Heat absorption of copper

LED Flashlights – General Info
1

I’ve ordered a cheap “10 watt” spotlight. Judging by the stated run time from the internal battery it seems to be running more in the 3 watt or 1 amp range. The heatsinking is probably almost non existent.

From the dimensions listed it would seem that I will be able to fit at least a 50mm long piece of 25mm copper bar inside. The problem is that it won’t have any air circulation to get rid of the heat and I would prefer not having to cut a hole in the back of the housing to extend the heatsink to the outside.

My question is how many watts in the form of heat can a 50x25mm copper slug absorb and for how long before the junction temperature would go into dangerous levels for common Led’s like an xml or xpg2.

I’m aiming for about 10 watts to the emitter so about 8 watts of heat that would need to be absorbed.

I’ve read that the Convoy s2 with a 7135x8 driver can be left on with almost no air movement without overheating, but this is still in open air and not an enclosed space.

Any help or advice would be greatly appreciated.

2

All the data that you need is here:

Calculate mass of copper bar, multiply by specific heat, times by Joules (Watt-seconds) and choose your upper temperature limit.

By the way, copper is one of the worst metals for absorbing heat. Aluminium is by far the best (except for Beryllium, Lithium, Magnesium, but don’t even think of going there), more than twice as good, mass for mass.

Nitecore do use Magnesium alloys, perhaps for good technical reasons.

Copper’s only virtue is it’s thermal conductivity, which ISTM is wasted in your application.

But it is shiney, and dense/heavy, meaning it does have its uses in high powered torches, in small pieces, where heat needs to be spread elsewhere.

3

Copper is very good at absorbing heat, 50% better than aluminum and nearly twice as good as magnesium. But that’s absorption by volume, not weight.

Do consider paraffin. While melting it will absorb lots of heat.

4

Yeah, OP is clearly volume-limited not mass-limited, so you should be calculating using the volumetric heat capacity.

5

Thanx for the link.

My reasoning for using copper was that it would conduct the heat away from the emitter better and warm up more of the bar rather than having the heat build up close to the emitter.

Probably not really an issue with a solid heat sink like that.

Anyway

21.9 g of copper

Q= (0.39KJ/Kg°C)(0.0219)((60°C)-(30°C))

=0.25623 KJ
= 256.23 watt hour

256/8 = 32 hours of 8 watt heat source before the bar would be at 60°C without any way of dissipating heat.

For aluminium

6.7 g of aluminium

Q= (0.91KJ/Kg°C)(0.0067)((60°C)-(30°C))

=0.18291 KJ
= 182.91 watt hour

183/8 = 22.9 hours before the bar would be at 60°C without any way of dissipating heat.

Does this look correct?

6

No, it doesn’t.

7

Don’t you mean watt-seconds, not watt-hours?

Edit: check your masses as well, your numbers look too light.

8

Indeed, mass seems wrong as well. click
321 seconds.

9

Thanx for being patient with me, I can be slow at times.

219g or 0.219 kg of copper

Q= (0.39KJ/Kg°C)(0.219)((60°C)-(30°C))

=2.566 KJ
= 0.7111 watt hour

0.7111/8 = 0.0888 hours before the bar would be at 60°C without any way of dissipating heat.

= 5.3 minutes minutes.

That seems more accurate

It looks like I’ll have to make a plan to get rid of some heat or run it at a little less current to make it usable.