Problem 26 - University Physics with Modern Physics

Question

(a) If an area measured on the surface of a solid body is \(A_{0}\) at some initial temperature and then changes by \(\Delta A\) when the temperature changes by \(\Delta T,\) show that $$\Delta A=(2 \alpha) A_{0} \Delta T$$ where \(\alpha\) is the coefficient of linear expansion. (b) A circular sheet of aluminum is 55.0 \(\mathrm{cm}\) in diameter at \(15.0^{\circ} \mathrm{C} .\) By how much does the area of one side of the sheet change when the temperature increases to \(27.5^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

(a) Derived formula is \( \Delta A = (2 \alpha) A_0 \Delta T \). (b) Area change is approximately 5.54 \( \mathrm{cm}^2 \).

Step 1: Understand the Problem

We're given a surface area of a solid body with an initial area \( A_0 \), which changes by \( \Delta A \) due to a temperature change \( \Delta T \). We need to prove the formula \( \Delta A = (2 \alpha) A_0 \Delta T \), where \( \alpha \) is the coefficient of linear expansion. Further, we must calculate the change in area of an aluminum sheet when its temperature changes.

Step 2: Linear Expansion Formula Derivation

The linear expansion of a solid is given by \( \Delta L = \alpha L_0 \Delta T \). For a two-dimensional surface, like an area, we multiply this effect by 2 (since area is \(L \times L\)), leading to the expression \( \Delta A = 2 \alpha A_0 \Delta T \). This shows that the change in area is proportionate to twice the linear expansion coefficient.

Step 3: Calculate Initial Area

For part (b), we have a circular aluminum sheet with a diameter of 55.0 cm. The initial area \( A_0 \) is calculated using the formula for the area of a circle: \( A_0 = \pi r^2 \). With a radius \( r = \frac{55.0}{2} \) cm, compute \( A_0 \).

Step 4: Temperature Change Calculation

The temperature increase is calculated as \( \Delta T = 27.5^{\circ}C - 15.0^{\circ}C = 12.5^{\circ}C \). This \( \Delta T \) will be used to find the area change.

Step 5: Apply Area Change Formula

Using the given coefficient of linear expansion for aluminum, \( \alpha \approx 23 \times 10^{-6} \ \text{per degrees Celsius} \), substitute \( A_0 \), \( \Delta T \), and \( \alpha \) into \( \Delta A = (2 \alpha) A_0 \Delta T \) to find the change in area.

Step 6: Compute Final Answer

Compute \( A_0 \) using the radius, then find \( \Delta A = (2 \times 23 \times 10^{-6}) \times A_0 \times 12.5 \). Substitute the calculated \( A_0 \) and solve for \( \Delta A \).

Key Concepts

Linear Expansion Coefficient

The linear expansion coefficient, commonly noted as \( \alpha \), is crucial in understanding how materials expand or contract with temperature changes. It measures the fractional change in length per degree change in temperature.

• For linear expansion, the change in length equation is \( \Delta L = \alpha L_0 \Delta T \).
• \( L_0 \) is the initial length, and \( \Delta T \) is the change in temperature.

In the context of linear objects, only the length change is considered. For area or volume, the effect is different. For instance, in a surface area, both dimensions expand, making area change doubly sensitive to the same linear coefficient. The area change formula becomes \( \Delta A = 2 \alpha A_0 \Delta T \). This highlights how two-dimensional expansion involves the linear expansion concept, factoring it twice because the area has two dimensions.

Temperature Change

Understanding temperature change \( \Delta T \) is essential for calculating thermal expansion. Thermal expansion occurs because molecules move more vigorously and distance between them increases as the temperature rises. Calculating \( \Delta T \) is simple:

• Subtract the initial temperature from the final temperature.

For example, if an aluminum sheet starts at \(15.0^{\circ}C\) and the temperature increases to \(27.5^{\circ}C\), the change is \( \Delta T = 27.5 - 15.0 = 12.5^{\circ}C\). This value is crucial for determining how much the area of an object, like the aluminum sheet, will increase due to temperature rise.

Aluminum Properties

Aluminum is a versatile and widely used metal known for its excellent thermal conductivity and lightweight nature. One of the key properties of aluminum in the context of thermal expansion is its linear expansion coefficient, approximately \(23 \times 10^{-6}\) per degree Celsius.

• It expands uniformly when its temperature increases, making it predictable and manageable in design and engineering.
• Its lightweight nature also means that large sheets can be accommodated without significant structural support.

Being aware of aluminum's expansion properties ensures structures or components remain within safety margins, especially under varying temperatures. Its reliability in expansion attributes makes it a choice material for many thermal application scenarios.

Area Calculation

Calculating the area change due to temperature variations relies on understanding both geometry and thermal expansion. For shapes like a circle, the initial area \( A_0 \) often needs calculating first:

• For a circle with diameter \(d\), the formula is \( A_0 = \pi \left( \frac{d}{2} \right)^2 \).

Let's consider a circular aluminum sheet with a diameter of \(55.0\) cm.
The radius \( r \) is \( \frac{55.0}{2} \) cm. Plug this into the circle area formula to find \( A_0 \).Once the initial area is known, apply the thermal expansion formula:

• Use \( \Delta A = 2 \alpha A_0 \Delta T \) to calculate the new area.

Simply substitute the appropriate values for \( \alpha, A_0, \) and \( \Delta T \) to yield the area change due to thermal effects. This approach effectively combines geometry with material physics.