Identify the material having the lowest coefficient of linear expansion.
The coefficient of linear expansion (\(\alpha\)) is a material property that describes how much a material expands or contracts in length per unit change in temperature. When a material is heated, its atoms vibrate more vigorously and move further apart, causing the material to expand. Different materials expand by different amounts for the same temperature change.
The change in length (\(\Delta L\)) of a material when its temperature changes by \(\Delta T\) is given by the formula:
\(\Delta L = L_0 \alpha \Delta T\)
where \(L_0\) is the original length of the material and \(\alpha\) is the coefficient of linear expansion. A lower value of \(\alpha\) means the material expands less for the same temperature change.
To find the material with the lowest coefficient of linear expansion among the given options (Brass, Iron, Copper, Lead), we need to compare their typical \(\alpha\) values. These values are usually determined experimentally and can vary slightly depending on the specific composition and temperature range.
Here are the approximate typical values for the coefficient of linear expansion (\(\alpha\)) for the given materials:
| Material | Approximate Coefficient of Linear Expansion (\(\alpha\)) in \((10^{-6} \text{ /}^{\circ}\text{C})\) |
|---|---|
| Brass | 19 |
| Iron (Steel) | 12 |
| Copper | 17 |
| Lead | 29 |
Looking at the table above, we can compare the approximate values:
Comparing these numerical values, \(12\) is the smallest number among 19, 12, 17, and 29.
Therefore, Iron has the lowest coefficient of linear expansion among Brass, Iron, Copper, and Lead.
The material having the lowest coefficient of linear expansion among the options provided is Iron. This means that for a given change in temperature, a piece of Iron will expand or contract less in length compared to equally sized pieces of Brass, Copper, or Lead.
| Term | Description | Formula |
|---|---|---|
| Linear Expansion | Change in length due to temperature change. | \(\Delta L = L_0 \alpha \Delta T\) |
| Coefficient of Linear Expansion (\(\alpha\)) | Property indicating length change per unit length per degree temperature change. | \(\alpha = \frac{\Delta L}{L_0 \Delta T}\) |
| Area Expansion | Change in surface area due to temperature change. | \(\Delta A = A_0 \beta \Delta T\), where \(\beta \approx 2\alpha\) |
| Volume Expansion | Change in volume due to temperature change. | \(\Delta V = V_0 \gamma \Delta T\), where \(\gamma \approx 3\alpha\) |
Thermal expansion is a fundamental property of matter. Besides linear expansion (change in length), materials also undergo area expansion (change in surface area) and volume expansion (change in overall volume) when heated or cooled.
Greater the value of _______ of a material, the more rapidly it will conduct heat.
Identify the material having low coefficient of volume expansion
Identify the material having high coefficient of volume expansion.
Thermal expansion of solids are:
A copper rod and a steel rod are to have lengths LC and LS, such that the difference between their lengths is the same at all ambient temperatures. If the coefficients of linear expansion of copper and steels are αC and αS respectively. The lengths are related to the coefficient of linear expansion as :
How much should the temperature of a brass rod be increased so as to increase its length by 1%?
Given: for brass α = 0.00002/°CA wooden wagon wheel has an outside diameter of 3750 mm. The iron tire for this wheel is deliberately made smaller so that it can be shrunk in place to be a tight fit. If the tire's inside diameter is 3737 mm at 20°C, the temperature to which it must be heated to fit over the wheel? The coefficient of linear expansion of the steel is 1.2 × 10-5/°C.
A cylinder of cross-sectional radius 1 cm and height 4 cm is heated from 0°C to 100°C. If the coefficient of linear expansion α = 4 × 10-4/°C, what will be the increase in the volume of the cylinder?