# Could a candle theoretically melt iron?

The title question is rather illustrative. I suppose the real question would be:

Is heat cumulative?

Put back into an example:

If I have a lit candle right beneath an iron bar, assuming the candle will remain lit indefinitely, and that the heat-losing rate is below the heat-getting rate. Will the bar eventually reach the needed temperature for it to melt?

Once the iron bar reached the max temp the candle can get it to. Where does all of the energy (heat) go after?

EDIT:

The question "is heat cumulative?" can be ignored as it is out of place and is misleading. Althrough the answer for it is "yes", it doesn't mean the answer for the general question is also "yes".

The point is not if it is actually possible to melt iron with a candle. Iron and candle are mere parts of the illustration, their properties are irrelevant.

A better phrasing of the main question would be:

Could a heating object contained in a perfectly closed system push the temperature of the system above its own temperature?

• what-if.xkcd.com/145 – Dan Henderson Aug 6 '16 at 23:02
• I feel like most answers focus on the wrong things: melting point of iron, steel wool or forges are irrelevant. The question is: In a closed system with a heat source at 100C, does the system ever go over 100C? If not, where does the energy go? – isanae Aug 8 '16 at 17:03
• @isanae a closed system would run out of oxygen and the candle would go out. If you pipe in new oxygen, it will cool down the system unless you add a new heat source. – EL_DON Aug 9 '16 at 15:44
• @EL_DON Again, focus on the wrong thing. Forget the candle, put radioactive stuff in it and use decay heat. Or whatever. – isanae Aug 9 '16 at 15:56
• Damn, so many upvotes and no one understanding the question. I feel with you, @isanae – Matsemann Aug 10 '16 at 17:57

I'll try a simple explanation. Assume that there are no phase transitions initially. As you heat a body, its temperature rises, and it radiates energy into the surrounding space according to $$P = A\varepsilon \sigma T^4$$ ($\sigma$ is the Stefan Boltzman constant, $A$ is the surface area, $T$ is the temperature, $\varepsilon$ is the emmissivity).

Obviously, $P$ increases as $T$ increases. So, if before the bar reaches its melting point, $P$ becomes equal to the power input (from the flame), there will be no net flow of energy across the surrounding-bar interface (for if there were, $P$ would be greater than the input power, and more energy will be lost than gained, taking it back to the stable equilibrium point). At this point, notice that the temperature is constant, because any energy supplied by the candle is equivalently emmitted by the body (sort of dynamic equilibrium). Of course, this is only possible if the bar doesn't melt before this temperature is reached.

• I thought we were talking about a perfectly insulated system, with no heat transfer out of the bar (except back towards the candle)? – immibis Aug 10 '16 at 6:20

No. The candle can only transfer heat into the iron as long as the candle is hotter than the iron. The temperature of a flame depends heavily on the chemicals being burned. A typical candle burns at about $1400^{\circ}\mathrm{C}$. Iron melts at $1538^{\circ}\mathrm{C}$. So, the iron would stop heating up at $1400^{\circ}\mathrm{C}$ and not melt.

However, you're not that far off! Probably you could find a hotter than normal candle (mix sawdust into the wax, maybe?) and use it to melt a tiny piece of iron, like the tip of an iron needle. You might find an iron alloy that melts at a lower temperature, too.

Keep in mind that the equilibrium temperature will be lower than the candle temperature since the iron loses heat to the environment through processes such as radiation, as mentioned in other answers (see @Lelouch's answer). These loses can be reduced by doing things like enclosing your system in a shiny insulated box to reflect radiated heat back.

• While the plasma in a candle flame may be locally as hot as 1400°C, I doubt you can actually get any iron object to much more than 1000°C in such a flame. The iron will radiate away extra heat, even the lightest suspension you can build conducts some, and a candle just doesn't have enough power to make up for it. – leftaroundabout Aug 6 '16 at 16:53
• @leftaroundabout I absolutely agree. My point is that if you can't even melt the iron in the ideal case, you certainly can't do it in the real world (unless you're allowed to be flexible with the definitions of candle and iron). – EL_DON Aug 6 '16 at 17:04
• This is the answer! The candle can't keep heating up the iron forever, even if the two are perfectly isolated from the environment. Once they're the same temperature, the candle sends energy to the iron at the same rate the iron sends energy back to the candle. – knzhou Aug 6 '16 at 20:11
• The problem likely isn't the candle, it's the lack of oxygen. I suspect you can get it well over 1400C by using pure oxygen. The problem with air is that you need to suppy 5 times as much air because it's only 20% oxygen, and that airflow carries away quite a bit of heat. – MSalters Aug 8 '16 at 7:44
• @MSalters: Yes I think you're right, but piping pure oxygen into the candle gets you halfway to a welding torch. I don't think it's really the point. – EL_DON Aug 8 '16 at 17:58

Is heat cumulative?

This is a bit of an ambiguous question, but yes it is "cumulative". If you put some heat into something, it will stay hot, unless that heat is taken out. Putting 10 units of heat at once is the same as putting 1 unit 10 times, again assuming that you can keep the heat insulated in the object.

The analogy here is a bucket with holes. Heat is like water you put in, temperature is the level of the water. It doesn't matter if you put the water in all at once, or in small bits, but as soon as you put in some water, it starts leaking out. So if you don't hurry up, you might never get enough water in, and the more water is in the bucket, the faster it leaks out.

Can I melt iron with a candle?

Probably not. I think the biggest problem is that heat flows from hot things to cold things. If the iron reaches the same temperature as the gas coming from the flame, it won't receive any further heat. If your target temperature (the melting point) is actually below this, you'll get stuck.

As it happens, the melting point of iron is about 1600C. The flame of a wax candle tops out at 1500C, but that is a tiny area in the center - the average temperature is 1000C.

You could solve this issue by using an exotic candle made of unusual materials. For instance, thermite burns hot enough to melt steel, acetylene torches can get you to 3500C. But when you say candle I assume you meant wax candle, and not a thermite "candle".

Even with a hot-burning candle, you still have the problem of small heat flux. You say:

the heat-losing rate is below the heat-getting rate

Sure, but that's easier said than done. If you had a magic iron that never loses heat, okay, but in real life the hot iron will heat up the air in the room (candle won't burn in vacuum), the hot air will float away, and in effect you will end up heating up the whole room because it's all in thermal contact. That's asking for a lot from a candle, but if you somehow had a candle with infinite fuel and had it burning indefinitely, then eventually you could heat the whole room above the melting point of iron (it has to be above because melting itself requires some heat input), and yes, it would melt - but we are now talking about a very extraordinary candle. Not to mention the mass of such a candle could act as a heatsink, or it could ignite in your firestorm of a room, so there is a whole host of practical problems.

So in conclusion, what kind of candle would it take to melt iron? You need:

• A fire hotter than 1500C.
• An insulated environment which can get to this temperature and stay there.
• A big enough candle to ensure that the rate of heat loss is always less than gain (remember that heat loss is faster when the temperature difference is larger).

This is kind of device is called a forge.

• Don't forget that exothermic reactions like combustion are still governed by equilibrium, and heat is a product. So a candle burning forever in a closed environment will eventually stop burning, because one of the products (heat) will be too abundant. So a magic candle in a sealed room may not necessarily melt iron, if the chemical reaction reaches equilibrium at a lower temperature than the melting point of iron. – Egor Aug 9 '16 at 17:38
• Everything you say makes sense, but as a counterpoint, flame temperature of butane in air is1760°F. Copper melts at 1981°F, but I can use a hand held butane torch to melt copper. I just tried it to make sure, lol! A jeweler might do something like that to make a ball on the end of a wire. How does the copper melt when the flame terperature is too low? – user18041 Jun 16 '18 at 1:09
• @user18041 The flame temperature of butane is 3578 F. – Superbest Jun 21 '18 at 1:35

Other answers are right about an iron bar, but it is possible to burn iron with a candle flame — in the right form. In particular, it's well known that steel wool will easily burn in self-sustaining fashion. But since it's mostly iron — only around 0.1% carbon by mass — and apparently iron wool will also work, it means that iron does indeed burn.

One difference from a bar is the proportionally greater surface area, allowing more interaction with air and hence more complete combustion. Note that there's a difference between melting and burning; you don't have to melt something before you burn it, and things can still burn even after they've melted.

Another difference is the smaller thermal mass of the heated element. Basically, the strands don't carry heat away quickly enough to stop the surface getting hot enough to react with the air.

Of course, @Criggie points out that iron powder will also burn. Indeed many materials that we don't normally think of as flammable (flour, for example) will burn when turned into dust.

• Powdered iron was going to be my suggestion, but yours is close enough to cover that. – Criggie Aug 9 '16 at 2:53

If I have a lit candle right beneath an iron bar, assuming the candle will lit indefinitely, and that the heat-losing rate is below the heat-getting rate. Will the bar eventually reach the needed temperature for it to melt?

Under those conditions, yes, the candle would melt the iron.

Ultimately, you're balancing heat gains with heat losses. If you assume that the heat gains are always higher than the heat losses, the temperature of the candle no longer carries any significance - it's effectively at infinite temperature, because that's all the meaning temperature has in the first place!

In reality, this is impossible, of course. It takes energy to make heat go against the heat gradient (that is, away from equilibrium). Even in an ideal setup, as the iron gets hotter than the candle flame, it will release more energy that gets in - this is a simple fact of thermodynamics.

However, let's say that we make the experiment in a magical environment like so:

• All the waste products of the burning are immediatelly magically turned back into oxygen (in a real environment, you'd need to remove the waste products and replace them with fresh air - this would of course make you lose heat from the system).
• The candle flame is 100% isolated from the fuel of the candle (in other words, we don't care about what happens to the candle itself as temperature gets hotter - in a real environment, it would eventually spontaneously combust).
• The box walls are 100% reflective and 100% thermal isolant. This means that all the radiation and hot air stays in. In a real environment, this is quite impossible, of course.

As the heat inside the box builds up, the temperature of the flame will also increase - the burning reaction will be far more favoured under higher temperature conditions. Eventually, the temperature of both the flame and the system as a whole will get higher than the melting temperature of the iron (though note that as temperature increases, so does pressure - so it will not be iron melting temperature at standard pressure anymore :)).

If you want to avoid increasing the temperature of the flame, you'd need to keep the candle out of the system entirely. Say, you would have a one-way window into the magical box, so that the candle's radiation will get in, but the heat couldn't get back out. This makes our magical box even more magical, of course - this is entirely un-thermodynamical :) But the main fun part at this point is that you can just remove the candle entirely - your lab room temperature will be quite enough to melt the iron eventually :)

If you want to keep to 100% realistic scenarios, it's not possible for a 1400 °C thing to heat something to 1600 °C; however, a candle under those conditions will no longer have a 1400 °C flame - as heat and pressure increases, so will the temperature of the flame, until your candle starts breaking down spontaneously :) In any case, the thing doing the heating must have a temperature higher than the melting point, otherwise, it would be heated in return.

• Would an air conditioner be an example of one of these magic boxes? – marshal craft Jan 29 '17 at 21:53
• @marshalcraft It could compensate for the losses, yes. But at that point, you're also putting more and more heat from the outside rather than having just the candle - you need to preheat the oxygen in the first scenario, or keep increasing the heat input in the second; in both cases, you're basically introducing an object with a much higher temperature to the system. It would be more like putting the candle in an industrial oven :) And just like with my original examples, the candle would disintegrate, since it would necessarily heat up to the same temperature (wax doesn't like 1400 °C :)). – Luaan Jan 30 '17 at 8:28

The bar has, depending on its composition, a specific temperature at which it will first become plastic, then a higher temperature at which it will eventually melt.

If you continue to put heat into the bar, by using say 100 similar candles, then, depending on the specific heat capacity of the metal, (that is the amount of heat needed to raise the temperature by 1 °C), the bar may melt, if enough heat is provided.

The specific heat capacity of solid aluminum (0.904 J/g/°C) is different than the specific heat capacity of solid iron (0.449 J/g/°C). This means that it would require almost twice as much heat to increase the temperature of a given mass of aluminum by 1°C compared to the amount of heat required to increase the temperature of the same mass of iron by 1°C.

This is a rough idea of how heat, increasing along the bottom line, raises temperature, and the horizontal lines are temperatures at which a phase change occurs.

Heat is a form of energy, when the bar heats up, and the candle burns out, the heat energy contained in the bar will be lost to the surrounding air, until everything in the system is at the same temperature.

There is one exception to this idea, called a phase transition. Say you continue to put heat into a solid bar, then just at the point where it goes to liquid form, you can put heat in, without raising the temperature.

The heat energy from the candles temporarily goes into breaking the chemical bonds that allow a solid to become liquid. After that the liquid metal will keep rising in temperature as long as you keep putting heat in.

Yes, heat is cumulative. The outer core of the Earth is molten iron, because the decay of radioactive minerals builds up heat. Heat is evolved and is conducted up from the ground and lost to space. The final temperature is determined by the balance of heat generation (near-constant) and loss (faster when the core-to-surface temperature difference is high).

But, a candle takes fuel, and air, at room temperature as inputs, and produces hot gasses as outputs. The maximum flame temperature for candle/cold-air isn't high enough to melt iron. That temperature is determined by the evolved energy and the heat capacity of the exhaust gas. One must use preheating of fuel and air (as in a blast furnace) to raise that temperature. Such preheating allows coke/air to smelt iron from ore.

A candle, burning in typical environments does not produce a flame with a temperature high enough to melt iron.

Placing a second candle next to it, does not change the temperature of either flame.

Placing many candles next to each other will then start influencing the environment. Warming up the air and the wax, the flame temperature will rise marginally.

However, the limiting factor in the combustion is the supply of oxygen. Medieval blacksmiths discovered that with the use of bellows blowing extra air into the fire they could accelerate the combustion, resulting in furnace temperatures hot enough to melt iron.

I think it is likely that you could do something similar with wax, boiling the wax, and then mixing the wax vapour with oxygen before feeding the combined vapour through a nozzle into the combustion chamber.

(I think most answers misses the point of the layman question which is more about conservation of the energy as I see it, and not specifically the iron bar and candle; and/or are too complex, so here is my attempt)

The iron bar will never heat more than the candle, so no, the heat is not "cumulative". If the candle burns with 1000C, then the 1000C is the maximum temperature the iron bar will ever get at, even with perfect thermal isolation (and thus your iron bar will not melt as it is below melting point of iron, as mentioned in other answers)

If it were cumulative, then you could use 5 candles to heat stuff up to 5000C which is simply not how it works (you could get your perfectly insulated iron bar up to 1000C 5 times faster that way, though).

And the heat will go wherever the heat would go if there wasn't your perfectly-isolated-iron-bar (imagine just burning the candle, without any iron bar: where does the heat go?). But if both iron bar and candle are at 1000C, and in perfectly insulated environment, there is no heat.

I think main confusion comes the name "heat/thermal energy". It might be more understandable to layman if you look at it more like "thermal difference energy". Because, when two objects are in equilibrium, ie. of the same temperature (whatever temperature that may be), there is no energy transfer involved, and hence no heat. Even if they are both at 1000C.

That is because heat is not what layman usually think of it (property of some object), but actually energy transfer resulting from different temperatures. Simple wikipedia should clarify things more.

(Note that this heavily simplified, but as I said, I try to explain it in simple-to-understand terms)

• You can't multiply temperatures in degrees C, but you can multiply Kelvins. 1000C * 5 = (1000+273) * 5 = 6365K = 6092C. – MSalters Aug 8 '16 at 8:06
• @MSalters so, are you saying that five candles would heat iron bar to over 6000C (provided you isolate it enough)? And 100 candles over 120 thousands degrees Celsius? – Matija Nalis Aug 8 '16 at 13:49
• No, not at all. I'm just pointing out that 5 * 1000°C = 6092°C. More importantly, $T^4$ (see Lelocuh's answer) means that $1000°C^4$ really is $1273K * 1273K * 1273K * 1273K$ – MSalters Aug 8 '16 at 15:26
• @MSalters If your candles have an electric charge of 1000 C each, you can easily multiply the charge of a single candle by the number of candles to get the total charge of 5000 C. Matija: The term "heat" doesn't mean what you think it means. – dasdingonesin Aug 9 '16 at 6:46
• @dasdingonesin: C in this context is degrees Celsius, not Coulomb. (Besides, 1000 Coulomb is unrealistic) – MSalters Aug 9 '16 at 9:52

Based on the comments of MSalters on this answer, the thing here is that the candle isn't just 'releasing' heat to be absorbed by other things — the candle is producing hot gasses.

The candle heats the iron bar through conduction; the hot gasses transfer their energy to the cooler iron.

As the candle burns, it produces more hot gasses, but not hotter gas; the temperature of these combustion products is the limit on how much the candle can heat the iron bar.

Similar to Matija Nalis I assume the OP is asking about conservation of energy here.

In that case here are my answers:

### No

If you have a trivial experiment where you put the candle below the bar in a wide open space, nothing spectactular will happen. In the best case scenario, the bar will get as hot as the candle, and that's it.

### Yes

If you put the candle and the bar in a box which is a perfect insulator and does not let any heat or light in or out (and does not heat up itself), and if you ignore the fact that the candle needs oxygen to burn and the fact that the candle consumes itself, yes, then you can melt the bar, or actually increase the temperature to any arbitrary amount. (Conveniently ignoring the fact that the heat will destroy the candle.)

Reason: the burning candle transforms energy stored in molecules into heat (and waste molecules / ash), which is just another kind of energy. This heat energy, since it cannot get out of the perfectly insulating box, will accumulate indefinitely. This kind of energy cannot spontaneously transform itself back into a candle, so it will stay as it is, and stuff inside the box will get ever hotter and hotter (until your fuel = oxygen/candle "stuff" runs out, but we decided to ignore that to concentrate on the "conservation of energy" question).

The bar will also start to glow brighter and brighter, but since our perfect box does not let that kind of energy out either (and does not heat up itself, based on our assumptions), this energy will also stay inside and become more and more.

• I'm not sure we can ignore the fact the candle consumes itself, because that too consumes heat -- the more candle in the box, the more non-iron substance you have to absorb the heat. – Hurkyl Aug 9 '16 at 11:05
• Yes, consider it a thought experiment. The point is that the heat energy increases in the closed box until "something gives" (obviously the candle will melt long before the iron bar, but that's beside the point), it is cumulative as the OP asked. – AnoE Aug 9 '16 at 11:42
• My point, though, is that heat != temperature. If you double the fuel available, you double the heat energy produced, but you also double the heat capacity of the combustion products! – Hurkyl Aug 9 '16 at 11:50

## protected by Qmechanic♦Aug 7 '16 at 12:46

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