4 = 5.3125 Ohms

So my resistor should be 5.3125 ohms and handle AT LEAST 21.25 W (probably much more for safety?)

I am just asking if my calculations is correct as I don't know if i am going into the right direction and doesn't have anyone to ask about it.

Thanks for any help :)

  • \$\begingroup\$ Hows about a peltier? \$\endgroup\$
    – Majenko
    Nov 19, 2014 at 23:13
  • \$\begingroup\$ I have limited budged which makes power resistor more attractive than peltier in price comparison. \$\endgroup\$
    – Ruoks
    Nov 19, 2014 at 23:50
  • 2
    \$\begingroup\$ Your calculations look fine. (I didn't check the numbers.) You'll also have to heat up the box holding the air. And you'll also be losing heat from the surface of the box. (I think that will be a bigger effect.) \$\endgroup\$ Nov 20, 2014 at 0:12
  • 1
    \$\begingroup\$ To have a uniform distribution of heat, It is recommended to use a combination of resistors (parallel and series to obtain the calculated value) instead of a sigle. \$\endgroup\$
    – GR Tech
    Nov 20, 2014 at 0:29
  • 1
    \$\begingroup\$ Since you didn't specify the thermal resistance and surface area, let me tell you I used a 25W soldering iron and 5 Watt fan to raise an insulated picnic box to 85'C for environmental testing on a shoestring. Whatever you do, use a fan to ensure rapid and homogenious temperature and a thermometer. \$\endgroup\$
    – user58601
    Nov 20, 2014 at 1:10

2 Answers 2


I am going to use 25 V source supply

If your supply voltage source is 25 volts then you need to use this to calculate the resistance needed.

\$Power = \dfrac{volts^2}{resistance}\$

Resistance = \$25^2 / 21.25 = 29.4\$ ohms

The error you made was to assume that your power supply will always supply 2 amps irrespective. In reality if the voltage is specified then hanging a load on the output will take precisely the current needed as defined by ohms law i.e. 0.85 amps in your case.

  • \$\begingroup\$ It could be a bench supply that has both constant-voltage and constant-current modes. Either way works. \$\endgroup\$
    – Dave Tweed
    Nov 19, 2014 at 23:33
  • \$\begingroup\$ Thanks a lot for pointing this out, yeah I just thought by somehow current going to be always 2A, being stupid...:) \$\endgroup\$
    – Ruoks
    Nov 19, 2014 at 23:43

I would work with what @user56801 suggested.

I would select a resistor that will use all of the 50W your PSU can supply and use a thermostat to control the temperature and it will be maintained at 60 degC much more accurately than hoping the external convection, radiation and conduction losses are going to be zero or even constant.

A 12.5 to 18 Ohm resistor rated at 60 to 100W with a simple electronic or mechanical thermo-switch would make a more reliable incubator, shoe dryer or curing oven. Having additional over temperature (70 to 110 degC) and over current (3 to 5 A) fuses would be prudent.


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