How warm does a Resistor get at 0.9 Watts?

Hi
I am using a 27 ohm, 25 watt (metal coated) resistor as simple heater for my project. The heater is inside a small box (maybe 20x20x10cm). As I am powering the project with 5 V, it would be handy to power the resistor with that sort of voltage which gives me an output of 0.9 Watts (if my calculations ore correct). Unfortunately, as I have not yet got the resistor (and other parts) I have no way of testing the setup. So how warm would you expect a 0.9 W resistor to get? Is 0.9 Watts “a lot” (would I burn my fingers touching it)? or would you hardly notice the heat?

"How warm" would depend on how fast heat is removed from the resistor, which depends on its surface area, the difference in temperature to the surroundings and the thermal conductivity to the surroundings.

Look up Fourier's Law of heat conduction.

V=IR, V/R = I. 5V/27ohm = .185A
P=V^2/R. 5V*5V/27ohm = 0.926W

Here are the high wattage parts I could find at Digikey
https://www.digikey.com/products/en/resistors/through-hole-resistors/53?FV=4044d%2Cffe00035&mnonly=0&newproducts=0&ColumnSort=-2&page=1&stock=1&pbfree=0&rohs=0&k=resistor&quantity=&ptm=0&fid=0&pageSize=25&pkeyword=resistor
https://www.digikey.com/products/en/resistors/chassis-mount-resistors/54?k=resistor&k=&pkeyword=resistor&pv1989=0&pv1=1101&FV=ffe00036&mnonly=0&newproducts=0&ColumnSort=0&page=1&stock=1&quantity=0&ptm=0&fid=0&pageSize=25

1W will be warm, but I don't think enough to burn a hand.
If you had a 7805 powered from 12V, with an output of 140mA, that's about what an Arduino driving some LEDs might draw. The regulator would get warm, I don't think hot enough to burn tho. Hook up your Arduino and try it.
A high wattage resistor would be similar.

This kind of thing is almost impossible to calculate... I'm going to guess "slightly warm".

If the box were perfectly insulated the energy would continue to go-in and never come-out and temperature inside would build-up to infinity. An expert could estimate the thermal resistance and make a calculation but in reality it will have to be determined experimentally.

At 25W it would to be too hot to touch and it would probably discolor over time. Obviously a 1/4W resistor would be too hot to touch and the resistor is going to burn-up (at 25W or 1W). The total energy is the same, but it's more concentrated with the smaller resistor... The air temperature inside the box would be the same either way, until the 1/4W resistor burned-up.

If you concentrate 25W at the tip of a soldering iron, you can melt solder. But, spread-out evenly over a 25W resistor the resistor isn't going to get that hot.

Note that you'll get a certain temperature rise inside the box compared to the outside. If the ambient temperature goes up 10 degrees, the temp inside the box will go up 10 degrees.

I have no way of testing the setup.

You might want to order a few different resistor values. Just don't exceed the power available from the power supply...

This kind of thing is almost impossible to calculate

Nonsense. See reply #1.

A 1/4 watt resistor will literally burn up if asked to dissipate 0.9 watts.

Which is why I am smart enough to use a 25 W resistor :slight_smile:

If the resistor is used to heat the inside of an insulated box, the power rating of the resistor is irrelevant, as long as it can safely dissipate the power.

Fourier's Law of Heat Conduction can be used to calculate the internal temperature rise of the box, provided you know the surface area and thermal conductivity of the walls.

I'd suggest you're asking the wrong question or at least addressing only a part of the system design. The power dissipated by the resistor defines the power into the system and is reasonably well constrained. The temperature in your chamber also depends on the power leaving the system, that is, how quickly the heat is conducted out of the chamber. For example this setup isn't going to do much to heat a room but would get quite hot were it sitting inside a small sealed vacuum flask.

Maybe a 1W light or LED would be better, then when you open the lid you can see what's going on.

bestanamnetnogonsin:
Which is why I am smart enough to use a 25 W resistor :slight_smile:

Then you should have asked, "how warm will a 25W resistor get when it is dissipating 0.9W?".

But the fact that it is 4% of its rated power should be a clue.

aarg:
Then you should have asked, "how warm will a 25W resistor get when it is dissipating 0.9W?".

First half of the first sentence in the first post. Read it. It's there.

But the fact that it is 4% of its rated power should be a clue.

Probably not as much as you think it does. Wattages that high are made to be used with heat sinks. When naked, their performance is probably much worse than you're assuming.

With all of this conjecture, I'm surprised nobody has asked for a part number for the resistor. If you're lucky, the manufacturer's datasheet will give you the thermal resistance of the package it's built into.

I’m surprised nobody has asked for a part number for the resistor.

Because the OP asked the wrong question. The point is to use the resistor to heat an enclosure. What matters is the temperature of the enclosure, given 0.9 Watts of heat input and unknown enclosure properties.

Jiggy-Ninja:
First half of the first sentence in the first post. Read it. It's there.

Yes but, along with "So how warm would you expect a 0.9 W resistor to get?", which made it a bit confusing about which resistor the question was about.

Have a read of this:-
http://www.physlink.com/education/askexperts/ae538.cfm

Grumpy_Mike:
Have a read of this:-
How is the measurement Watts related to temperature in terms of energy?

I'm not sure that's the best source to link to. Thermal capacity will affect how quickly the temperature can change, but it will have no affect on the equilibrium temperature. The temperature will stabilize at the point were the net heat flux is 0 (ie. when heat energy going in equals heat energy going out). That's affected by thermal conductivity and temperature gradients, not heat capacity.