Question

A rod is measured to be \(20.05 \mathrm{~cm}\) long using a steel ruler at a room temperature of \(20^{\circ} \mathrm{C}\). Both the rod and the ruler are placed in an oven at \(270^{\circ} \mathrm{C}\), where the rod now measures \(20.11 \mathrm{~cm}\) using the same rule. Calculate the coefficient of thermal expansion for the material of which the rod is made.

Short answer

The coefficient of linear thermal expansion for the material of rod at the given conditions is \(1.2 \times 10^{-5}~K^{-1}\).

Step-by-step solution

Calculate the change in length and temperature

The change in length \(\Delta L\) of the rod is the final length subtracted by the initial length: \(\Delta L = L_{final} - L_{initial} = 20.11~cm - 20.05~cm = 0.06~cm\). The change in temperature \(\Delta T\) is also the final temperature subtracted by the initial temperature: \(\Delta T = T_{final} - T_{initial} = 270^{\circ} C - 20^{\circ} C = 250^{\circ} C\).

Conversion of units

Although not strictly necessary, to maintain unit consistency, it is better to convert temperatures from Celsius to Kelvin. In this case, because we're dealing with a temperature difference, the numeric value remains the same, since degrees Celsius and Kelvin have the same magnitude. It is just \(\Delta T = 250 K\). Furthermore, convert the length from cm to m, so \(\Delta L = 0.0006~m\) and \(L_{initial} = 0.2005~m\).

Calculate the coefficient of linear thermal expansion

The coefficient of linear thermal expansion \(\alpha\) can be calculated from the equation \(\alpha = \frac{\Delta L}{L_{initial} \cdot \Delta T} = \frac{0.0006~m}{0.2005~m \cdot 250~K} \)

Key concepts

Coefficient of Linear Thermal Expansion

When a material is subjected to a change in temperature, it tends to expand or contract depending on the temperature change. This change in dimension is characterized by the coefficient of linear thermal expansion, denoted by \(\alpha\). This coefficient measures the fractional change in length per degree change in temperature. It helps in predicting how much a material will expand or contract with temperature variations. The equation to find \(\alpha\) is given by:

• \(\alpha = \frac{\Delta L}{L_{initial} \cdot \Delta T}\).

Here, \(\Delta L\) is the change in length, \(L_{initial}\) is the initial length, and \(\Delta T\) is the change in temperature. It's an essential parameter in engineering and construction, ensuring that buildings and machines function safely across temperature changes.

Temperature Change

Temperature change, denoted as \(\Delta T\), is a crucial variable when determining thermal expansion. It is calculated by subtracting the initial temperature from the final temperature. In this exercise, the initial temperature is \(20^{\circ} \mathrm{C}\). The final temperature is \(270^{\circ} \mathrm{C}\). Thus, the temperature change is:\

• \(\Delta T = 270^{\circ} C - 20^{\circ} C = 250^{\circ} C\)

Remember, when calculating temperature change for thermal expansion, both Celsius and Kelvin scales can be used interchangeably, as the interval is the same. Using Kelvin might be preferred in scientific contexts for consistency, though \(\Delta T\) remains the same numerically: \(250 K\).

Length Change

Length change, denoted as \(\Delta L\), is the difference between the final and initial length of a material after it experiences a change in temperature. This measure helps determine how much a material expands or contracts due to temperature changes. In the exercise, the rod's length changed from \(20.05\, \mathrm{cm}\) to \(20.11\, \mathrm{cm}\). Calculating this gives:

• \(\Delta L = 20.11 \, \mathrm{cm} - 20.05 \, \mathrm{cm} = 0.06 \, \mathrm{cm}\)

For accurate calculations, especially in larger structures or precise engineering applications, converting measurements to consistent units, such as meters, is often necessary. Thus \(\Delta L = 0.0006 \, \mathrm{m}\).

Units Conversion

Proper unit conversion is vital for consistency in calculations, especially in scientific equations. In this exercise:

• Length is initially measured in centimeters, but converting to meters ensures calculations align with standard units where 1 meter = 100 centimeters.
• \(\Delta L = 0.0006 \, \mathrm{m}\) and \(\text{initial length} = 0.2005 \mathrm{m}\).
• Temperature changes in this context permit both Celsius and Kelvin, as they share the same interval scale. Although converting to Kelvin is generally preferred in scientific work, here \(\Delta T = 250 K\) remains the same as \(250^{\circ} C\).

Attention to unit conversion avoids potential errors and ensures that results are accurate and meaningful within the context of the problem.