The two ends of a cylindrical rod of 1025 steel \(75.00 \mathrm{~mm}\) long and \(10.000 \mathrm{~mm}\) in diameter are maintained rigid. If the rod is initially at \(25^{\circ} \mathrm{C}\), to what temperature must it be cooled to have a \(0.008-\mathrm{mm}\) reduction in diameter?

Understanding the linear thermal expansion formula is crucial when calculating how much a material will expand or contract with temperature changes. In scientific terms, thermal expansion refers to the tendency of matter to change in shape, area, and volume in response to a change in temperature.

One of the fundamental formulas used in thermal expansion is: \( \Delta L = L_0 \alpha \Delta T \), where:\

• \(\Delta L\) is the change in length,
• \(L_0\) is the original length,
• \(\alpha\) is the coefficient of linear expansion of the material,
• \(\Delta T\) is the change in temperature.

Materials expand or contract at different rates, which is determined by their linear expansion coefficient, \(\alpha\). This coefficient is a measure of how much a material expands per degree of temperature increase. To solve problems involving thermal expansion, such as the diameter change of a rod, we adapt this formula for diametrical changes, utilizing the same principles as with linear changes.

The behavior of materials under temperature variations is not uniform across all substances. Thermal properties of materials, such as thermal conductivity, specific heat, and thermal expansion coefficient, play a pivotal role in determining how a material interacts with heat.

In the context of thermal expansion, the coefficient of linear expansion, \(\alpha\), is of particular interest. It is a material-specific value that represents the fractional change in length per degree of temperature change. For instance, 1025 steel has its own coefficient of linear expansion. These properties are critical for engineers and designers to consider when constructing objects that will experience temperature changes to avoid structural issues such as thermal stress or deformation.

Calculating the temperature change for a material involves understanding the relationship between the material's size change and the corresponding temperature differential that caused it. To find the new temperature of a material after heating or cooling, one must know the initial temperature and calculate the change in temperature using the formula: \( \Delta T = \frac{\Delta D}{D_0 \alpha} \)

In the exercise presented, a cylindrical rod of steel is subjected to temperature changes that affect its diameter. By rearranging the linear thermal expansion formula and solving for \(\Delta T\), we can deduce the necessary temperature shift to achieve a specific dimensional change. The result gives us a clear answer regarding how cool the rod must be to shrink its diameter by 0.008 mm, illustrating the practical application of temperature change calculations in real-world scenarios.