Some Effects of Temperature Help (page 2)
Temperature can affect the volume of or the pressure exerted by a sample of matter. You are familiar with the fact that most metals expand when they are heated; some expand more than others.
Temperature, Volume, And Pressure
A sample of gas confined to a rigid container will exert more and more pressure on the walls of the container as the temperature goes up. If the container is flexible, such as a balloon, the volume of the gas will increase. Similarly, if you take a container with a certain amount of gas in it and suddenly make the container bigger without adding any more gas, the drop in pressure will produce a decrease in temperature. If you have a rigid container with gas in it and then some of the gas is allowed to escape (or is pumped out), the drop in pressure will chill the container. This is why, for example, a compressed-air canister gets cold when you use it to blow dust out of your computer keyboard.
Liquids behave a little more strangely. The volume of the liquid water in a kettle and the pressure it exerts on the kettle walls don’t change when the temperature goes up and down unless the water freezes or boils. Some liquids, however, unlike water, expand when they heat up. Mercury is an example. This is how an old-fashioned thermometer works.
Solids, in general, expand when the temperature rises and contract when the temperature falls. In many cases you don’t notice this expansion and contraction. Does your desk look bigger when the room is 30° C than it does when the room is only 20°C? Of course not. But it is! You don’t see the difference because it is microscopic. However, the bimetallic strip in the thermostat, which controls the furnace or air conditioner, bends considerably when one of its metals expands or contracts just a tiny bit more than the other. If you hold such a strip near a hot flame, you actually can watch it curl up or straighten out.
Standard Temperature And Pressure (stp)
To set a reference for temperature and pressure against which measurements can be made and experiments conducted, scientists have defined standard temperature and pressure (STP). This is a more or less typical state of affairs at sea level on the Earth’s surface when the air is dry.
The standard temperature is 0°C (32°F), which is the freezing point or melting point of pure liquid water. Standard pressure is the air pressure that will support a column of mercury 0.760 m (just a little less than 30 in) high. This is the proverbial 14.7 pounds per inch squared (lb/in 2 ), which translates to approximately 1.01 × 10 5 newtons per meter squared (N/m 2 ).
Air is surprisingly massive. We don’t think of air as having significant mass, but this is because we’re immersed in it. When you dive only a couple of meters down in a swimming pool, you don’t feel a lot of pressure and the water does not feel massive, but if you calculate the huge amount of mass above you, it might scare you out of the water! The density of dry air at STP is approximately 1.29 kg/m 3 . A parcel of air measuring 4.00 m high by 4.00 m deep by 4.00 m wide, the size of a large bedroom, masses 82.6 kg. In Earth’s gravitational field, that translates to 182 pounds, the weight of a good-sized, full-grown man.
Thermal Expansion And Contraction
Suppose that we have a sample of solid material that expands when the temperature rises. This is the usual case, but some solids expand more per degree Celsius than others. The extent to which the height, width, or depth of a solid (its linear dimension ) changes per degree Celsius is known as the thermal coefficient of linear expansion .
For most materials, within a reasonable range of temperatures, the coefficient of linear expansion is constant. This means that if the temperature changes by 2°C, the linear dimension will change twice as much as it would if the temperature variation were only 1°C. However, there are limits to this, of course. If you heat a metal up to a high enough temperature, it will become soft and ultimately will melt or even burn or vaporize. If you cool the mercury in a thermometer down enough, it will freeze. Then the simple length-versus-temperature rule no longer applies.
In general, if s is the difference in linear dimension (in meters) produced by a temperature change of T (in degrees Celsius) for an object whose linear dimension (in meters) is d , then the thermal coefficient of linear expansion, symbolized by the lowercase Greek letter alpha (α), is given by this equation:
α = s /( dT )
When the linear dimension increases, consider s to be positive; when it decreases, consider s to be negative. Rising temperatures produce positive values of T; falling temperatures produce negative values of T .
The coefficient of linear expansion is defined in meters per meter per degree Celsius. The meters cancel out in this expression of units, so the technical quantity is per degree Celsius, symbolized/°C.
Effects of Temperature Practice Problems
Imagine a metal rod 10.000 m long at 20.00°C. Suppose that this rod expands to a length of 10.025 m at 25.00°C. What is the thermal coefficient of linear expansion?
This rod increases in length by 0.025 m for a temperature increase of 5.00°C. Therefore, s = 0.025, d = 10, and T = 5.00. Plugging these numbers into the preceding formula, we get
α = 0.025/(10 × 5.00)
= 0.00050/°C = 5.0 × 10 −4 /°C
We are justified in going to only two significant figures here because that is as accurate as our data are for the value of s .
Suppose that α = 2.50 × 10 −4 /°C for a certain substance. Imagine a cube of this substance whose volume V 1 is 8.000 m 3 at a temperature of 30.0°C. What will be the volume V 2 of the cube if the temperature falls to 20.0°C?
It is important to note the word linear in the definition of α. This means that the length of each edge of the cube of this substance will change according to the thermal coefficient of linear expansion.
We can rearrange the preceding general formula for α so that it solves for the change in linear dimension s as follows:
s = α dT
where T is the temperature change (in degrees Celsius) and d is the initial linear dimension (in meters). Because our object is a cube, the initial length d of each edge is 2.000 m (the cube root of 8.000, or 8.000 1/3 ). Because the temperature falls, T = −10.0. Therefore,
s = 2.50 × 10 −4 × (−10.0) × 2.000
= −2.50 × 10 −3 × 2.000
= −5.00 × 10 −3 m = −0.00500 m
This means that the length of each side of the cube at 20°C will be 2.000 − 0.00500 = 1.995 m. The volume of the cube at 20.0°C is therefore 1.995 3 = 7.940149875 m 3 . Because our input data are given to only three significant figures, we must round this off to 7.94 m 3 .
Practice problems of these concepts can be found at: Temperature, Pressure, And Changes Of State Practice Problems
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