Science desk
< December 29	<< Nov \| December \| Jan >>	December 31 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

December 30

salt vs. freshwater evaporation rates

Hi. I know this is probably an obvious question (and it's not homework). Say you have two bodies of water. They can be lakes or puddles, for example. The two bodies of water are the same shape, depth, volume, surface area, etc, and are placed under the same temperature with the same amount of sunlight and the same initial conditions. One body of water is freshwater, while the other is saltwater. They both have the same amount of water entering and exiting via rivers, if you are considering lakes rather than puddles. Assuming no other factors (and assuming that the initial conditions either don't change or change at the same rate), which body of water will evaporate faster, and in the case of puddles, leave a dry basin sooner? What about brackish, or hypopthetical stratified lakes with freshwater on top and saltwater on the bottom? How do they compare? Thanks. ~A H 1^(TCU) 00:06, 30 December 2008 (UTC)[reply]

The fresh water will evaporate faster. See colligative property and especially Raoult's law. Basically, the vapor pressure of water vapor over a body of water is proportional to the relatve amount of actual water in that body of water. Since salt water is less "water" than fresh water is, it will evaporate slower, since it has a lower vapor pressure. --Jayron32.talk.contribs 00:21, 30 December 2008 (UTC)[reply]

Following EC:

Going by Evaporation: Concentration of other substances in the liquid (impurities): If the liquid contains other substances, it will have a lower capacity for evaporation. Thus saltwater should evaporate more slowly and increasingly so since the concentration of salt increases in relation to the remaining water. Pretty much the same as Jayron said. Lisa4edit (talk) 00:24, 30 December 2008 (UTC)[reply]

Arguing to the other side, and varying the stated initial conditions somewhat: I have observed that if ice is on a sidewalk, if salt or calcium chloride icemelter is added. it seems that changing the ice to water with chemical additions, causes it to evaporate, leaving dry sidewalk, compared to ice covered sidewalk where no salt/icemelter was added. Water ice does not appear to sublimate (ice phase to vapor phase) as rapidly as melted ice (water with dissolved solids) evaporates. Edison (talk) 04:32, 30 December 2008 (UTC)[reply]

Solid ice has a MUCH lower vapor pressure than does liquid water; in this case because you are looking at different phases (solid versus liquid) and thus different modes of molecular organization, you are really comparing apples to oranges. --Jayron32.talk.contribs 04:35, 30 December 2008 (UTC)[reply]

Fine. Then I will build a nice little machine: Ice is melted with salt. Then water evaporates into a vertical column (height h) above it. The partial pressure of water vapor will be the vapor pressure of the salt-water mixture (at some temperature below 0 degree Celsius) at the bottom of the column and will be lowered by a factor of exp(-m*g*h/(k*T)) at the top of the column (the mass of a water molecule is about 18 g/N_A; the factor is about 0.9999 for T = 270 K ad h = 1 m). If the column isn't very high then the partial pressure of water vapor is still larger than the vapor pressure of pure ice. I will place nucleation sites at the top of the column so ice crystals can grow. While the salt-water solution at the bottom has cooled during evaporation, the ice crystals will get a little bit warmer due to the heat of sublimation being regained. When the ice crystals are large enough, I'll drop them down and use some mechanism to extract energy from the falling. Then they'll meet the salt-rich (due to evaporation) solution at the bottom and melt, and I can restart the whole process ;-)

Icek (talk) 11:24, 30 December 2008 (UTC)[reply]

Something similar to the device you describe already exists and is widely available in a – non-energy producing – prototype form. It uses an exotic fluid called 'seawater', which falls as a crystalline solid known as 'snow'. This remarkable technology is entirely solar powered and carbon neutral.

Somewhat less facetiously, I assume that you're going for a tongue-in-cheek description of a perpetual motion machine (of the first kind): a device which extracts free energy from the environment. As with all such devices, the challenge is in locating the 'leaky' assumption. Here, the trick is that the gravitational potential energy of the ice crystals at the top of the column doesn't come for 'free'. That energy is pulled out of its surroundings, from the kinetic energy of the air in the column (heat). Since you're stealing that energy out of the system (by making the falling ice do work), everything cools down. If the column isn't fitted with an external heat source, then eventually it will be chilled to the point where it stops working.

In principle, you could put the device in a sunny location; that's how the Earth makes its snow, after all. But there are much more efficient ways to extract energy from sunlight. TenOfAllTrades(talk) 16:35, 30 December 2008 (UTC)[reply]

It is of course tongue-in-cheek, but my real point is that it's still a perpetual motion machine of the second kind. When starting the machine, the temperature may be equal in all parts of the machine. Yet mechanical energy can be extracted while it cools without heat exchange with the environment. Icek (talk) 17:11, 30 December 2008 (UTC)[reply]

If the temperature is equal in all parts of the machine, then why would ice crystals grow? Wouldn't they reach thermal equilibrium with the water vapour, adding and sublimating molecules at the same rate? Franamax (talk) 19:23, 30 December 2008 (UTC)[reply]

My description is to be read in the context of the 2 preceding comments by Edison and Jayron32. Icek (talk) 20:05, 30 December 2008 (UTC)[reply]

As nobody seems to care about disproving my perpetual motion machine, I will explain: TenOfAllTrades was of course correct that it doesn't generate energy from nothing, because the crystallization process takes out water vapor molecules with higher-than-average potential energy and if the remaining vapor equilibrizes (is that a word?) and reaches a Maxwell–Boltzmann distribution again, it will be colder. But if it is true that the vapor pressure of water above a liquid solution of water and salt at a temperature below the melting point of pure water is higher than the vapor pressure above pure water ice at the same temperature, then my machine is still a perpetual motion machine of the second kind. Whenever two substances, A and B, mix spontaneously, and the vapor pressure of substance A (salt in our case) is practically zero above the solution, then the vapor pressure substance B above the solution cannot be higher than the vapor pressure above pure B, or the second law of thermodynamics is violated. Icek (talk) 07:12, 31 December 2008 (UTC)[reply]

Collisions in an ideal gas

What is the frequency of particle collisions in an ideal gas, given the standard parameters (particle number, pressure, volume, temperature)? Is this substantially different in a solid? Thanks, *Max* (talk) 06:21, 30 December 2008 (UTC).[reply]

Frequency of particle collisions in ideal gas is zero. Ideal gas is a gas of non-interacting particles. It is, of course, an approximation; no gas is truly ideal. You have to specfy the chemical componsition and the thermodynamic parameters of the real gas to be able to determine the collision frequency. And, by the way, the "collision frequency" in a solid is a very poorly defined quantity. There are electron-electron, electron-phonon, phonon-phonon, electron-impurity collisions, and so on. Collisions may be with or without umklapp if the solid is ordered. Effective collision frequency for longitudinal momentum, transverse momentum, and energy transfer may be quite different; etc. Your question is very good; it is just not answerable in simple terms. Please be more specific. All the best, --Dr Dima (talk) 07:57, 30 December 2008 (UTC)[reply]

Why did this atom get together with that atom?

I mean to ask this on two levels. Re: the formation of the universe. What attracted the first atoms in existence to interact with the other atoms? Secondly,-today. If I point to a random object or even thin air, and in my direct geometrical line lies two atoms. One here, one there. One touches the other one. Why?--Mark L. Dowry (talk) 06:48, 30 December 2008 (UTC)[reply]

I don't know why about the second question, but my guess for the first is that it is still gravity which attracts too objects together. The very first atoms were likely attracted in this way. —Cyclonenim (talk · contribs · email) 09:22, 30 December 2008 (UTC)[reply]

It's not gravity that causes atoms to join to form molecules. There are many types of chemical bonds. Ionic bonds, for example, are electromagnetic attractions between positive and negative ions. StuRat (talk) 15:22, 30 December 2008 (UTC)[reply]

What attracted them? Well, the first atom looked across the rom in science class, saw the 2nd one's hair glistening in the sun, her eyes just sparkled so, and she looked at him with the most beautiful smile...sorry, I just had to say that after reading the title :-)Somebody or his brother (talk) 13:52, 30 December 2008 (UTC)[reply]

The OP may find Cosmogony somewhat interesting. Questions of the behavior of matter during the first instants after the Big Bang are quite difficult to answer, and our understanding of a small, hot universe during that time period is that it was a VERY weird place indeed. As far as what holds atoms together NOW (which is kind like asking what caused them to form molecules in the first place) is pretty simple. Atoms in all chemical compounds are held together by electrostatic forces, see electromagnetism. Its ALL held together, at the molecular level, by positive and negative charges attracting each other. Ionic bonds, covalent bonds, metallic bonds, all of it is simply the positive bits and the negative bits sticking together. Inside of the nucleus of the atom, forces such as the strong nuclear force and the weak nuclear force are at work. Depending on which model you are working from, gravity either is a fictional force, or it is simply too weak to be able to explain the strengths of the interactions at the molecular and sub-atomic scales. --Jayron32.talk.contribs 18:01, 30 December 2008 (UTC)[reply]

PROOF

09:29, 30 December 2008 (UTC)harshagg hi,

actually one asked in misc section that less of school teachers know how to prove like charges repel and unlike attract.Can anyone prove this.I don't think so that simply saying its because property of charges.There must be some trick behind that —Preceding unsigned comment added by 122.163.44.220 (talk) 09:29, 30 December 2008 (UTC)[reply]

I think it is a theorem in quantum field theory. See Chapter I.5 of Quantum Field Theory in a Nutshell by A. Zee, available here. In ~~classical~~grade-school physics there's no reason for it to be true, so you can't fault grade-school teachers for failing to give an explanation. -- BenRG (talk) 12:10, 30 December 2008 (UTC)[reply]

Be a smart ass and say there are no proofs outside of logic and mathematics.--droptone (talk) 12:33, 30 December 2008 (UTC)[reply]

It's probably sufficient to say that centuries of consistent empirical observations lead us to believe that there are two classes of electric charge - positive and negative - and that objects with net positive charge attract objects with net negative charge, etc. You can go on to say that more detailed observations explain the subatomic structures (protons and electrons, generally) which contain this charge. To try and prove that decades or centuries of consistent observations are all correct is more of a philosophy question like Droptone implied. Nimur (talk) 13:46, 30 December 2008 (UTC)[reply]

What's wrong with getting two objects with the same general charge and seeing if they repel and doing likewise with objects of opposite charges? You'd have to assume, of course, that the objects are in fact electrically charged. --Bowlhover (talk) 17:31, 30 December 2008 (UTC)[reply]

See below. -- BenRG (talk) 18:55, 30 December 2008 (UTC)[reply]

The problem of induction.--droptone (talk) 18:41, 30 December 2008 (UTC)[reply]

True, but I would consider the fact that all known like charges repel to be an acceptable proof, especially since all electric charges are due to the same particles and those particles are exactly the same. --Bowlhover (talk) 23:46, 1 January 2009 (UTC)[reply]

The OP is looking for a strange definition of "proof". Its a meaningless question. How do you prove that the sun will actually rise tomorrow? How do you prove that when I let go of my pencil, it will drop to the floor? How do you prove that fire is hot? All of these things, and the entire scope of scientific thought, are "proven" in the same manner as electric charge dynamics. The explanation of "like charges repeling and unlike changes attracting" is consistant with all known observations, and the theory can be used to accurately predict future behavior with something like 100% accuracy. That's a proven as anything in the entire world can be. How do you prove it? Because it always happens that way, and it never happens the other way. What more proof do you need?!? If you can get two negatively charged objects to attract to each other, there are people in Sweden who would like to give you a shiny medal. --Jayron32.talk.contribs 17:53, 30 December 2008 (UTC)[reply]

Nonsense. There are lots of proofs in physics. Newton proved Kepler's laws from his laws of dynamics and gravitation. Before that Kepler's laws were raw empirical facts, but afterward they were necessary consequences of more fundamental laws. Kepler couldn't explain why orbits aren't square, but Newton could. Hundreds of years later general relativity showed that the exponent in the law of gravitation, which was an adjustable parameter in Newton's theory, can only be 2. In quantum field theory, like charges have to repel in a vector field theory—you can't have a theory that's like electromagnetism except that like charges attract. I think that's also true in classical electromagnetism (Maxwell's equations). These are all theorems in the mathematical sense. Proofs in physics rest on axioms that may or may not be right, but proofs in mathematics also rest on axioms that may or may not be right, and mathematicians aren't actually much more certain of their axioms than physicists are. They believe in the basic structure but there's a lot of uncertainty about the details. The accumulated knowledge of biology is to a large extent a brute collection of facts, but in physics it's largely a collection of theorems. -- BenRG (talk) 18:55, 30 December 2008 (UTC)[reply]

I disagree that, for example, Einstein's formulation of gravity "proved" the adjustable parameters any more than Newton's estimate for that parameter. The mathematics may be more elegant, but there is still a fundamental level at which we must say, "General Relativity seems consistent with every observation we have made thus far." The mathematics of Relativity may dictate a parameter of 2 - I don't dispute this. But it is an equally strong statement as Newton's assumption, because it is fundamentally saying that this equation can be used to model reality. The level of elegance of the derivation is irrelevant. Science is the process of modeling the real world, and we can develop any model we like - and any model we contrive can be useful if it fits our observations and makes our predictions more accurate. The notion of "proving" that the theory fits the experiment is only a matter of building trust through repetition. Nimur (talk) 22:37, 30 December 2008 (UTC)[reply]

Which is why, in the end, Popper embraced falsification, not verification, as the mechanism of creating reliable knowledge. It is easy to come up with verifiable claims that are not falsifiable; it is hard to come up with falsifiable claims that can be verified. --98.217.8.46 (talk) 04:37, 2 January 2009 (UTC)[reply]

imortal

If you are biological imortal. What would happen to you, would you age ? would your body still deteriorate. And what are your chances of liveing to 2200

Seeing as being immortal (as a human) is impossible it's an impossible question to answer. If science could develop far enough to make a person immortal it would only be worthwhile if it could ensure that people maintained a decent age of body (say full adulthood, mid 20s - though anything up to maybe 60 could have a market value - being immortal and being in the body of a 110 year old is not likely to be appaeling to consumers). Being immortal body-function wise is different to being immortal life-wise. You might be shot or run-over, or die in a freak magic trick - essentially it's one thing to make people's bodies not die 'naturally', another thing entirely to make someone immune from death itself. 194.221.133.226 (talk) 12:57, 30 December 2008 (UTC)[reply]

Considering that we're talking about make-believe here, any number of scenarios could happen. You might grow older and more decrepit with each passing decade, but never die. Or you might look like you're 25 forever. Perhaps you could be killed by violence or poison or a disease, but your body would never fail on its own. But you might as well wonder if a man who can fly can fly to Mars. That depends entirely on the internal rules of the fiction. -- Captain Disdain (talk) 14:25, 30 December 2008 (UTC)[reply]

Oh dear, an edit conflict ate my post. The links I wished to provide, if some kind soul would like to weave them together coherently then by all means: The Picture of Dorian Gray Vlad the Impaler Programmed cell death Free-radical theory cancer mutation solar radiation Ship of Theseus how many times can the same person step in the same stream? 98.169.163.20 (talk) 15:01, 30 December 2008 (UTC)[reply]

If biological immortality were possible, I'd say it would take the form of constant replacement of body parts (with replacements grown in a lab) as they wear out, so you would have some "new" parts and some "old" parts at any given time. The problem with this is the replacement of the brain. Perhaps it could be replaced a few cells at a time, so as to not wipe out your memories. StuRat (talk) 15:18, 30 December 2008 (UTC)[reply]

We have a very detailed article on immortality. What you asking is, however, a bit vague. If you refer to this reality, then there is no magic, and no immunity from death. You can expect the science to advance far enough in the next 50-100 years to allow replacement of any tissue damaged by aging or disease, without affecting the person's mind and memory too much. Severe enough trauma would still kill anyone; there is no way around it. You cannot make "backups" of people like you make backups of your files; that was only possible in Star Trek reality, if at all. Now, in fantasy books there are many kinds of immortality, too. Probably the most common one is where some races (as in LoTR) or some individuals (as in Juuni Kokki) do not age but can die by the sword. Another type of immortality is that of Koschei, who can only die if the inanimate object holding his soul is destroyed. Third type is immortality of the "living dead", often employed, for example, in vampire or shinigami fiction. Fourth type is immortality of non-biological or purely spiritual beings, but that's not what you are talking about. Ultimate type is the immortality of God; some religions describe "dissolution" in God as an ultimate fate of the righteous people. I hope this helps in any way. All the best, --Dr Dima (talk) 21:02, 30 December 2008 (UTC)[reply]

Turritopsis nutricula is an immortal jellyfish, it reverts to its polyp stage after becoming an adult. However, I don't know if if its cells deteriorate over time. ~A H 1^(TCU) 22:24, 30 December 2008 (UTC)[reply]

To say living things cannot (in principle) be "backed up" is vitalism in my humble opinion. —Tamfang (talk) 06:05, 5 January 2009 (UTC)[reply]

Velocity of an Object under Gravitational Influence

What is the velocity of an object under gravitational influence at any time t?(Reminder:Velocity is dependent on acceleration which is dependent on distance which is dependent on velocity......(I attempted to solve this question a few tens of times but always failed(I even used calculus(I prefer calling differential calculus fluxions and integral calculus fluents because thats how Newton, the first discoverer of calculus called it) and other complex methods to try to solve this question)))----~~The Successor of Physics~~ 14:56, 30 December 2008 (UTC) —Preceding unsigned comment added by Superwj5 (talk • contribs)

I'm not going to solve this, but will make some comments. First, I assume you want to neglect air resistance ? The velocity is given as v = gt, where g is the acceleration due to gravity. But, as you said, the value of g changes with distance from the Earth (or other massive object). In such a scenario we need to determine if the mass of the object is trivially small, such that the acceleration of the massive object towards it can be ignored. If not, then we need to decide what we mean by the velocity of the object. Do you mean the velocity towards the massive object, which is the sum of both accelerations, or do we mean the acceleration of the smaller object alone ? StuRat (talk) 15:06, 30 December 2008 (UTC)[reply]

I mean something more like the acceleration of the smaller object alone.----~~The Successor of Physics~~ 14:39, 31 December 2008 (UTC)

Hmm. So, as is described in Newton's law of universal gravitation, the time derivative of the velocity of an object in a gravitational field is proportional to the strength and direction of the gravitational field. The gravitational field is determined by the distribution of masses. Therefore in order to figure out what the velocity of an object in a gravitational field is at any time, you must know the distribution of other masses in the system, as well as the initial velocity of your mass. There is unfortunately no way to write an algebraic expression for the velocity at time t without specifying the distribution of mass creating the gravitational field. Do you have a specific system in mind in which you wish to know the velocity of a mass? If so, we might be able to help you solve that specific differential equation and determine the velocity at any time. --Bmk (talk) 22:32, 30 December 2008 (UTC)[reply]

I would like an uniform sphere mass model and I would want it to allow for tidal forces too. Also, I would like to include the initial velocities of the objects. I hope I could get something for as much objects as possible, but if it doesn't work a system of two objects would do.----~~The Successor of Physics~~ 14:39, 31 December 2008 (UTC)

It looks like you're trying to solve a complicated version of the n-body problem. Unless you know advanced mathematics and would accept a very slowly-converging infinite series as a solution, and unless one mass is much greater than that of all the others, it is only possible to solve a two-body problem. The solution for a two-body problem is the Kepler orbit; spheres can be treated as point masses located at their centers. Just make sure the distance between them is greater than the sum of the radii and this simplification and this simplification would be realistic. As for tidal forces, I'm really not sure how those can be taken into account, but they have an extremely small effect in reasonably short (with millions of years) timescales for planet- or moon-sized bodies at the usual astronomical distances. --Bowlhover (talk) 07:27, 1 January 2009 (UTC)[reply]

Thank you! It helped a lot!!!----~~The Successor of Physics~~ 09:26, 1 January 2009 (UTC)

I'd suggest a point mass model first. Then maybe try a uniform sphere mass model. StuRat (talk) 03:37, 31 December 2008 (UTC)[reply]

Whatever you might be trying to calculate, remember that only two variables can be in any equation. If you have more, find a function expressing one of the variables with respect to another. For example, in the equation dv/dt = Gm1/r^2, v and r are related by the conservation of energy. If r is written as a function of v, the equation can be solved by integrating. --Bowlhover (talk) 01:54, 31 December 2008 (UTC)[reply]

Motion of an aircraft

If an aircraft travels against the Earth's spin, does it take a larger force for it to reach the same speed as if it were to take off in the direction of the Earth's spin. According to Newton's first law, shouldn't the aircraft taking off in the direction of the Earths spin have a higher initial speed? Thanks in advance. Clover345 (talk) 17:10, 30 December 2008 (UTC)[reply]

Interesting question, but I'd lean toward no. What gets the aircraft moving is the tires pushing against the tarmac and the turbine blades pushing against air molecules. Initially the force vectors resulting from earth's spin affect all those factors. It's not a case where earth spins fast enough for the forward force vector getting strong enough to overcome the aircraft's inertia and flinging it off the surface. Lisa4edit (talk) 18:22, 30 December 2008 (UTC)[reply]

Not tires pushing against the runway- airplanes do not have drive wheels. What matters for takeoff is airspeed. The earth and its atmosphere are rotating, so this is probably pretty close to being ignorable. Friday (talk) 18:28, 30 December 2008 (UTC)[reply]

Lisa, I recently watched a Mythbusters episode where they dragged a carpet opposite the direction of the airplane, and sure enough it took off like normal (much to the surprise of the pilot). Graphical confirmation of what Friday is saying - all the wheels do is support the plane, within a few percent. Franamax (talk) 01:05, 31 December 2008 (UTC)[reply]

Evidently not Albatross airlines. (My flight instructor had it wrong, too.) Thks. for putting the facts straight. Lisa4edit (talk) 01:44, 31 December 2008 (UTC)[reply]

Aircraft speed is more influenced by wind speed, but centrifugal force is important when putting rockets into orbit. Less fuel is needed if the rocket is launched at the equator and ends up going in the direction of the earth's spin. Dmcq (talk) 18:27, 30 December 2008 (UTC)[reply]

You might also want to read up on the jet stream: "Within North America, the time needed to fly east across the continent can be decreased by about 30 minutes if an airplane can fly with the jet stream, or increased by more than that amount if it must fly west against it."--Shantavira|^{feed me} 18:54, 30 December 2008 (UTC)[reply]

Fictitious forces due to the Earth's non-inertial reference frame on a jet plane are very small (less the one percent of a g) - see rotating frame of reference. Since the atmosphere mostly moves with the earth, aside from changing wind patterns, an eastward flying plane has no advantage over a westward plane; they are both pushing through a medium which is (on average) at rest relative to the earth's surface. --Bmk (talk) 22:44, 30 December 2008 (UTC)[reply]

And when I said fictitious forces, I really meant fictitious accelerations. --Bmk (talk) 03:14, 31 December 2008 (UTC)[reply]

Since the Earth, the plane, and the air are all moving at the same speed initially, due to the Earth's rotation, there is no net effect on the velocity as measured relative to the Earth. If you measured relative to a stationary object (let's say the Moon, as it's relatively stationary compared with the Earth), then it would, indeed make a huge difference which way you went. If flying East near the equator, you'd need to go around 1000 miles per hour (relative to the Earth) just to become stationary (relative to the Moon). StuRat (talk) 03:33, 31 December 2008 (UTC)[reply]

I think you've missed the point of the question, all speeds are measured relative to the earth. —Preceding unsigned comment added by 92.8.100.245 (talk) 11:41, 31 December 2008 (UTC)[reply]

No, I didn't miss the point of the Q. The whole point of my answer is that it doesn't matter if the jet was already moving in one direction, as that movement in one direction doesn't exist when measured relative to the Earth and air, which are all moving the same velocity as the jet initially. I then expanded on the answer to point out that the premise is only correct if viewing the jet's motion relative to a stationary observer (off the Earth). StuRat (talk) 07:59, 1 January 2009 (UTC)[reply]

But you are discussing simple Galilean mechanics, which doesn't work here as the earth is a non-inertial reference frame, the question is regarding the increment in centripetal force caused by the planes motion, and you cant explain this away by changing reference frame as the plane is accelerating, not moving, towards the centre of the earth, and is so in all inertial reference frames. —Preceding unsigned comment added by 92.1.148.100 (talk) 12:46, 1 January 2009 (UTC)[reply]

helium

If you inflate a rigid tank to it's maximum pressure with helium, will the tank get lighter? Thanks Bill —Preceding unsigned comment added by 208.100.237.48 (talk) 20:43, 30 December 2008 (UTC)[reply]

That would depend on what was in there before. If it was a vacuum inside, then it will be heavier (since helium weighs more than nothing). If there was already something in there, and you replaced it with helium, then it would depend on the densities involved. - Akamad (talk) 20:48, 30 December 2008 (UTC)[reply]

If the tank is a K bottle at 2500 psi of helium and room temperature then the tank will contain about 1.5 kilograms of helium. If it were full of air at standard pressure, it would contain about 50 grams of air. The answer is no, the pressure of the helium would be sufficient to make it more dense than air. In general, the answer will depend on whether you can sustain a pressure in the tank such that the helium (pressurized) is denser than air at standard pressure; that would yield

{\frac {P_{helium}}{P_{atmospheric}}}={\frac {molarmass_{air}}{molarmass_{helium}}}

...or a pressure of about 7.5 atmospheres (110 psi) - not a very strong tank, if that's the maximum pressure it can withstand. Nimur (talk) 22:54, 30 December 2008 (UTC)[reply]

Passive solar building design

Why are west facing windows bad in passive solar building design? It is mentioned a few times in the article but no full explanation is given. All the other aspects I understand. 41.243.38.111 (talk) 21:31, 30 December 2008 (UTC) Eon[reply]

It's due to the fact that the sun sets on the west. For example, in the middle of summer, at 3pm, when it's hot enough as it is inside your house as it is, you don't want the added heat from solar radiation. - Akamad (talk) 22:21, 30 December 2008 (UTC)[reply]

Thanks, but that still doesn't quite explain it for me, because in the winter you'll want that extra solar radiation. Also, in a climate with very hot summer you might want to avoid the morning sun (east) just as much if enough heat enters the building through conduction. My problem is that when averaged out over 24 hours of earth rotation, I don't see what can be achieved by leaving out west facing windows that doesn't hold true for leaving out east facing windows. 41.243.38.111 (talk) 06:47, 31 December 2008 (UTC) Eon[reply]

Just don't forget that the air is colder in the morning than in the evening because it cools the whole night. Icek (talk) 07:00, 31 December 2008 (UTC)[reply]

Sorry if this is starting to become a discussion. If heat conduction from the air itself is the concern, then (ignoring solar radiation) does the direction that the window is facing make a difference at all? I don't know if it is realistic to assume that cold morning air gets trapped on the east side of the house for the entire day, but even if we did assume this, then preferring east facing windows only offers an advantage as a method of keeping the building cool on hot days and it would be a disadvantage during a cold winter. This is never mentioned in the article. So I'm still in the dark about this one. 41.243.38.111 (talk) 08:38, 31 December 2008 (UTC) Eon[reply]

Well, it's really not just the air that is colder, but also the house itself, and my point was that you generally need more heat in the morning as the temperatures are lower. Icek (talk) 09:58, 31 December 2008 (UTC)[reply]

From what you're saying it does make sense to me that, given the thermal delay surrounding day/night temperatures, east facing (as opposed to west facing) windows will help even out the 24 hour temperature profile. What it does not seem to do in my mind is to even out seasonal variations. Is my understanding correct? —Preceding unsigned comment added by 41.243.38.111 (talk) 16:18, 31 December 2008 (UTC)[reply]

I think it is. For correcting seasonal variations, one simple approach is to have large (vertical) south-facing or north-facing (in the northern or southern hemisphere, respectively) window panes. Then the amount of sunlight entering through these windows will be larger in winter, because the altitude of the Sun will be lower. Icek (talk) 16:35, 31 December 2008 (UTC)[reply]

Thanks, that all makes sense to me now. 41.243.38.111 (talk) 23:41, 31 December 2008 (UTC) Eon[reply]

in the northern hemisphere, in summer, the sun sets north of west, and in the winter, it set south of west. The total difference is large. Therefore, west-facing windows are a major disadvantage in summer and a trivial advantage in winter. -Arch dude (talk) 16:44, 31 December 2008 (UTC)[reply]

Thanks but, again, couldn't the same be said for sunrise and east facing windows? 41.243.38.111 (talk) 23:41, 31 December 2008 (UTC) Eon[reply]

I've wondered the same thing myself, and from my reading I eventually formed this opinion: The inside of your house gets hot by conduction and by radiation from outside. Whilst most of the heat gain is due to radiation (the sun roaring in through your window), this is either contributed to (or offset) by the conductive heat gain (or loss). In the morning, the ambient air temperature is lower, and offsets the radiant heat from the sun (the cooler air absorbs radiant heat energy). In the afternoon, the air is hotter so this offset is lost. That, plus what Arch dude said. Mattopaedia (talk) 03:40, 1 January 2009 (UTC)[reply]

Now I'm totally in the dark again. Looking at some average daily temperature profiles I see daily thermal lag isn't so high. The air temperature is lowest just as the sun rises. Due to conduction air temperature inside a building will tend to rise along with outside air temperature in the first half of the day to reach a maximum a few hours after noon. By allowing sun to also heat the building during the morning upward slope (with east facing windows), higher peak temperatures will be reached. West facing windows sound more optimal to me! They would also capture solar energy before sunset, allowing some stored solar energy (as heat) inside the building to help keep the temperatures slightly warmer at night. So no, I'm not convinced at all :( 41.243.38.111 (talk) 11:12, 1 January 2009 (UTC) Eon[reply]

In winter. It could well be better to have west facing windows, since this will warm your house in the evening. (However, it's more important to have east facing windows, since the morning is when you really need the solar radiation.) But I guess the argument is that the benefits you get in the winter are more than offset by the detriments you get in the summer (of course, your location has a big part to play in this). I am speaking from experience: my house has floor to ceiling windows and it get horrendously hot in the afternoon during summer. The benefits during the winter are not that great. - Akamad (talk) 11:27, 2 January 2009 (UTC)[reply]

Maybe it depends on where you live. I live in a subtropical climate, and I can tell you for certain the afternoon sun really heats the place up, which is nice in winter, but not so much in summer. Mattopaedia (talk) 22:39, 2 January 2009 (UTC)[reply]

Waiting for the water to turn hot

In my house, the water heater is on one side of the building and the kitchen is on the opposite side. When I turn on the hot water tap, I have to wait for up to 30 seconds before the water runs hot. If I'm wanting to waste as little water as possible (let's assume it's not possible to save it in a container), does it make any difference whether I have the water just trickling out while I'm waiting, or have it on greater pressure? Does this depend on the type of heater? I'm assuming that all the cold water in the pipes, between the heater and the tap, has to be let go first before any of the heated water can arrive, so if it makes no difference whether it flows out slowly or quickly, I may as well have it on full pressure and at least save some time. Am I thinking right? -- JackofOz (talk) 22:51, 30 December 2008 (UTC)[reply]

The more quickly the water moves from the heater to the sink, the less time it will have to lose heat through the walls of the pipe. If you reduce the problem to an absurd level, if you open the tap only very slightly, the water might move so slowly toward the sink that it will have lost all heat before it gets there. Get yourself some pipe insulation; you're throwing money out the window as we type. --Milkbreath (talk) 23:04, 30 December 2008 (UTC)[reply]

If you want to read up on this, the length of pipe between the heater and the tap is known by plumbers as a "dead leg", and the insulation as "lagging" (it's a noun and a verb). BrainyBabe (talk) 17:54, 31 December 2008 (UTC)[reply]

You can actually buy devices that continually send the cold water back to the tank rather than running it down the drain - these are used in some hotels where a room may be a LONG distance from the water heater. However, they put the cost of heating the water up because the insulation of the pipes is not perfect. A pretty decent alternative is to use a 'flash heater' that sits under the sink and heats the water up right there under the sink as you need it. The efficiency of those devices isn't as good as the big heaters in your water tank - but because it only heats exactly what you need and does it instantly, there is almost zero wasted water and no energy is wasted in keeping the tank hot just in case you might need it at 4am! I had one of those in our house in the UK - along with a shower fitting that did the same thing. Turning off the big immersion heater made HUGE savings. Sadly, when I built my house here in the USA, I couldn't get the 220 volt wiring to the bathroom in any manner that would pass the local building codes and 110volts isn't enough to get the water hot enough quickly enough to make the system practical. Yet another case where stupid laws make us waste energy. SteveBaker (talk) 23:05, 31 December 2008 (UTC)[reply]

Thanks for those helpful suggestions about how to improve the efficiency of the system. What I'm wanting to know, though, is about how it operates right now, today. Is it a better strategy to have greater pressure or less pressure while waiting for the cold water in the pipes to empty? My theory is that if I can't save any of that cold water, I can at least save some time by letting it out quickly as opposed to slowly. That's what I want confirmation about. -- JackofOz (talk) 01:44, 1 January 2009 (UTC)[reply]

More pressure: The water moves through the pipe faster, so there's less heat lost to the pipe. Although, I suspect the reality is you're only talking about a time saving of a few seconds. This I know because of the time difference to get hot between my pressure regualted bathroom taps versus my unregulated laundry taps. Either way, you're going to lose at least the same initial volume of water that's in the pipe & cold, but theoretically the less heat lost to conduction, the sooner the water gets hot at the tap, and the less water wasted. Mattopaedia (talk) 04:37, 1 January 2009 (UTC)[reply]

Thanks, Matto. -- JackofOz (talk) 04:59, 1 January 2009 (UTC)[reply]

No wuckers, Jacko! Mattopaedia (talk) 05:11, 1 January 2009 (UTC) ;-)[reply]

I have the same problem, Jack. I've noticed the water stays ice cold for 30 seconds, then gets warmer instantly, but only slowly gets hot after that. I believe that the 30 seconds was for the water in the pipes to be dumped out. After that, I believe I'm waiting for the pipes themselves to warm up. With the metal pipes I have, even hot water is only warm when it gets to the end of the pipe. So, if I get in the shower when it's in the warm phase, it then gets hotter and I burn my kibbles and bits. Therefore I have to wait several minutes. I'm also concerned about wasting all that water, so have been taking baths instead. The cold water that pours out at first then mixes with the warm and eventually hot water to make for a just-right bath. Too bad every tub ever made is apparently only long enough for munchkins, though. To answer your original Q, fast would be good for the first phase, but then slower is better to allow the pipes time to heat. However, if you have plastic PVC pipes, then the "warming the pipes" phase may not apply to you, as they don't hold much heat. StuRat (talk) 07:46, 1 January 2009 (UTC)[reply]

I very much doubt that you are saving water by taking a bath instead of a shower. You should take the scientific approach:

When pipes and water are cold - run the water into a bucket until it gets hot enough to shower in. Count how many bucketfulls you'd waste. Now run the shower (again, into a bucket) for the typical duration for which you shower. Count how many bucketfulls you use usefully.

Now run a bath to the depth you normally use - and empty it (using the same bucket) into a nearby sink or toilet - counting the bucketfuls.

I can almost guarantee that the shower will use less water (and hence also less energy to heat the water).

SteveBaker (talk) 17:01, 1 January 2009 (UTC)[reply]

Oh - and remember that when you take a bath, some of the energy in the water goes into warming up the bathtub itself. SteveBaker (talk) 17:03, 1 January 2009 (UTC)[reply]

One big factor that will throw off all your calcs is that I shave in the shower/bath, which adds a good 20 minutes to the timetable. I wonder if I'm the only one who does this ? If seems to work a lot better for me as the used shaving cream, bits of hair, and inevitable blood that falls off my face goes down the drain instead of on the edge of the sink, floor, my clothes, towels, etc. There's also a temperature concern. I leave it cold in my bathroom, but it gets nice and warm in the bath/shower, from the bath water/shower water. I have an electric space heater I could use to heat the entire bathroom, but that's got to cost more as it's expensive electrical heating and I'd be heating a much greater volume of air that way. Also note that, after I finish a bath, I don't drain the bath water immediately, but leave it to add heat and humidity to the cold, dry winter air inside my house, first. StuRat (talk) 16:52, 2 January 2009 (UTC)[reply]

I don't know about the shaving thing (I have a beard which comfortably - and naturally - circumvents that whole thing). An electric shaver seems like it would be more efficient (and certainly shouldn't take 20 minutes or involve large quantities of blood splatter!) But as for the warming & humidifying of the bathroom - I would suspect that the shower does that better because you have an aerosol spray that's mixing with the air rather than water just sitting there in a puddle. Much of the heat from your bathwater is almost certainly being conducted away through the plumbing. But in any case, heating water in order to subsequently heat the air can't possibly be energy-efficient when compared to a heater that's designed to heat air. So I think you're deluding yourself - by the time the bath water has given up it's heat, you're going to be dressed and outta there - so even if you do warm the bathroom that way, the benefit is slow in coming. But, again, a scientific approach is required. Take a thermometer into the bathroom - shower one day - bath the next - plot graphs - publish results. SteveBaker (talk) 19:46, 2 January 2009 (UTC)[reply]

I have to disagree with "heating water in order to subsequently heat the air can't possibly be energy-efficient...". Heating water and using it to subsequently heat the air is hydronic heating, which is quite common. The shower puts too much humidity in the air, so that it condenses on the windows and soaks the window sills. Bath water, on the other hand, adds heat and humidity slowly, so that it has a chance to distribute itself evenly throughout the house. As for electric razors, I've never found one that can give me a smooth shave; my skin feels like sandpaper after using one. StuRat (talk) 07:24, 3 January 2009 (UTC)[reply]

The Californian Energy Commission [1] says: "If your home was built before 1992, chances are your showerheads put out about five gallons of water per minute (gpm). Multiply this by the number of minutes you're in the shower, and the water adds up fast! An average bath requires 30-50 gallons of water. The average shower of four minutes with an old shower head uses 20 gallons of water. With a low-flow shower head, only 10 gallons of water is used. To test the amount of water used in a shower vs. a bath is to put the plug in the bath next time you take a shower (but not a stand-alone shower as you might spill over the lower shower wall). After you've showered, see how much the tub filled up."

That's useful data. If you have a half inch diameter water pipe going between hot tank and shower - and (let's guess) your hot tank is no more than 30' (360") from the shower (that would be a very large and spectacularly badly designed house!) - then the amount of cold water in the pipe is something like 90 cubic inches - which is only about a third of a gallon. If that takes 30 seconds to flow through your pipes and you have to wait another couple of minutes for the pipes to warm up - then you're probably wasting no more than a couple of gallons in letting the water heat up - and that's pretty negligable compared to the 30 to 50 gallons in a bathful. But that doesn't quite gel with a showerhead putting out 5 gallons per minute - because it's being fed from the same water source at the same speed. So I suspect that if your water takes 30 seconds to start getting hot - then your showerhead must be a more modern lower-flow head. Hence, I estimate that you're wasting 2 gallons waiting for the water to get hot - then using 10 gallons more for your shower for a total of 12 gallons...versus a minimum of 20 gallons for a bath. So showering is VASTLY more efficient...even if you have a long wait for the water to get hot. SteveBaker (talk) 19:44, 1 January 2009 (UTC)[reply]

I think it'd be fair to say that most Australian household taps and shower heads are low flow - since 80-odd % of the country's been in drought for the last 5 or so years. Now, to me, a gallon is about 4.5 litres. A showerhead that puts out 22.5L/min seems absolutely insane to me, but that's because we've had some pretty severe water restrictions here (140L/day/person) where a showerhead like that would use up your day's water allowance in a little over 6 minutes. My showerheads use 9L/min, which is pretty common, but you can by ones that use 6. I reckon it's not unreasonable to have more than 30' of pipe between your hot water service and the tap, when you allow for the course the pipe has to take from the HWS under the floor and/or through the roofspace and walls. But, for the sake of the discussion, I'll call it 10m (about 33') of 12mm pipe (Volume =πr²h, or 3.1415926536x0.6²x1000 = 1130cm³ or1.3L (~0.3 Gal)). So that would take a little under 9 seconds to empty the cold water from the pipe at 9L/min. That would mean the water volume needs to be replaced about another 2 times for the water to heat the pipe & become hot. You're still only talking about losing 3.9L though. If you use a 9L/min showerhead, then the water saving compared to a bath is enormous - a 4 minute shower would use no more than 45L (10gal) allowing for time for the water to get hot, probably at least half what you'd use in a bath. Mattopaedia (talk) 23:37, 2 January 2009 (UTC)[reply]

More pressure of course. If you can't get at most of the pipes, at least insulate where the hot pipe comes out of the heater for as long a distance as you can. That hot bit wastes a lot of energy. Polypipe Wrangler (talk) 04:00, 3 January 2009 (UTC)[reply]

Since all of the water that you use that comes from the hot water tank has to be replaced - and therefore heated, the amount of energy you consume is directly proportional to the amount of water you take from the hot tap. Insulation will help only in that it allows you to run the water for a shorter amount of time before jumping into the shower (but we've already decided that this is a small part of the problem compared to the time you spend showering or the depth of the water in your bath). What's more beneficial is that insulating the pipe will make the water arrive at your shower head at a higher temperature which allows you to mix in more cold water to get the temperature bearable. However, the driving factor is still the amount of water you use - and the shower beats the bath hands-down in that regard. SteveBaker (talk) 13:35, 3 January 2009 (UTC)[reply]

Depth and water pressure

If I went down an oceanic trench for thousands of meters aboard a sub, and all of a sudden found a cave on a wall of the trench, could I go in, come out the sub, and scuba dive inside the cave without dying? 190.157.120.42 (talk) 22:52, 30 December 2008 (UTC)[reply]

Why would it matter if there is a cave? The water pressure would still be at ambient (at thousands of meters depth, that would be a really high pressure, like tens of thousands of psi). Your scuba gear would probably be crushed - a normal scuba tank is only at 2000 psi, so air would not flow out of the tank (you'd be lucky if water didn't rush INTO the tank). Also, you would need some kind of hatch on the sub that wouldn't suffer the same problem - which is not likely. While exploring Challenger Deep [2], the Navy was not even able to use air for ballast, (it was not feasible to blow the tanks with air at lower pressure than the ambient water!). Instead they used gasoline or hydraulic fluid for ballast, to overcome the pressure problem. The crew compartments of DSV Alvin or Trieste are far below ambient water pressure, and you'd have a hard time pushing the door outward... Even if the hypothetical cave did exist, and hypothetically, if there were a trapped air space, that air would still be at ambient pressure and you'd have the same problems. Human biology is not well-suited for high pressures, as are found in the deep ocean. While scuba-diving, you cope because the air you breathe is kept at ambient water pressure via a pressure regulator, and so you are not crushed by the water. This limits you to only a shallow dive, though. A bathyscaphe or other deep-sea vehicle protects and isolates the humans from the pressurized water environment, by keeping them inside a hull which is kept at survivable pressure levels. Nimur (talk) 23:10, 30 December 2008 (UTC)[reply]

But doesn't the height of water determine its pressure? In other words, "the pressure at any point in a still body of water is the sum of the pressure due to the weight of the atmosphere and the pressure due to the weight of the column of water directly above that point" [3] So out in the open that would certainly be true, but if I get into a cave, don't I reduce the pressure from the column of water above any point inside the cave? 190.157.120.42 (talk) 23:27, 30 December 2008 (UTC)[reply]

Pressure of a fluid is due to the pressure of anything above it (or horizontally exposed to it, or below it), including pressurized air in a cave (or solid material "pushing back" on the water) (or the water outside the cave squeezing on the water inside the cave). Take a look at Hydrostatic equilibrium - sorry that this article is a little bit technical - but the idea is fairly simple. The water must be in equilibrium with the other water around it. This means that the water in the cave will be at the same pressure as the water outside the cave. Pressure acts in all directions - up, down, horizontally - and the unique case of a column of water results in increasing pressure with depth because of the weight of the water above. Nimur (talk) 00:13, 31 December 2008 (UTC)[reply]

Thank you for your replies. 190.157.120.42 (talk) 00:30, 31 December 2008 (UTC)[reply]