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 lengthversustemperature 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
Problem 1
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?
Solution 1
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 .
Problem 2
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?
Solution 2
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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