91Ó°ÊÓ

Linear expansion refers to the change in dimensions of a material when its temperature changes. As the temperature increases, most solids will expand in size. This occurs because the particles within the material move more vigorously at higher temperatures:

• Each particle requires more space.
• The overall dimensions of the object increase.

\[ \Delta d = \alpha \times d_i \times \Delta T \]
This is the formula to calculate the change in dimensions, where \( \Delta d \) is the change in length or diameter, \( \alpha \) is the coefficient of linear expansion, \( d_i \) is the initial size, and \( \Delta T \) is the temperature change.

In the exercise, both stainless steel and copper experience linear expansion due to heating, resulting in a change in diameter. The degree of expansion is dictated by each material's unique coefficient of linear expansion.

The coefficient of linear expansion \( \alpha \) is a crucial factor measuring how much a material expands per degree of temperature change.

Each material has a unique coefficient, influencing how it responds when heated:

• A high \( \alpha \) implies greater expansion per degree Celsius.
• Conversely, a low \( \alpha \) means less expansion.

In our case:

• Stainless Steel: \( \alpha = 17.3 \times 10^{-6} \text{ K}^{-1} \)
• Copper: \( \alpha = 17.0 \times 10^{-6} \text{ K}^{-1} \)

These coefficients dictate how much each material's diameter will change with the same temperature increase.
The slightly higher coefficient for stainless steel results in a greater change in its diameter compared to copper.

The temperature change \( \Delta T \) is the difference between the final temperature and the initial temperature. It directly affects how much a material will expand. In our problem, the temperature change is significant:

\[ \Delta T = T_f - T_i = 610^{\circ} \text{C} - 22^{\circ} \text{C} = 588^{\circ} \text{C} \]
For both materials, the same \( \Delta T \) was applied, yet due to their different coefficients of linear expansion, the extent of change in their diameters varies.

Understanding \( \Delta T \) helps predict the amount of expansion and is critical in designing materials and structures that will be exposed to temperature variations.
By calculating the difference in the material's initial and final temperature, we can better plan for and accommodate thermal stresses in engineering and manufacturing applications.