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Imagine that you had two boxes with of steam and water of equal mass, both at 100 degrees Centigrade, and you decide to expose your left land to the steam and your right to the water for a short period of time. Which hand would 'feel hotter' (i.e if you were to take your hand out and measure its temperature, which would be higher ?

The common answer is that the steam but that answer assumes the steam condenses, which would result in a large heat transfer during the phase change. If your hand and the steam or the water are allowed to reach thermal equilibrium then it would make sense for the steam to transfer more heat since it would transfer extra heat equal to its heat of vaporization. But in a scenario where you are not in contact with the water or the steam for very long, the steam would not condense(thermal equilibrium is not reached).

The heat of vaporization of water is quite large, so it should take a decent amount time for all that heat transfer to take place (time that is not available in this scenario). Taking the rate of heat transfer between the steam and your hand and the water and your hand into account, will the heat transferred by the steam and the water be any different ?

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  • $\begingroup$ have you googled this? there seems to be some info already scienceline.ucsb.edu/getkey.php?key=1322 Even if you use "short" time of contact, vapor molecules' energy is higher than that of water molecules' $\endgroup$ – aaaaaa Apr 13 '17 at 3:17
  • $\begingroup$ yes, I tried google but I couldn't find an answer. The link that you linked to says this "...where it would decrease in temperature (to your skins temperature) but would not have to go through a phase change ...."; the answer is again assuming that the person is in contact with the water/steam for long enough for the system to reach thermal equilibrium. My whole question was "what if they are in contact for a short period of time, short enough that thermal equilibrium isn't reached" $\endgroup$ – E7_82_8E Apr 13 '17 at 3:23
  • $\begingroup$ ok, then just imagine two molecules of $H_2O$, in either liquid of gas state, getting in contact with skin. Gas will have higher energy, hence possible more damage $\endgroup$ – aaaaaa Apr 13 '17 at 3:27
  • $\begingroup$ I'm not sure I follow. Why would the gas have higher energy if they are the same temperature? They both have the exact same average kinetic energy (and are composed to the exact same molecules), so wouldn't they have the same energy? the only difference between the two would be distance between each molecule right? $\endgroup$ – E7_82_8E Apr 13 '17 at 3:31
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    $\begingroup$ by energy I'll assume you meant potential energy. I'm still not sure why the higher potential energy of the gas would matter if they are not allowed to reach the same temperature as the skin. The extra potential energy that the gas has shouldn't make a difference if the water isn't allowed to reach thermal equilibrium(because they are in contact for only a short period of time). $\endgroup$ – E7_82_8E Apr 13 '17 at 5:10
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I assume this is a thought experiment. Your question is more complicated than you think because of the ill-defined parameters of this situation. For one, the word 'amount' here could be taken to mean mass, volume or number of particles and each would be a different scenario. This question also raises a common myth that most people still believe which is 'steam burns are a lot more dangerous/serious than hot water burns because you need to consider the latent heat of vaporisation'. In everyday life, this is almost always not true.

Now, first things first; $1$kg of steam at $100°C$ does have more heat in it than $1$kg of water at $100°C$. That being said, steam "usually" does not pose a source of danger that is significantly more hazardous than boiling water does. This is because of many things. Firstly, if your steam is not being forcibly convected or circulated at a high speed, the rate of heat transfer in steam is much lower than in boiling water. This is because steam is a gas and water is a liquid. The great distance between particles in a gas just does not allow for efficient heat flow. Boiling water almost always boils through nucleate boiling and that itself lends a great increase in the convection of boiling liquid water. In other words, boiling water creates bubbles that make the water churn and roll. Unless of course, you're assuming that the water is kept exactly at $100°C$ with just enough heat to maintain its temperature but no more to trigger boiling but even then water wins in passing the heat by virtue of being a liquid. Barring the odd case of superheated, high-pressure steam, you're not conducting heat well enough.

Secondly, not many people often interact with "real" steam in their everyday lives. And by "real", I mean steam that is transparent, hot, dense and highly saturates the air. The white mist most people perceive as steam is not steam, it is "wet steam". It is usually a mix of steam, air, water droplets and steam that has cooled down but remains in a gaseous state i.e. water vapour. Let me be clear, water vapour is not steam. This has to do with phases of water that vary with temperature and pressure. You can easily picture this mentally by observing the fact that clouds are not boiling hot fluffs of water. Unless you're working at a factory of sorts, you're not going to encounter pure, hot, compressed steam.

Additionally, there are other things to consider. Let's say you get steam that condenses and releases all of its energy into your skin to form a water droplet at say, $50°C$. Assuming no strong currents in the steam, that droplet of water is now actually protecting your skin from further steam burns, a sort of reverse-Leidenfrost effect. The water droplet will have to rise to $100°C$ first and then vaporise before any new steam can touch and burn your skin. I assume near constant pressure and no significant movement of air/steam that would "wick away" the droplet or evaporate it to expose more skin. The water droplet would not conduct the heat energy from the steam to your skin any more efficiently than boiling water would conduct heat from the surrounding water to your skin.

If you're going by number of particles, you'd need a lot more steam than you would hot water, volume-wise. If you're going by mass, you'd also need a lot more steam than you would water, volume wise. I assume the time of exposure here isn't nearly enough for those huge volumes of steam to fully transfer heat, nor is the scenario feasible. You're not going to compare dunking your hand in a pot of boiling water to standing in a pressurised room of $100°C$ steam, are you ? The only case to consider here is if you take 'amount' to mean equal volumes of steam vs water and even then the math is against you:

Given $Q=mc \Delta T$ and $Q=mL_f$ and $\rho =\frac{m}{V}$

$1m^3$ of steam at $100°C$ to $50°C$ water = $(1m^3\times0.645kg/m^3)(50K\times4.2kJ/kgK+2257kJ/kg)$ = $1591kJ$ of energy

$1m^3$ of water at $100°C$ to $50°C$ of water = $(1m^3\times970kg/m^3)(50K\times4.2kJ/kgK)$ = $203 700 kJ$ of energy

(I'm estimating here but you get the point)

But isn't the heat of vaporisation of water absurdly bigger than the specific heat capacity of water ? Yes, it is. In fact, making $1$kg of steam from $1$kg of $100°C$ water requires more than five times the amount of energy than taking $1$kg of water from $0°C$ to $100°C$. But here's the thing: Do you know how hard it is to boil a kettle of water empty ? You'd have to boil that kettle for a solid half-hour, potentially longer. You'd never run into anything nearly like that much steam.

Both my parents are doctors and have been for decades. Neither one of them have treated close to any bad steam burns but lots of hot water burns and spills. I practically lived in the clinic equipment room because I had afternoon school and my parents had to take me to work in the mornings. While my parents were seeing patients one wall away, I was told to be quiet and locked in that room with a table, chairs, boxes of stuff, my books and a huge, silver cylindrical autoclave.

The autoclave was used to sterilise surgical equipment and it was built like a vault. The pressure needed to sterilise equipment with steam was so high that the turning door-lock sometimes got stuck and the nurses had to call my dad to crank it open. One day I tried messing around with it and took a shot of steam to my hand: first degree burns that only needed a plaster. Years later, I remember a much more intense pain, bigger blisters, more bandages and antibiotic creams because I spilt spaghetti water over my hand trying to help my mum in the kitchen. Still not convinced ? Ask any cook whether boiling or steaming is a faster cooking method. Unless you've got some kind of space-tech steamer, boiling's going to be faster with a pressure cooker being the fastest. Picture pouring a kettle of boiling water and the steam from that wafting up on your fingers. Okay now picture pouring that same water onto your fingers. Which would you choose ?

Pure steam isn't easy to get. Steam dissipates super quick because steam expands a lot. The expansion of steam is the main force that drives steam turbines around the world. In fact, steam so wants to get "up and out" that it's an effective lifting gas. Lifting gas !

Of course given the right conditions, steam can and will burn worse than water but the pressure needed is industrial standard. You'd have to break a steam-carrying pipe or hold your hand over a steam vent for a silly amount of time. Given a short amount of time, hot water will burn you worse. High speed, pressurised movement of the steam is the dominating factor here.

TLDR; if you had a box of $100°C$ steam and a box of $100°C$ water and you had to put your hand in one of them for 10 seconds, choose the box of steam.

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  • $\begingroup$ To be fair, the times you would be at risk for "steam burns" are when you are dealing with pressurized or high moving steam anyways. That's different than being burned by vapour, which is what you are more likely to encounter on any given day if you don't work in industrial settings. In those cases, quick exposure to steam is likely far worse than quick exposure to water (i.e. busted steam pipe spraying steam vs. busted hot water pipe spraying water). $\endgroup$ – JMac Apr 13 '17 at 11:42
  • $\begingroup$ I agree, you're right. However, I took OP's question to mean non-moving steam or a more everyday, realistic scenario. $\endgroup$ – HsMjstyMstdn Apr 13 '17 at 19:39
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equal amounts of steam and water, both at 100 degrees Centigrade

I think it boils down (sorry) to the density and this notion of "equal amounts".

Water is dense, and a stream of boiling water is going to do a lot of damage to your skin simply because there's so many more molecules hitting you per second than a vapor ordinarily would, as vapor is not very dense.

Now if you're talking about steam in equal amounts (meaning the same mass per second hitting you), then that's probably going to be worse than the water. This is because to get steam on you at that rate would require very high pressure steam, and would strike me a something that could literally burn the flesh from you, if not rip it from you.

Both of these, lest there be children reading, are things that should not be tried at home (or anywhere else). :-)

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  • $\begingroup$ Well put, I think you are right for concentrating on defining "equal amounts" and in differentiating between vapour and steam. +1 for pun $\endgroup$ – HsMjstyMstdn Apr 13 '17 at 19:42
  • $\begingroup$ By equal amounts I meant equal mass. I'll edit the clarification in. And by "a more severe burn" I really just meant "more heat is transferred" and not necessarily the actual damage(like that flesh getting ripped off because of pressurized steam) $\endgroup$ – E7_82_8E Apr 13 '17 at 21:19
  • $\begingroup$ So you mean equal mass in equal time or over different times ? I suspect a sustained but lengthy exposure to a steam flow (at low pressure) would be worse than the equivalent mass in boiling water in some cases, as heat would be able to build up in the skin. $\endgroup$ – StephenG Apr 13 '17 at 21:27
  • $\begingroup$ Yes, I mean you have the same amount of mass in a box(the steam would be under pressure) and you expose you hand to each for an equal but short interval of time. $\endgroup$ – E7_82_8E Apr 13 '17 at 22:09
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It's because of something called the heat of vaporization. It takes 40.65 kJ/mo to get the water from it's boiling temperature as a liquid to become a gas.

Water has a Specific Heat of 4.18j per cubic centimeter degree Celsius. So if you have 1 cubic centimeter of water and you raise it from room temperature at 23 deg Celsius to boiling at 100 deg Celsius, you have to add 103.86 joules of thermal energy to do so. Water has a density of 1g/cc, and has 18g/mol, so that means there are 0.055 mol/cc at STP. That means you have to add 2235.75joules of energy to then take that one cubic centimeter of liquid water and then vaporize it. So you have to add nearly 22 times the amount of thermal energy to vaporize it than to just bring it to boil

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  • $\begingroup$ could you expand please, now it doesn't really answer the question $\endgroup$ – aaaaaa Apr 13 '17 at 4:22
  • $\begingroup$ you can edit your answer to include info there. otherwise it can be flagged for not being an answer and/or downvoted $\endgroup$ – aaaaaa Apr 13 '17 at 5:08
  • $\begingroup$ Sorry, I didn't realize I could edit it. $\endgroup$ – Bryan.Russell.Potts Apr 13 '17 at 5:30
  • $\begingroup$ I think you are assuming that the two systems(the steam and the hand, the water and the hand) reach thermal equilibrium when in the question I said they are not in contact for long enough to reach thermal equilibrium. Read the comments on the question, aaaaaa mentioned the same thing in the comments $\endgroup$ – E7_82_8E Apr 13 '17 at 5:45
  • $\begingroup$ They do not have the same energy even given that they are the same temperature. The reason for the difference in energy levels is the fact that one is a gas and one is a liquid. There are Van der Waals forces that hold the molecules in closely when they are in a liquid state. The energy goes into overcoming those Van der Waals forces, and therefore allowing the molecules to get a larger distance between them. Seeing as how the energy goes into this, the thermal energy contained in the molecules themselves remains the same, so to speak, but the overall energy greatly increases. $\endgroup$ – Bryan.Russell.Potts Apr 13 '17 at 7:03

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