Thermal Expansion in Two and Three Dimensions
Thermal Expansion in Two and Three Dimensions
Objects expand in all dimensions, as illustrated in the figure below. That is, their areas and volumes, as well as their lengths, increase with temperature. Holes also get larger with temperature. If you cut a hole in a metal plate, the remaining material will expand exactly as it would if the plug was still in place. The plug would get bigger, and so the hole must get bigger too. (Think of the ring of neighboring atoms or molecules on the wall of the hole as pushing each other farther apart as temperature increases. Obviously, the ring of neighbors must get slightly larger, so the hole gets slightly larger).
Thermal Expansion in Two Dimensions
For small temperature changes, the change in area \(\text{Δ}A\) is given by
\(\text{Δ}A=2\alpha A\text{Δ}T,\)
where \(\text{Δ}A\) is the change in area \(A\), \(\text{Δ}T\) is the change in temperature, and \(\alpha \) is the coefficient of linear expansion, which varies slightly with temperature.
In general, objects expand in all directions as temperature increases. In these drawings, the original boundaries of the objects are shown with solid lines, and the expanded boundaries with dashed lines. (a) Area increases because both length and width increase. The area of a circular plug also increases. (b) If the plug is removed, the hole it leaves becomes larger with increasing temperature, just as if the expanding plug were still in place. (c) Volume also increases, because all three dimensions increase.
Thermal Expansion in Three Dimensions
The change in volume \(\text{Δ}V\) is very nearly \(\text{Δ}V=3\alpha V\text{Δ}T\). This equation is usually written as
\(\text{Δ}V=\mathrm{\beta V}\text{Δ}T\text{,}\)
where \(\beta \) is the coefficient of volume expansion and \(\beta \approx 3\alpha \). Note that the values of \(\beta \) in this table are almost exactly equal to \(3\alpha \).
In general, objects will expand with increasing temperature. Water is the most important exception to this rule. Water expands with increasing temperature (its density decreases) when it is at temperatures greater than \(4\text{º}\text{C}(\text{40}\text{º}\text{F})\). However, it expands with decreasing temperature when it is between \(+4\text{º}\text{C}\) and \(0\text{º}\text{C}\) to \(\text{32}\text{º}\text{F})\). Water is densest at \(+4\text{º}\text{C}\). (See the figure below.) Perhaps the most striking effect of this phenomenon is the freezing of water in a pond. When water near the surface cools down to \(4\text{º}\text{C}\) it is denser than the remaining water and thus will sink to the bottom.
This “turnover” results in a layer of warmer water near the surface, which is then cooled. Eventually the pond has a uniform temperature of \(4\text{º}\text{C}\). If the temperature in the surface layer drops below \(4\text{º}\text{C}\), the water is less dense than the water below, and thus stays near the top. As a result, the pond surface can completely freeze over. The ice on top of liquid water provides an insulating layer from winter’s harsh exterior air temperatures. Fish and other aquatic life can survive in \(4\text{º}\text{C}\) water beneath ice, due to this unusual characteristic of water. It also produces circulation of water in the pond that is necessary for a healthy ecosystem of the body of water.
The density of water as a function of temperature. Note that the thermal expansion is actually very small. The maximum density at \(+4\text{º}\text{C}\) is only 0.0075% greater than the density at \(2\text{º}\text{C}\), and 0.012% greater than that at \(0\text{º}\text{C}\).
Making Connections: Real-World Connections—Filling the Tank
Differences in the thermal expansion of materials can lead to interesting effects at the gas station. One example is the dripping of gasoline from a freshly filled tank on a hot day. Gasoline starts out at the temperature of the ground under the gas station, which is cooler than the air temperature above. The gasoline cools the steel tank when it is filled. Both gasoline and steel tank expand as they warm to air temperature, but gasoline expands much more than steel, and so it may overflow.
This difference in expansion can also cause problems when interpreting the gasoline gauge. The actual amount (mass) of gasoline left in the tank when the gauge hits “empty” is a lot less in the summer than in the winter. The gasoline has the same volume as it does in the winter when the “add fuel” light goes on, but because the gasoline has expanded, there is less mass. If you are used to getting another 40 miles on “empty” in the winter, beware—you will probably run out much more quickly in the summer.
Because the gas expands more than the gas tank with increasing temperature, you can’t drive as many miles on “empty” in the summer as you can in the winter. (credit: Hector Alejandro, Flickr)
Example: Calculating Thermal Expansion: Gas vs. Gas Tank
Suppose your 60.0-L (15.9-gal) steel gasoline tank is full of gas, so both the tank and the gasoline have a temperature of \(\text{15}\text{.}0\text{º}\text{C}\). How much gasoline has spilled by the time they warm to \(\text{35}\text{.}0\text{º}\text{C}\)?
Strategy
The tank and gasoline increase in volume, but the gasoline increases more, so the amount spilled is the difference in their volume changes. (The gasoline tank can be treated as solid steel.) We can use the equation for volume expansion to calculate the change in volume of the gasoline and of the tank.
Solution
1. Use the equation for volume expansion to calculate the increase in volume of the steel tank:
\({\text{Δ}V}_{\text{s}}={\beta }_{\text{s}}{V}_{\text{s}}\text{Δ}T.\)
2. The increase in volume of the gasoline is given by this equation:
\({\text{Δ}V}_{\text{gas}}={\beta }_{\text{gas}}{V}_{\text{gas}}\text{Δ}T.\)
3. Find the difference in volume to determine the amount spilled as
\({V}_{\text{spill}}={\text{Δ}V}_{\text{gas}}-{\text{Δ}V}_{\text{s}}.\)
Alternatively, we can combine these three equations into a single equation. (Note that the original volumes are equal.)
\(\begin{array}{lll}{V}_{\text{spill}}& =& ({\beta }_{\text{gas}}-{\beta }_{\text{s}})V\text{Δ}T\\ & =& [(\text{950}-\text{35})×{\text{10}}^{-6}/\text{º}\text{C}](\text{60}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{L})(\text{20}\text{.}0\text{º}\text{C})\\ & =& 1\text{.}\text{10}\phantom{\rule{0.25em}{0ex}}\text{L}.\end{array}\)
Discussion
This amount is significant, particularly for a 60.0-L tank. The effect is so striking because the gasoline and steel expand quickly. The rate of change in thermal properties is discussed in Heat and Heat Transfer Methods.
If you try to cap the tank tightly to prevent overflow, you will find that it leaks anyway, either around the cap or by bursting the tank. Tightly constricting the expanding gas is equivalent to compressing it, and both liquids and solids resist being compressed with extremely large forces. To avoid rupturing rigid containers, these containers have air gaps, which allow them to expand and contract without stressing them.
This lesson is part of:
Temperature, Kinetic Theory, and Gas Laws