Why does a plastic food container shrink after microwaving?

Lou Bloomfield Leave a comment

After I microwave food in a plastic container and remove the container from the oven, the container begins to shrink. It seems like there is a vacuum inside. The lid and all the sides of the container are squeezed inward. Why does this happen? — U.S.

You’re right about the vacuum; the shrinkage you’re seeing is caused by a decrease in pressure inside the plastic container. When you heated the food, you also heated the air inside the container. Additionally, you converted some of the liquid water in the food into water vapor. Both of these effects, heating the air and adding water molecules to it, increased the gas pressure inside the container. Though you might not have noticed, the pressurized gas then leaked out of the plastic container through tiny gaps between its body and lid.

When you then took the container out of the microwave oven and exposed it to cooler room air, the temperature of the container and its contents decreases. The air cooled and some of the moisture in that air condensed back into liquid water. Both of those changes caused the gas pressure inside the container to decrease. While you might expect air to then leak back into the container, modern plastic containers and their lids tend to act as one-way valves — they leak air out, but not in. That’s because when the pressure inside the container is lower than atmospheric, the surrounding higher air pressure pushes the lid onto the container and improves the seal between those two items. So the container & lid leak air and water vapor during heating, but they then seal during cooling.

With low pressure inside the container and atmospheric pressure outside, there are substantial inward forces on the surfaces of the container and the container shrinks inward. The squeezing effect you observe is real — you have created a partial vacuum inside the container and it’s being squished by the surrounding air.

That same effect occurs whenever you cool the gas inside a sealed plastic container — for example, a plastic bottle that you filled with hot water and then closed tightly. The dropping temperature lowers the pressure of the gas and may even cause some of that gas to condense into liquid. With less pressure inside the container than outside the container, the container experiences inward forces that dent it inward. Ultimately, the container dents inward until the net forces on its surfaces become zero. It reaches that new equilibrium situation (new balance between forces) because compressing the gas inside it increases the pressure of that trapped gas and because the bottle or container itself has some elasticity that fights the denting. Either way, the container ends up looking squished.

Why do things float better in salt water than in fresh water?

Lou Bloomfield Leave a comment

A floating object is displacing fluids that would otherwise fill the space it occupies. For example, a ball floating motionless on water is displacing the water and air that would normally be where the ball is. If we remove the ball, water and air will fill its space and soon everything will be motionless again.

Just because that ball-shaped portion of water and air is motionless doesn’t mean that it’s weightless. It does have a weight! But its weight is supported by the water and air that surround it. Because of the earth’s gravity, the pressure of stationary water or air decreases steadily with altitude, so pressure exerted on the bottom of this ball-shaped portion is greater than the pressure exerted on its top. This unbalanced pressure produce a net upward force on the ball-shaped portion of water and air.

That upward force is known as the buoyant force and it’s evidently just strong enough to support the weight of the ball-shaped portion of water and air. If it weren’t the ball-shaped portion would accelerate up or down.

When we put the real ball back where it was and let it again float motionless on the water, the surrounding water and air continue to exert the same buoyant force on the real ball that they exerted on the ball-shaped portion of water and air. So the ball experiences an upward buoyant force that’s equal in amount to the weight of the water and air it displaces. That observation is known as Archimedes’ principle.

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Which brings me to your question. Here are two identical balls floating motionless on fresh water (left) and on salt water (right). In each case, the ball is experiencing a buoyant force that exactly cancels its weight. To obtain that exact buoyant force, the ball must displace a portion of water and air that weighs exactly as much as the ball weighs.

Salt water is denser than fresh water, meaning that salt water has more mass per volume (more kilograms per liter) than fresh water. A liter of salt water consequently weighs more than a liter of fresh water. Displacing a liter of salt water therefore produces a stronger upward buoyant force than displacing a liter of fresh water. That’s why the ball is floating higher on the container of salt water than it does on the container of fresh water.

The ball doesn’t need to displace as much salt water to obtain a buoyant force that supports its weight, so it rises higher on the salt water than it does on the fresh water. In each case, the ball finds just the right mix of water and air so that it displaces exactly its own weight in those two fluids.








Why do greenhouse gases warm the earth?

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The earth receives heat from the sun at an incredible rate and, if it didn’t get rid of that heat, it would get hotter and hotter. To maintain a steady average temperature, the earth must radiate away heat just as fast as it receives that heat from the sun. In other words, the thermal radiation that the sun emits into empty space must be equal to the thermal radiation the earth receives from the sun.

Though they’re equal in amount, these two thermal radiations have quite different spectrums. Because the sun is extremely hot, its thermal radiation spectrum is largely visible light. The earth, on the other hand, is relatively cool and its thermal radiation spectrum is almost entirely invisible infrared light. While the sun’s thermal radiation is brilliantly visible, we can’t see the earth’s thermal glow or where it’s coming from, which is important because some of it comes from our atmosphere.

If the earth had no atmosphere, its average surface temperature would be approximately -18 °C. At that temperature, the earth’s surface would emit just enough thermal radiation to balance the thermal radiation it receives from the sun. If the earth were hotter, it would radiate away more heat than it receives and cool down. If the earth were cooler, it would radiate away less heat than it receives and warm up.

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But the earth does have an atmosphere and that atmosphere contributes to these exchanges of thermal radiation. Although it’s relatively transparent to visible light, the atmosphere is able to emit and absorb significant portions of the infrared spectrum. As a result, a substantial fraction of the earth’s thermal radiation originates in its atmosphere.

Because of the atmosphere’s contribution to the earth’s thermal radiation, the average altitude at which the earth’s thermal radiation originates is not ground level. Instead, it’s 5 kilometers above sea level, the altitude at which the air temperature is about -18 °C. So the earth’s atmosphere shifts the -18 °C average radiating surface from sea level to an altitude of 5 kilometers above sea level.

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If you’ve ever traveled up into the mountains and felt the air during your trip, you’ve probably noticed that air’s temperature decreases with altitude. The earth’s atmosphere has a natural temperature gradient of about -6.6 °C per kilometer upward. A simple explanation for that temperature gradient is that at higher altitudes, air must commit more of its overall energy to gravitational potential energy (energy stored in the force of gravity), leaving it with less for thermal energy (energy associated with temperature).

Since the air temperature increases by 6.6 °C each time you descend 1 kilometer, the air temperature at an altitude of 4 kilometers is -11.4 °C (-18 °C + 6.6 °C), at an altitude of 3 kilometers is -4.8 °C (-11.4 + 6.6 °C), …, and at sea level is 15 °C. Sure enough, the average historical air temperature at sea level is about 15 °C.

Which brings us to the question itself: “Why do greenhouse gases warm the earth?”

The 5 kilometer average altitude of origin for the earth’s thermal radiation actually depends on the atmosphere’s chemical composition. Some molecules in the air, notably nitrogen and oxygen, are remarkably transparent in the infrared and barely contribute to the earth’s thermal radiation. Other molecules, notably water, carbon dioxide, methane, and nitrogen oxides, interact strongly with infrared light and contribute significantly to the earth’s thermal radiation. Those thermally radiating gases are collectively known as “greenhouse gases.” The higher the concentration of those greenhouse gases in the atmosphere, the more the atmosphere contributes to the earth’s thermal radiation and the higher the average altitude of origin for the earth’s thermal radiation.

As human-produced greenhouse gases accumulate in the earth’s atmosphere, the average altitude of origin for the earth’s thermal radiation increases. If that average altitude were to rise from 5 kilometers to 6 kilometers, the average temperature at sea level would increase by another 6.6 °C to 21.6 °C. A rise of that magnitude would be catastrophic or even apocalyptic.

Alas, the average altitude of origin for the earth’s thermal radiation has already risen significantly since the industrial revolution and with it, a rise in the average temperature at sea level. The rate of temperature rise is alarming and the task of halting it or at least slowing it substantially cannot be put off for another generation. Even with serious international effort, it’s likely that many areas of the world will become uninhabitable by the next century, either because they are too hot for human survival or because they are under water as the result of rising sea levels.








Why is easier to keep a bicycle upright when it is moving faster?

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Why is it easier for a bicycle to stay upright when it is moving faster? And hard to stand up straight when it is moving slowly? What forces are acting on it that depend on its speed?

A bicycle is an example of an object that is unstable at rest but stable in motion. That it is unstable at rest means that it has an unstable equilibrium: when it is exactly upright, it is at equilibrium (zero net force and therefore inertial). As soon as it tips even a tiny bit, however, forces and torques arise that push it away from that equilibrium. Since it spontaneously accelerates away from equilibrium, given a chance, that equilibrium is termed “unstable.” In contrast, a tricycle has a stable equilibrium — when tipped, forces and torques arise that push it back toward equilibrium and it spontaneously accelerates toward equilibrium.

Even though an equilibrium is unstable, it is possible to keep the object at equilibrium is some other mechanism acts to return it to that equilibrium. In the case of a broom balanced on your hand, the mechanism is you — you can keep the broom in its vertical unstable equilibrium if you move your hand around properly. The bicycle does that return-to-equilibrium trick for you automatically whenever the bicycle is moving forward . The bicycle steers itself automatically back to the the unstable equilibiurm. Taster it is moving forward, the more effective its automatic steering mechanism becomes.








How can an object spin at constant angular velocity when its parts are accelerating?

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How can a spinning object keep constant velocity with the direction of its parts changing at every instant?

When you consider an object as rotating, you normally stop thinking of its parts as moving in their own independent ways and treat the whole assembly as a single object. While it’s true that the various parts of that object are accelerating in response to internal forces those parts exert on another, the object as a whole is doing a simpler motion: it’s rotating about some axis. This ability to focus on a simple motion in the midst of countless complicated motions is an example of the beautiful simplifications that physics allows in some cases.








When you push on a rotating object, when are you doing work?

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When you push on a rotating object, when are you doing work?

That’s an interesting question and requires two answers. First, if you push on a part of the rotating object and that part moves a distance in the direction of the force you exert, then you do work on it. In principle, it is possible to identify all the work that you do on the rotating object via this approach.

However, it is also possible to determine the work you do entirely in terms of physical quantities of rotation. If you exert a torque on the rotating object and it rotates the an angle in the direction of your torque, you again do work on the object. That’s the rotational version of the work formula: whereas force time distance is the translational work formula, torque times angle is the rotational work formula.

An important complication arises, however, in that you must measure the angle in the appropriate units: radians. The radian is the natural unit of angle and is effectively dimensionless (no units after it). When you multiple the torque times the angle in radians, the resulting units are those of work and energy. If you use a non-natural unit of angle, such as the degree, then you’ll have to deal with presence of the angle unit in your result.








Rotating “up” or “down” — is that like clockwise and counter-clockwise?

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When an object is rotating, both the “up” and “down” directions point along the vertical axis. Do they correspond to clockwise and counterclockwise?

Yes. Distinguishing between the two opposite directions of rotation using words alone requires that everyone agree on what to call those two directions. It also requires that everyone have an artifact that they can use to identify which direction is which. When something is spinning about a vertical axis, a carousel or merry-go-round, for example, then physicists name the two possible directions of rotation “up” and “down” and use a right-hand rule to identify which is which. Since most people have a right hand and know which hand it is, the necessary artifact is built-in.

In more common language, the two directions might be called “clockwise as viewed from above” and counter-clockwise as viewed from above”. In this case, the artifact is an old-fashioned analog clock and is probably more of a remembered artifact than one that is in the room with you. Nonetheless, that common naming convention is fine; it’s just wordier than the physicist’s version.








When you push someone on a swing, why don’t the forces cancel?

Lou Bloomfield Leave a comment

If you push someone on a swing with 50 N of force and they push back with 50 N of force, then why does the person still move? Shouldn’t they stay motionless if all the forces are cancelled out?

You’re struggling with the most common misconception about Newton’s third law of motion — the law stating that for every force object A exerts on object B, there is an equal but oppositely directed force exerted by object B on object A. The pair of forces described in Newton’s third law always act on different objects and therefore never cancel one another. Object A’s force on object B acts only on object B and can cause object B to accelerate. Object B’s force on object A acts only on object A and can cause object A to accelerate.

When you push someone on the swing, your force on the person affects that person and will affect their swinging motion. The person does indeed push back equally hard on you, but that force affects you! If you are wearing roller skates, you will accelerate backward and drift away from the swing.

 








Is deceleration something different from acceleration?

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Is deceleration something new or just acceleration in the opposite direction?

Deceleration is simply a special case of acceleration. An object decelerates by accelerating in the direction opposite its current velocity. For example, if your car is heading toward Washington at 60 mph (100 km/h) and you push on the brake pedal, your car will begin to accelerate in the direction pointing away from Washington and your forward velocity (toward Washington) will decrease with time. Since an object that accelerates in the direction opposite its velocity always slows down, it has become conventional to say that it is decelerating.








How can a basketball weigh 7.5 to 8.5 pounds when blown up but much less when deflated?

Lou Bloomfield Leave a comment

How can a basketball weigh 7.5 to 8.5 pounds when blown up but much less when deflated? What is it filled with when deflated?

I will answer your question in two parts. First, the actual weight of a basketball is dominated by its skin and, which weighs about 22 ounces (about 1.4 pounds). The air inside a properly inflated basketball weighs only about 0.03 pounds. Of that 0.03 pounds of air, only about 0.01 are measurable on a scale because buoyant effects due to the surrounding air support the other 0.02 pounds of air. That’s because the first 0.02 pounds of air put into the basketball simply fill it so that it’s spherical– air has gone from outside the basketball to inside the basketball and the scale won’t notice this change in location. Once you pump extra air into the ball, packing the air more tightly than normal and stiffening the ball’s surface, that additional air will appear on the scale’s weight measurement. A properly inflated basketball has about 0.01 pounds of extra air in it, so it’ll weigh an extra 0.01 pounds on a scale.

So what is the 7.5 to 8.5 pounds that a basketball is supposed to contain? Or the 13 pounds that a football is supposed to contain? Those aren’t weights at all. In fact, they are careless abbreviations for a different physical quantity: pressure. They should actually be written “7.5 to 8.5 pounds-per-square-inch” and “13-pounds-per-square inch” respectively.

Fluids such as air have pressures — the forces they exert on each unit of surface area they contact. For example, air that is listed as having a pressure of 7.5 pounds per square inch exerts a force of 7.5 pounds on each square inch of surface it touches. That means that the air in a properly inflated basketball pushes outward with a force of 7.5 to 8.5 pounds on each square inch of the inner surface of that ball. That outward push stretches the ball tight and gives it its feel and bounciness. Similarly, a properly inflated football has a pressure of 13 pounds per square inch and thus the air inside it exerts an outward  force of 13 pounds on each square inch of surface inside the ball. Again, this outward push stretches the ball taut and gives it its bounciness and feel.

An underinflated basketball or football weighs just slightly less than a properly inflated ball because its skin hasn’t changed and the weight of the air it contains is so insignificant. But the decrease in outward forces on the skin of the ball significantly changes its feel and bounciness.