I’m helping on a lesson plan for grades 3-12 where students make ice cream. Addi…

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I’m helping on a lesson plan for grades 3-12 where students make ice cream. Adding salt to the ice makes the ice colder. I’m having trouble explaining why we put salt on the roads to melt ice, but in making ice cream the salt actually lowers the temperature of the ice. — N

These two observations—that salt melts ice and that salt makes ice colder—are actually consistent with one another. When you add salt to ice, you make a relatively ordered mixture—pure crystalline ice and pure crystalline salt. This orderly arrangement is looked on unfavorably by nature; given a chance, nature tends to maximize randomness. There is a much more disorderly arrangement available—salt water—and nature tends toward disorderly arrangements. When you put the salt and ice together, nature’s tendency toward randomness begins to drive the system to rearrange. The ice begins to melt so that the salt can dissolve in it. Although the melting of ice requires energy, the randomness this melting and dissolving produces makes this process take place. The energy needed to melt the ice is extracted from the remaining ice and that ice gets colder. When you’re making ice cream, some of the energy needed to melt the ice also comes from the ice cream mix, so that it gets colder, too. If there is enough salt around, the ice will melt completely to form very cold salt water—the desired result with salt on a slippery sidewalk. The salt water remains liquid well below the normal freezing temperature of water because forming ice crystals would require the salt and water to separate from one another—an orderly and therefore unlikely event. In short, nature’s trend toward disorder causes salt to melt ice, even though that melting lowers the temperatures of everything involved well below the freezing temperature of pure water.

What is the relationship between gravitational force and electromagnetic force?

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What is the relationship between gravitational force and electromagnetic force? — TPC, Foster, OK

As yet, there is no direct relationship between those two forces. Our best current understanding of gravitational forces is as disturbances in the structure of space itself while our best current understanding of electromagnetic forces involves the exchanges of particles known as virtual photons. However, physicists are trying to develop a quantum theory of gravity that would identify gravitational forces with the exchange of particles known as gravitons. How closely such a quantum theory of gravity would resemble the current quantum theory of electromagnetic forces (a theory called quantum electrodynamics) is uncertain. It’s also uncertain whether those two quantum theories will be able to merge together into a single more complete theory. Only time will tell.

How would I go about making a camera that’s more than just a pinhole camera?

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How would I go about making a camera that’s more than just a pinhole camera? — JL, Longview, WA

While a pinhole will project the image of a scene on a piece of film, it doesn’t collect very much light. That’s why a pinhole camera requires very long exposures. A better camera makes use of a converging lens. If you hold a magnifying glass several inches away from a white sheet of paper, you will see that it forms a real image of anything on the other side of it—particularly bright things such as light bulbs or well-lighted windows. A typical camera uses a converging lens that’s not unlike a magnifying glass to form an image of this sort. You could use a magnifying glass to build a camera, but I’d suggest that you start with a camera and rebuild it yourself. Go to a company that processes film and see if they will give you any used disposable cameras. These cameras are of essentially no value to them and they either discard them or recycle them. If you ask around, you should find a photo shop that will give you a couple. You can then disassemble them. You’ll find a very nice lens, a shutter system, a film advance mechanism, and so on. You can use a toothpick or small screwdriver to turn the exposure dial backward so that the camera behaves as though it still has film left. You can then “advance the (non-existent) film” by turning the film sensing gears in the back of the camera with your fingers until the shutter cocks. Finally, you can press the shutter release and watch the shutter open the lens to light. Disposable cameras are great because if you break something in your experimenting, you can just throw away your mistake.

How does an overhead projector work?

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How does an overhead projector work? — SR, Hartford, CT

An overhead projector uses a converging lens and a mirror to project a real image of your transparency onto a screen. A lamp brightly illuminates the transparency and a special surface under the transparency (actually a Fresnel lens) directs the light from the transparency through the projector’s main lens. This lens bends the light rays in such a way that all of the rays spreading outward from one point on the transparency bend back together and merge to one point on the screen. For example, if you make a green dot on the transparency, light rays spread outward from that green dot and some of them pass through the main lens. The lens bends these rays back together so that they form a single green dot on the screen. There is a single point on the screen for the light rays from each point on the transparency.

The pattern of light that forms on the screen is called a real image because it looks just like the original object—in this case the transparency—and it’s real, meaning that you can touch it with your hand. Real images are usually upside-down and backward, but the overhead projector uses its mirror to flip the image over so that it appears right side up. Because of this vertical flip, the side-to-side reversal is a good thing—the right side of the transparency becomes the left side of the screen image (as viewed by the same person) and the screen image is readable.

What path does sunlight follow for you to see a mirage?

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What path does sunlight follow for you to see a mirage? — XF

The first step in explaining a mirage is to understand why the sky is blue, or why it has any color at all. If it weren’t for the earth’s atmosphere, the sky would be black and dotted with stars. That’s how the moon’s sky appears. But the earth’s atmosphere deflects some of the sunlight that passes through it, particularly short-wavelength light such as blue and violet, and this scattered light (Rayleigh scattering) gives the sky its bluish cast. When you look at the blue sky, you’re seeing particles of light that have been scattered away from their original paths into new paths so that they reach your eyes from all directions.

The blue light from the sky normally travels directly toward your eyes so that you see it coming from the sky. But when there is a layer of very hot air near the ground in the distance, some of the blue light from the sky in front of you bends upward toward your eyes. This light was traveling toward the ground in front of you at a very shallow angle but it didn’t hit the ground. Instead, its entry into the hot air layer bent it upward so that it arced away from the ground and toward your eyes. When you look at the ground far in front of you, you see this deflected light from the blue sky turned up at you by the air and it looks as though it has reflected from a layer of water in front of you. This bending of light that occurs when light goes from higher-density cold air to lower-density hot air is called refraction, the same effect that bends light as light enters a camera lens or a raindrop or a glass of water. Whenever light changes speeds, it can experience refraction and light speeds up in going from cold air to hot air. In this case, the light bends upward, missing the ground and eventually reaching your eyes.

I read a recent article about the FCC requiring all TV stations to switch to dig…

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I read a recent article about the FCC requiring all TV stations to switch to digital signals instead of analog ones by 2006. How are digital signals different from analog signals, and will they work with our current TV’s? — JP

Current video signals use continuous physical quantities to represent the brightness and color of the spots on a television screen. For example, the current in a video cable can take any value and that value is used to represent the brightness and color of the spots. This use of a continuous physical quantity (such as current) to represent a continuous physical quantity (such as brightness) is called analog representation.

In a digital video signal, a physical quantity first represents numbers and then these numbers represent the brightness and color of the spots. The physical quantity representing the numbers doesn’t have to be continuous. For example, a current that’s on could represent the number 1 while a current that’s off could represent the number 0. A certain pattern of on and off currents could represent larger numbers and these numbers could then represent brightness and color. This use of a continuous or non-continuous physical quantity (such as magnetization, charge, or current) to represent numbers and then these numbers to represent a continuous physical quantity (such as brightness) is called digital representation.

One advantage of digital representation is that it’s relatively immune to noise. In analog representation, any disturbance in the continuous physical quantity representing the information leads directly to a disturbance in the recovered information. For example, if the strength of a radio wave is representing brightness and color on your television (the current technique), then any disturbance of the radio wave leads directly to a damaged image on your television. But in digital representation, small changes in the physical quantity that’s carrying the information won’t change the numbers that are obtained from that physical quantity and will thus have absolutely no effect on the recovered information. For example, if the strength of a radio wave is representing numbers in digital format, using binary (base two) encoding, then a small disturbance of the radio wave will not affect the binary numbers that are recovered from the radio wave. To see why that’s true, imagine representing the number 1 as a powerful radio wave and a 0 as no radio wave at all. It’s pretty easy to tell a powerful radio wave from an absent one so that, even if there is some radio interference around, it’s unlikely to confuse the receiver. Moreover, even if noise does occasionally confuse the receiver about a number or two, the digital scheme can include redundant information that allows the receiver to identify errors and to fix them! That’s why a compact disk is so immune to noise—even if there is a flaw or dirty spot on the disk, there is enough redundant digital information to reproduce the music flawlessly.

The other advantage to digital representation is that digital compression techniques become possible. A typical video signal contains lots of unnecessary and duplicated information. For example, when two people are standing in a room and the only things that are changing with time are the images of those two people, there is really no reason to keep sending an image of the room itself from the broadcast station to your home. Digital compression can identify redundant information and remove it from the transmission. In doing so, it can use the communication channel more efficiently.

By adopting a digital transmission scheme, the FCC has recognized that broadcasters will be able to send much clearer, more detailed images using digital representations than with the current analog representations, while still occupying the same portions of the electromagnetic spectrum. However, there is a cost—current televisions will not work directly with these new digital signals. To fix that shortcoming, there will be inexpensive converters that receive the new digital signals and recreate the analog signals needed for current televisions. This conversion will allow older televisions to keep working, but the new digital televisions will be designed to make better use of the enhanced details in the transmissions. The new transmissions will contain about 4 times the detail of current transmissions so that the images will be sharper as well as more immune to noise than the current transmissions.

Why does water freeze at very low pressure? I saw an experiment in which a small…

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Why does water freeze at very low pressure? I saw an experiment in which a small amount of water first boiled and then froze solid when exposed to a vacuum. — BLG, Old Bridge, NJ

Water molecules are always leaving the surface of liquid water and when they do, they carry away more than their fair share of the water’s thermal energy. Placing the water in a vacuum speeds this process because (1) it prevents those gaseous water molecules from returning to the liquid water, in which case they would return the thermal energy, and (2) it makes it possible for bubbles of water vapor to remain stable inside the liquid water even at low temperature, so that the water can boil. Overall, the main effect of putting the water in a vacuum is that its molecules leave rapidly and don’t return. Since each leaving water molecule takes away more than its fair share of thermal energy, the water molecules that remain behind become cooler and cooler. You experience this effect when evaporating water from your skin makes you feel cold. In the present case, this cooling is so effective that the remaining water cools all the way to water’s freezing point and the water begins to crystallize into ice. Water molecules continue to leave the surface of ice, a process called sublimation, so that even the ice gradually gets colder in the vacuum.

How much steam is required to produce a unit of power?

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How much steam is required to produce a unit of power? — DKB, Dubai

There is no easy answer to this question, but for an interesting reason. First, “power” is a measure of energy per time (e.g. joules per second or BTUs per hour) so any answer would have to involve the amount of steam per time (e.g. kilograms per second or cubic meters per hour). But even recognizing that requirement, I can’t answer the question. First, I’d need to know the temperature of the steam. The hotter the steam, the more thermal energy it contains and the more energy it could provide. For more complicated reasons, I’d also have to know the pressure of the steam. But there is a fourth issue: even knowing the amount of steam involved and the temperature and pressure of that steam, the amount of useful energy that can be extracted from that steam depends on the existence of a colder object. You can’t turn thermal energy—the type of energy that steam contains—directly into useful work or into electric energy in a continuous manner. You must use the steam in a “heat engine”, converting a fraction of its thermal energy into work as that thermal energy flows as heat from the hot steam to a colder object. This requirement is established by the laws of thermodynamics and there is no way to get around it. The hotter the steam and the colder the object, the larger the fraction of the steam’s thermal energy you can convert to work. However, there is no way to convert all of the steam’s thermal energy into work continuously.

How does a standard water pump work?

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How does a standard water pump work? — ML, Wilmington, NC

The water pumps in most cars are centrifugal pumps. These pumps work by spinning water around in a circle inside a cylindrical pump housing. The pump makes the water spin by pushing it with an impeller. The blades of this impeller project outward from an axle like the arms of turnstile and, as the impeller spins, the water spins with it. As the water spins, the pressure near the outer edge of the pump housing becomes much higher than near the center of the impeller. There are many ways to understand this rise in pressure, and here are two:

First, you can view the water between the impeller blades as an object traveling in a circle. Objects don’t naturally travel in a circle—they need an inward force to cause them to accelerate inward as they spin. Without such an inward force, an object will travel in a straight line and won’t complete the circle. In a centrifugal pump, that inward force is provided by high-pressure water near the outer edge of the pump housing. The water at the edge of the pump pushes inward on the water between the impeller blades and makes it possible for that water to travel in a circle. The water pressure at the edge of the turning impeller rises until it’s able to keep water circling with the impeller blades.

You can also view the water as an incompressible fluid, one that obeys Bernoulli’s equation in the appropriate contexts. As water drifts outward between the impeller blades of the pump, it must move faster and faster because its circular path is getting larger and larger. The impeller blades do work on the water so it moves faster and faster. By the time the water has reached the outer edge of the impeller, it’s moving quite fast. But when the water leaves the impeller and arrives at the outer edge of the cylindrical pump housing, it slows down. Here is where Bernoulli’s equation figures in. As the water slows down and its kinetic energy decreases, that water’s pressure potential energy increases (to conserve energy). Thus the slowing is accompanied by a pressure rise. That’s why the water pressure at the outer edge of the pump housing is higher than the water pressure near the center of the impeller.

When water is actively flowing through the pump, arriving through a hole near the center of the impeller and leaving through a hole near the outer edge of the pump housing, the pressure rise between center and edge of the pump isn’t as large. However, this pressure rise never completely disappears and it’s what propels the water through the car’s cooling system.

If one accepts the existence of black holes, would it be plausible to assume tha…

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If one accepts the existence of black holes, would it be plausible to assume that a “white hole” exists on the opposite end due to captured light by the black hole?

I think not. Depending on your frame of reference, the passage of material into a simple black hole—one that isn’t spinning very fast and that doesn’t have a great deal of electric charge in it—has one of two results. If you are traveling with the material, things proceed more or less normally as you pass the point of no return—the so-called “event horizon” from which even light can’t escape. You accompany the material all the way to the center of the black hole—its “singularity”—and are crushed to infinite density. If instead of traveling with the material, you remain outside the black hole looking in toward it, you see the material approach the event horizon but without ever quite entering its surface. In fact, all of the material that went into forming the black hole in the first place, plus all the material that has fallen into the black hole since its formation, appear to reside forever on the event horizon surface. In effect, the material never quite gets to the black hole. Since the material never quite gets to the black hole, there is no need for it to reemerge elsewhere from a “white hole.”

However, there are more complicated black holes—ones involving angular momentum and electric charge—that have more complicated structures. In falling into one of these black holes, it is apparently possible to miss the singularity. There is some discussion of such material reemerging from the “other end” of one of this black holes but I believe that there are serious problems with such two-ended interpretations of the equations governing such black holes.