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Posts Tagged heat

Infra-red and heat Marcus Wilson Jul 19

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Here’s a bit of physics that’s coming up in my lectures – what’s the connection between heat and infra-red?  You’ve probably seen imagery from ‘thermal imagers’ or infra-red (IR) cameras, usually on police shows, taken from a helicopter as it follows a suspect fleeing down some alley-way at night.  You’ll see that ‘hot’ things (like the bonnet of a car or a person) will be emitting strongly, whereas colder things emit less. Click here for a more cute example.

Infra-red is part of the electromagnetic spectrum, and is the region with wavelengths a bit longer than visible light (about 0.7 micrometres to about 14 micrometres). Our eyes aren’t sensitive to it, but it is not so difficult to build equipment that is. We ’see’ because our eyes pick up light that has been reflected from an object (outdoors in the daytime, it will be sunlight that has been reflected). An infra-red camera will ’see’ because it picks up infra-red light that has been emitted by an object AS WELL AS light that has been reflected (e.g. from the sun).  The sun emits a truck load of infra-red light, just as it does visible light. 

Why do everyday objects emit IR but not visible?  It’s because of their temperature. A blackbody (an object that doesn’t reflect light) will emit energy across a broad spectrum, and just where this spectrum is centred depends on its temperature. Hotter objects have a higher fraction of their emissions at shorter wavelengths than cooler objects. The sun (about 6000 Kelvin) emits a lot of energy in the visible spectrum (400 – 700 nanometre wavelengths), but an object much cooler (such as me) will emit virtually zero energy in the visible spectrum – instead it will be all in the infra-red region (and longer wavelengths still). An object somewhere between the two in temperature (e.g. a log on a fire) will  emit a bit of visible light, but this will be at long-wavelengths, so will look red or perhaps orange – it will never get hot enough to start looking green or blue.  

So an infra-red camera used at night will tend to see warm or hot things. It’s also perfectly possible to use it during the daytime; it will still ’see’ warm objects, but interpreting the image will be complicated by the fact that the sun also irradiates the scene with infra-red. So in that sense, IR is not quite synonomous with heat – a cold object that happens to be quite reflective and have the sun on it could look quite bright to an infra-red camera.

Also, bear in mind that there are other forms by which heat can move. With infra-red (and visible) we are talking about radiation – remember that heat can pass by convection (e.g. circulation of warm air) and conduction (flow through a solid such as the bottom of your frying pan). And, while we’re at it, remember that visible (and ultraviolet) light carries heat energy that can heat an object as well – bathe yourself in u.v. and you’ll start burning nicely. So we need to be a bit careful using the words infra-red and heat synonomously.

 

It’s cold outside… Marcus Wilson Jul 01

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…Well, it was this morning. Those unfortunate people like us who have two cars and a lot of stuff and only a double garage, meaning one car has to sit uncovered on the drive, will have noticed that the ice on the car windscreen is generally thicker than the ice on the side windows. Why is that? Surely the temperature of the air is the same on all surfaces of the car?

The reason is that it isn’t just the temperature of the air that controls the temperature of a surface. Convection isn’t the only form of heat transfer. In this case we have radiation too. When you sit in front of a fire, you feel nice and warm, but if someone put a screen between you and the fire you would immediately feel the difference. It’s because heat is transferring from the fire to you radiatively (as well as through heating the air which moves around the room – convection).

In the case of a clear winter’s morning, the opposite is happening. The sky has very little means of radiation. [Amendment 2 July 2010 - The NIGHT sky has little means of radiation - the day sky has lots of scattered sunlight...] There’s a bit of air in it, which will radiate some energy, mostly the longer infra-red wavelengths, but this air is both cold and thin, meaning there isn’t a lot of heat coming from it. Meanwhile, your windscreen will happily radiate energy in the infra-red being glass (it’s partly emissive and partly reflective to infra-red), and the net effect is that it will cool down to a lower temperature than the the ambient air around it.

However, your side windows are facing more horizontally, and are receiving energy from the ground, hedges, nearby buildings, which are all going to be at a rather higher temperature than the sky. Consequently, they are going to be a bit warmer.  Parking in a covered carport helps, even though the air might be below zero, because now the windscreen faces the inside roof of the carport, which is going to be radiating a whole lot more infra-red than the clear night sky will.

Incidently, when I lived in Bedford, in the UK, we had a winter where the temperature failed to go above zero for three weeks (the river froze – you could tell because there was a shopping trolley in the middle of it with a set of footprints across to it – rather stupid if you ask me – but someone did it and survived). I used to measure the temperatures in the mornings in ‘minutes’ – being the time it took to get my car into a state where I could see out of the windscreen.  I think the lowest temperature was 12 minutes of frost. (N.B. A blanket over the windscreen overnight didn’t help much – as soon as you took it off in the morning the frost would start forming…)

Why you need to proof read Marcus Wilson May 28

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I’ve just supervised a test for a group of second year students. On looking at their answers afterwards, it was rapidly clear that there was a problem with one of the questions. Specifically, I had given the value of Boltzmann’s constant as 1.38 times 10 to the power 23  Joules per Kelvin, instead of 1.38 times 10 to the power minus 23 Joules per Kelvin.   Just one little minus sign in an exponent – it had slipped my proof reading, but it makes a difference of 10 to the power 46.   Not exactly a trivial mistake, and using the wrong value (as all my students did) leads to answers that are clearly wrong.

Maybe it did confuse some students. But not a single one was prepared to comment that their answers didn’t look right.  Not a single one seemed to spot that the value given was wrong.   (Here’s a question – should a 2nd year physics or electronic engineering student be expected to know the values of constants like charge on electron, Boltzmann’s constant, Planck’s constant etc.?) Of course sticking your hand up in the middle of a test and telling the lecturer that he’s made a mistake takes a lot of courage, but writing a note on your manuscript that the answer doesn’t look right isn’t so difficult.  So I wonder whether it is a case that no-one could see there was a problem (I hope not) or that no-one was prepared to comment on it.  I’ll have to ask.

Mistakes like this do happen, quiet frequently.  Usually, like this one, they are trivial (I’m not penalising students for using the wrong value). Even after proof-reading things slip though the net. I remember proof-reading this test and checking the values of the constants.  But I think that often, when you look at your own work, you see what you intended to write, not what you actually wrote. Sometimes, mistakes can have really big consequences (e.g. in the case of the  Mars Climate Orbiter, about 300 million US dollars worth of consequence). As scientists and engineers, we need to have a culture of saying when we think others have made mistakes. And when that person is more senior than you,  it’s very difficult.

What goes up… must come down Marcus Wilson Mar 26

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Yesterday morning while driving into work I was reminded that this week is ‘Balloons over Waikato‘ – the annual hot air balloon festival.  It was hard to miss; I counted 20 balloons making their way gracefully over south-east Hamilton and drifting slowly towards Morrinsville. (NB: I counted the balloons AFTER I had parked the car, not while driving, in case you are worried.)

Then this morning I was treated to the sight of a balloon landing on the University sports fields. I say ‘landing’, but don’t get the impression this was a smooth touchdown. There was a fair wind blowing – in fact I was surprised the balloons were flying at all –  and the balloon came down at about 45 degrees at a fair pace – hit the ground, did a short bounce, hit again, whereupon the basket was tipped onto its side and dragged for several metres before the balloonists manged to deflate the balloon a bit and reduce some of that lateral pull from the wind.  I’m sure it’s the kind of landing that could break bones if you’re not prepared for it.  (Not that I know – I’ve never been in a balloon – and this sight doesn’t encourage me to).

Being the physicist that I am, I think a few estimates are in order.

Wind speed – probably aroud force 3, so about 15 km/h or about 4 metres a second. Cross-sectional area of balloon – maybe about 10 metres by 10 metres, or 100 metres squared. Then, I’ll assume the balloon intercepts in one second a lump of air of size 100 metres squared times 4 metres, that’s  400 metres cubed, which weighs about 400 kg  (density of air is about 1 kg per metre cubed).  At 4 metres per second, the momentum transfered in one second is then 400 kg times 4 m/s or 1600 kg m/s.   The sideways force on the balloon, being rate of change of momentum, is then 1600 kg m/s /s, or 1600 newtons. 

In context, that’s the same as the force of gravity on a mass of 160 kg.   I’m not sure how mass the basket has, but I’d imagine with a couple of people and some propane tanks it could be around 200 or 300 kg or so. So the lateral force of the wind on the balloon is fairly close to the force of gravity on the basket.  (And, while there’s still buoyancy in the balloon – the net downward force on the basket would be rather less than its weight, bringing them closer still.)

I don’t know what the coefficient of friction is for short grass, but probably not desparately high – so I’d expect that sideways force to be able to move the basket across the ground. Which is exactly what happened.

I have no idea what injuries, if any, were sustained by the crew (I was watching at a distance), but I saw one guy hop out without difficulty before I left to go to my office.

Mobile phone physics Marcus Wilson Mar 19

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Just occasionally, I have a crazy thought regarding a physics demonstration.   This is one that I’m thinking about inflicting on my third year electromagnetism class.  

We’ve been discussing the way electromagnetic waves travel (or rather, do not travel) through electrical conductors. Basically, conductors allow electric currents to flow in response to an applied electric field (in simple terms this just means applying a voltage). Electromagnetic waves such as visible light, radio and X-rays contain electric fields, so when one hits a conductor electric currents flow. Flowing currents heat up a material. Where does this heat energy come from? From the wave. In other words, conductors suck out energy from an electomagnetic wave, and, broadly speaking,  the wave can only penetrate so far into the conductor. This distance is what’s known as the ’skin depth’.

Skin depth depends importantly on two things – the conductivity of the material and the frequency of the wave. The higher the conductivity, or the higher the frequency, the smaller the skin depth.  Thus, if you consider the waves to/from a mobile phone (frequency of around 1000 MHz) travelling through aluminium (a very good conductor) the skin depth turns out to be small indeed – microns in size.  That means wrapping a phone in aluminium foil will prevent it from picking up a signal. I’ve already shown this in class.

But – here’s the crazy thought – what about water?   Distilled water is a pretty non-conductive, but what comes out of the tap is loaded with dissolved salts and has a moderate conductivity, albeit several orders of magnitude below aluminium foil.   What’s its skin depth for  mobile phone frequencies?  I’ve done some quick back-of-the-envelope, and I reckon something of the order  few centimetres.  So….I predict that if we put the phone in just a few millimetres of water (YES, it needs waterproofing first!) it will still receive a signal, but suspend it in the middle of a swimming pool and there’s going to be no reception at all.

 I reckon that getting my class to estimate how much water would be required to shut out the signal, and then design an experiment (that might or might not need to include ‘borrowing’ the university swimming pool for a short while) would be a great way to get them to think about the various issues themselves.  There’s plenty of literature to back up that assertion – e.g. Etkina et al., American Journal of Physics 74(11), p979  (2006). The best thing is that I can’t be tempted to tell them the answer –  because I don’t know it – I haven’t done the experiment myself. Though I have found this YouTube…

Copper Conducting Considerable Current Marcus Wilson Mar 12

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The latest kitchen acquisition (no, we don’t spend all our money on buying things for the kitchen) is decent frying pan. We’ve spent too long with frying pans that are about as flat as the Southern Alps.  It’s a copper-based pan, which probably accounts for its expense, with a stainless steel surface.   The reason for the copper is that it conducts heat extremely well, meaning that the surface of the pan will respond nice and evenly and quickly to the heat from below. 

Copper is probably better known, however, for conducting electricity rather than heat.  Its electrical conductivity is extremely high, and, coupled with its ease of working, there is no surprise that electrical wiring accounts for a huge amount of copper. 

These two facts (high thermal conductivity and high electrical conductivity) are not unrelated.  This is because the processes by which they occur are very similar. Electricity is carried by movement of electrons, and electrons are also a major carrier of heat. In copper, there are a lot of electrons that are highly mobile, and hence it has both high electrical and thermal conductivity.

In fact, in metals, the two follow (approximately) a simple relationship – the ratio of the thermal conductivity to the electrical conductivity is approximately proportional to temperature.  This is called the Wiedemann-Franz law. One can ‘derive’ this relationship using some fairly simple hand-wavy physics arguments, though to do it properly is not so easy. I’ll be doing the hand-wavy approach (for those that want to know, it’s the Drude theory) with my second year students soon.

 

Fallstreak cloud Marcus Wilson Mar 10

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Those of you who check out the NZ metservice website frequently, may remember last week’s ’photo-of-the-week’:

fallstreak cloud seccombe reduced.JPG

It’s of fallstreak cloud, and this example was spotted by my mother-in-law, Barbara Seccombe, off the coast from New Plymouth recently.  (Photo credit to my father-in-law, Wally Seccombe, used with permission).

It’s not something you see everyday, so I asked my brother (Damian Wilson), who is a meteorologist in the UK, for an explanation.  (N.B. – would-be meteorologists take note – you should be studying physics…)  Here is Damian’s explanation, used with permission (It’s great when you can get other people to write your blog for you ;-)  thanks guys!)

Very pretty. So you want to know what’s going on?
 
The layer of stratocumulus cloud you can see across the whole picture is composed of droplets of water, but it’s temperature is actually a few degrees below zero (probably around -5C to -10C). Water droplets can happily exist at these temperatures without freezing because there is no nucleation site for ice to start forming. If you think about ice on a car on a cold morning (I know, you’re still in the tail end of summer) you’ll recall there can be lots of fern like pattens – that’s because the ice couldn’t spontaneously form from the water, it needed somewhere to start forming, and it does this on molecular scale defects on the surface. But once a crystal of ice starts growing the molecules of water around it now have something to attach themselves to (the ice crystal itself) and will readily freeze to it. The result is several large crystals covering your car. Now, for a cloud droplet to freeze there must be something for it to nucleate on, and such particles are few and far between in the atmosphere - however, if you cool the cloud, the particles there are do become more effective at nucleating. And once you get nucleation at these temperatures, you’ll get ice crystals. And once you’ve got some ice crystals a little bit of thermodynamics that I won’t go into will ensure that the ice crystals grow further by "deposition" of water molecules from the air and the liquid water drops around them evaporate into the air. In effect, the ice crystals steal the molecules of water from the water droplets. Because there are so few nucleation sites, there are many fewer ice crystals than water droplets, with the result that they are much larger. And larger particles fall faster than small ones (because of air resistance).
 
 So, what’s happened in the cloud in the photograph is that there has been an area of nucleation that has taken place. This is probably because something has caused the air to rise further than in the surrounding air - maybe a thermal of some sort. As the air rises, it expands (because the pressure gets less), it therefore cools and in this case has cooled to a temperature where substantial nucleation is able to start. And hence it has grown lots of ice particles, which are falling out, and are what you can see in the picture. These eventually evaporate (or melt) below the cloud.
 
 Eventually you will end up with a large circular hole in the cloud.
 
  Unless of course, this cloud is entirely above freezing, in which case there’s another method that will work, involving water droplets coalescing to form drizzle. This is a positive feedback type of process, once it gets going it will continue, with the effect that you can get regions of drizzle forming in areas that are otherwise drizzle free. But this picture looks more cloudy than drizzly, which is why I think it is ice.

Heat transfer within edible objects Marcus Wilson Mar 08

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The veggie-juicer in our kitchen will happily take fruit, such as apples and oranges. Apparently, in the case of the orange, it works best if the fruit is cold (but not frozen) throughout. So here’s the question my wife asked me last week: If I have an orange at room temperature, and want to cool it to fridge temperature throughout as quickly as possible, how should I do it?

Putting it in the fridge for several hours will do the trick. But it’s quite slow. Putting it in the freezer would cool it quicker, but we don’t want the outside frozen while the inside remains warm. So what about a combination of the above – putting it in the freezer for a while, then transferring it to the fridge.

Now, I have to say I don’t know the answer to this. The combination freezer-fridge method seems on the face of it to have some merit – get the outside really cold as quickly as possible, then let that cold help bring down the temperature of the inside, while, at the same time, letting the outside rise a bit.

But a second thought says that the distance heat (and cold) penetrates into a substance in a given time depends very much on the thermal diffusivity of that substance.   To penetrate a few centimetres of orange is going to take a given time (approximately the distance squared divided by the thermal diffusivity), no matter what you do to it (except drilling holes in it and pouring liquid nitrogen inside).

So I don’t have an answer. (It sounds like a nice investigation for a science fair project to me – get a nice big orange and stick thermometers into different parts of it – N.B. please don’t ram mercury thermometers into anything – you DON’T want mercury all over your kitchen bench).  But I do know that the question isn’t as trivial as it might sound.  I remember several years ago reading a research article looking at the mathematical modelling of the penetration of burns/scalds into the skin.  What matters here is things like the temperature of the hot object, how long it was in contact with the skin, the area of contact, and how long it was before the patient got the burn under running cold water (and how long they held it there for).  I think the idea of the article was that knowledge of these things would help medical staff make decisions on treatment options.

I could, I suppose, do some mathematical modelling of heat transfer in spherical volumes of water (i.e. oranges), which isn’t going to be too taxing for a theoretical physicist.  But I’ll leave it to those who like experimenting to give it a shot and tell me the answer…

Heads I win, tails you lose Marcus Wilson Jan 25

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The comment on my previous entry raises a few  issues with the way we feel heat.  (NB for those who normally read this blog on http://www.sciblogs.co.nz , you’ll need to go onto physicsstop to see the comment – http://sci.waikato.ac.nz/physicsstop ) 

How hot we feel has more to do than just what the temperature is.  Anyone who has stood outside in a gale will know that it feels much colder than what the thermometer reads. That’s the windchill.  The temperature is the same, but the rate at which heat leaves your body is much higher when the airflow past you is greater. That’s because in still air, your body heats up the air around your skin, so unsurprisingly it feels warmer to you (because the air next to your skin really is warmer). But in a strong wind, that warm air is just blown straight away.

Humidity plays a key role too. Water requires energy to evaporate, and it takes that energy from what it is in contact with. So when sweat evaporates, it cools the skin. It evaporates more readily in low humidity conditions, so here it takes energy from you at a greater rate than in high humidity. Thus a dry heat might feel more tolerable than a wet heat.

Evaporation is what makes you feel cold the second you step out of a swimming pool, especially in nice sunny weather. All that water on you starts evaporating, and it sucks heat out of your body. So there’s a vicious irony here – when you jump in the swimming pool to start your swim, it feels cold (water is better at carrying heat away from you than air), and when you get out of the pool at the end of your swim, it also feels cold.  Why can’t it feel warmer both ways?

Can you feel the cold? Marcus Wilson Jan 21

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Writing the last piece about fridges has reminded me about a comment I heard from a fellow student while I was an undergraduate. I can’t remember the exact circumstances, but it quite possibly had something to do with objects in liquid nitrogen.  Anyway, the comment was something along the lines of ‘The temperature’s so low you can feel the cold radiating from it’.

Hmmm. Yes, we know what you mean, but it’s not quite right, is it?  It is heat that radiates. Hotter things radiate more. Cold is the lack of heat.   If we hold our hand close to something very hot, we can feel the heat radiation. But the hot thing isn’t the only thing that’s radiating, our hand radiates heat as well. What matters is the difference between the heat it receives and the heat it gives off. In my office at the moment, my hands feel to me neither cold nor hot, because what they are pointing at (namely the keyboard) is pretty well the same temperature as my hands themselves.  The amount of radiation arriving on them is roughly balanced by the amount leaving.

So when we ‘feel the cold radiating from something’, what we are feeling is that not enough heat arrives on our hands to balance the heat that is leaving. (Plus probably we are feeling the cold air too, due to convection currents).

But, before I treat my fellow undergraduate too harshly here, I should point out that physicists are quite adept and speaking about the absence of something as being something itself.   When we describe semiconductors (e.g. silicon, as in chips), we talk about n-type and p-type material. ‘n’ stands for negative, and in n-type silicon we consider electrical conduction happening because electrons move. That is a conventional way to think about electrical conductivity.  But in p-type (p is for positive) the mechanism for conduction is slightly different – we talk about the moving of positively charged holes.   A hole is really the absence of an electron, but we can treat a hole as an entity itself – even to the point of assigning it a mass. It’s a bit like those slidy puzzles – slide one tile into the square gap to create a gap where the tile was – the gap (hole) appears to move, though of course it is really the tiles (electrons) that do the moving. When you’ve worked with semiconductors for a while you can forget that a ‘hole’ isn’t a real thing.

So is it then really wrong to say you can feel the cold radiating off something?

 

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