A piece of metal under the hot sun...

Recommended Posts

Sixty Celsius is about the maximum you can handle without burning your hands...

That piece of metal left under the sun, or a car bumper, or the beach sand can burn your skin but the ambient temperature is only 30 Celsius.

What is going on ? How does the metal gets way hotter ? It is not only the thermal conductivity from the material-to-skin; it is truly hotter. Or seems to be

Share on other sites

It is to do with how much energy the material absorbs from the sun. The air is transparent and most of the sunlight passes through (warming a bit on route). The metal bumber actually absorbs more heat and is indeed hotter. The metal then slowly releases said heat and warms the air slowly, but yes - it is hotter, because it absorbs more radiation.

PS - it's why you can fry an egg on the pavement on a hot day - or burn your feet on the sand.

Share on other sites

yeah, the metal can absorb much more heat than the air around it. It can also radiate a lot more heat than the air (or transfer it into your hand)

Share on other sites

Because of the thermal conductivity of metals, the part you touch is not immediately cooled down; instead, heat from a large portion of the metal is transferred to your skin. Consider if you touched very hot foam, the part you touched would immediately cool and you wouldn't get burned.

Share on other sites

Nevermind the rate of heat conduction of the metal and your hand. A thermometer reading the air temperature would read 30C and one reading the temperature of the metal would read 60C. The metal feels hotter because it is hotter. As DrP says, it's because the metal absorbs the radiation, but the air doesn't.

Share on other sites

Thanks.

OK, it is truly hotter, got it. Yes, it has to be, feet at 37 Celsius get blisters on pavement, and fried blisters on a steel plate. That can reach a guessing 70 Celsius

Now, the metal mass has nothing to do with the temperature after reaching plateau, right ? (am not saying amount of heat!)

What material index, property, constant... tells which metal will be hotter under the same sun?

Stainless steel, plain iron, aluminium, brass, lead... which will be at the higher temperature ?

And if all of the above receive the same flat black coating of paint, would they all be at the same temperature or still differ ?

Edited by Externet

Share on other sites
Nevermind the rate of heat conduction of the metal and your hand. A thermometer reading the air temperature would read 30C and one reading the temperature of the metal would read 60C. The metal feels hotter because it is hotter. As DrP says, it's because the metal absorbs the radiation, but the air doesn't.

Ah, but this effect should not be ignored. The ability of a material to transfer heat is very important in this scenario. The metal will feel hotter even if it is at the same temperature as the air. Consider this: would you place your hand in an oven that was at 175 ºC (350 ºF)? Would you grab a casserole dish or cookie sheet that was in that oven?

Thanks.

OK, it is truly hotter, got it. Yes, it has to be, feet at 37 Celsius get blisters on pavement, and fried blisters on a steel plate. That can reach a guessing 70 Celsius

Now, the metal mass has nothing to do with the temperature after reaching plateau, right ? (am not saying amount of heat!)

What material index, property, constant... tells which metal will be hotter under the same sun?

Stainless steel, plain iron, aluminium, brass, lead... which will be at the higher temperature ?

And if all of the above receive the same flat black coating of paint, would they all be at the same temperature or still differ ?

In equilibrium, they will be at the same temperature if they are blackbodies. A material with a high heat capacity will take longer to get to equilibrium.

Share on other sites

"Ah, but this effect should not be ignored."

Failing to ignore it is going off topic, you might find youself talking about dishes in ovens.

Share on other sites
"Ah, but this effect should not be ignored."

Failing to ignore it is going off topic, you might find youself talking about dishes in ovens.

It's a combination of the two main effects listed above, neither is negligible, so we need to discuss both.

Share on other sites

I think there are at least three factors at work:

1) The metal absorbs radiation very readily, so a lot of heat builds up.

2) It has a very low specific heat capacity (the amount of energy needed to raise a given mass a given temperature), so it gets to a very high temperature from relatively little absorbed radiation. To contrast, water might absorb as much energy, but it takes ten times as much energy to raise the temperature the same amount. It also stores energy very well, which is why deserts have extreme temperature fluctuations, and islands have moderate climates.

3) It conducts heat very well. As Mr. Skeptic and SwansonT mention, you're helping a large piece of metal reach equilibrium by absorbing a great deal of heat. A good insulator probably wouldn't burn you, even if it was very hot, because you're only absorbing heat from the immediate point of contact.

Share on other sites

Thanks.

Let's see if I understood...

Specific heat capacity (from Wikipedia) :

Iron = 0.45

Aluminium = 0.897

Copper = 0.385

This means that plates of the same dimension exposed to same sun, all kept say on top of a piece of insulator; the lead will increase its temperature the highest, and aluminium the least.

and for thermal conductivity :

Aluminium = 237

Iron = 80

Copper = 401

Means that if all are at the same high temperature, touching copper will likely burn skin faster because transfers (conducts) its heat faster than lead.

Is that right ?

Share on other sites

Don't forget to take into account the densities of the above metals!

Share on other sites

I think that's about right, yes, although a few things come to mind.

First, I'm not sure how the absorption/reflection works with different materials. Probably if you painted them all the same color it would mitigate it.

Second, we're ignoring rate of re-radiation into the air, here, which I think is closely related to thermal conductivity. As in, leaving the plates outside all day won't give you nearly as dramatic a temperature difference as the numbers suggest, since they're giving off heat the whole time. If you give them all an instantaneous burst of energy, however, their temperatures would be exactly inversely proportional to their heat capacities (I think).

Also, it seems like you have to consider heat capacity and conductivity together. For example, copper is more conductive, but aluminum at the same temperature would have more heat to conduct. So if you put one hand on each plate, the copper one will burn faster at first, but will be "used up" quicker.

EDIT: And the densities! (Thanks Skeptic.) That does complicate things. 1 gram of aluminum is a lot bigger than 1 gram of lead, so it's going to be intercepting a lot more radiation, and the heat has to be conducted farther to reach your had. Oy. You know what? Somebody should just test it...

Share on other sites

The final temperature of the metal is not determined by density or specific heat.

the final temperature is determined by the equilibrium between the solar radiation (or insolation - the sun shines on the surface, mostly visible light) and thermal radiation (infrared from the metal). Insolation from the sun is about 1000 W/m2 on a nice day. Since metals can be up to 80% reflective, only 20% gets absorbed. So, we can say that:

$W_{in}=0.2\cdot{1000}=200W$

The second effect is the thermal radiation. This is the infrared that is emitted by the metal. It increases a lot when it gets hot (it is a 4th order relation to the temperature!!):

$W_{out} = \epsilon(T) \cdot \sigma \cdot{A}\cdot T^4$

$\sigma$ = Stefan–Boltzmann constant = 5.670400×10−8 W·m-2·K-4

$\epsilon(T)$= a correction factor because the metal is no perfect "black body".

$A$=the surface area (1 m2, because we looked at the sun's radiation in W/m2)

$T$=Temperature - the parameter we're trying to determine

$W_{in}$=incoming power (insolation from the sun)

$W_{out}$=outgoing power (infrared radiation from the metal)

If we neglect the convective heat transfer (air moving because of temperature differences) - which is quite wrong in fact, but makes explaining a little easier now... we can say that after enough time, the temperature does not increase anymore.

So: $W_{in}=W_{out}$

$W_{in}=0.2\cdot{1000}=W_{out} = \epsilon(T) \cdot \sigma \cdot A \cdot T^4$

Which enables the calculation of the temperature. Note that there is no specific heat ($C_P$) or density here!!

Heat transfer into your hand (when you're stupid, and touch the metal):

This is where the mass, specific heat and density come into play... and mostly the conductivity.

It goes a bit too far to explain all the theory of heat transfer today. Have a read at the website of wikipedia - it's quite brief, but it's a good start for those who want to learn:

http://en.wikipedia.org/wiki/Heat_transfer#Conduction

Edited by CaptainPanic
fixing latex

Share on other sites

If there is no heat capacity, no thermal conductivity, no mass and no density in the formula; I would say the plateau temperature reached calculates to the same figure no matter which material is under the sun.

That does not make any sense to my ignorance. Where am I goofing?

Does it mean wood and iron will have the same temperature but touching the wood will not burn because its amount of heat conducted to the skin due to poor conductivity will be dissipated sooner and not 'burn' ?

[ The 1m^2 exposed surface is a good number for the analysis, all the different material plates under the same sun being that area, each being thicker or thinner to yield same mass, all painted same flat black and resting on the same insulation ]

Miguel

Share on other sites
If there is no heat capacity, no thermal conductivity, no mass and no density in the formula; I would say the plateau temperature reached calculates to the same figure no matter which material is under the sun.

That does not make any sense to my ignorance. Where am I goofing?

The formula assumes the energy is absorbed, which is not necessarily the case for a gas that is mostly transparent to the radiation. It's also included in the blackbody "correction factor" CaptainPanic mentioned; it's called the emissivity. (However, one would not include both it and the effect of reflection, since the latter is included in the former)

Does it mean wood and iron will have the same temperature but touching the wood will not burn because its amount of heat conducted to the skin due to poor conductivity will be dissipated sooner and not 'burn' ?

[ The 1m^2 exposed surface is a good number for the analysis, all the different material plates under the same sun being that area, each being thicker or thinner to yield same mass, all painted same flat black and resting on the same insulation ]

Miguel

Basically, yes. You could have temperature differences because of different emissivities, but if they are at the same temperature, something with a higher heat capacity has more energy that it can transfer for each degree it drops in temperature. The bottom line is the ability to burn is not just a function of the temperature, as in the oven example I gave previously.

Create an account

Register a new account