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The area \(A\) of a rectangular plate is \(a b=1.4 \mathrm{~m}^{2}\). Its coefficient of linear expansion is \(\alpha=32 \times 10^{-6} / \mathrm{C}^{\circ} .\) After a temperature rise \(\Delta T=89^{\circ} \mathrm{C},\) side \(a\) is longer by \(\Delta a\) and side \(b\) is longer by \(\Delta b\) (Fig. 18-61). Neglecting the small quantity \((\Delta a \Delta b) / a b\), find \(\Delta A\).

Short Answer

Expert verified
\( \Delta A \approx 0.00797 \text{ m}^2 \).

Step by step solution

01

Understand the problem

We are given a rectangular plate with an original area of 1.4 m² and a coefficient of linear expansion, \( \alpha = 32 \times 10^{-6} / \text{C}^\circ \). After a temperature increase, we are to find the increase in area, \( \Delta A \), neglecting the small term \((\Delta a \Delta b) / ab\).
02

Apply the formula for linear expansion

When the temperature rises, a side \( x \) will change in length by \( \Delta x = \alpha x \Delta T \). The same formula applies to both sides \( a \) and \( b \), meaning:\[ \Delta a = \alpha a \Delta T \] \[ \Delta b = \alpha b \Delta T \].
03

Express the change in area

The change in area \( \Delta A \) is given by:\[ \Delta A = (a + \Delta a)(b + \Delta b) - ab \].
04

Simplify using approximations

Expanding the expression from Step 3 we get:\[(a + \Delta a)(b + \Delta b) = ab + a\Delta b + b\Delta a + \Delta a \Delta b\]. Neglect the term \( \Delta a \Delta b \), so:\[ \Delta A = a\Delta b + b\Delta a \].
05

Substitute the expressions for \( \Delta a \) and \( \Delta b \)

Substitute \( \Delta a = \alpha a \Delta T \) and \( \Delta b = \alpha b \Delta T \) into the equation:\[ \Delta A = a(\alpha b \Delta T) + b(\alpha a \Delta T) \]\[ \Delta A = \alpha \Delta T (ab + ba) \].
06

Substitute known values

Substitute the given values \( ab = 1.4 \, \text{m}^2 \), \( \alpha = 32 \times 10^{-6} / \text{C}^\circ \), and \( \Delta T = 89 ^\circ \text{C} \):\[ \Delta A = 32 \times 10^{-6} \times 89 \times (1.4 \times 2) \].
07

Calculate \( \Delta A \)

Simplify and calculate:\[ \Delta A = 32 \times 10^{-6} \times 89 \times 2.8 \]\[ \Delta A \approx 7.9744 \times 10^{-3} \text{ m}^2 \]\[ \Delta A \approx 0.00797 \text{ m}^2 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Expansion Coefficient
The linear expansion coefficient is an important concept when studying how materials change when heated. It's denoted by \( \alpha \) and represents how much a specific material expands per degree change in temperature. This coefficient is constant for a given material and describes the fractional change in length per unit temperature change. For example, a higher \( \alpha \) indicates more expansion under the same temperature conditions.
In the equation \( \Delta x = \alpha x \Delta T \), \( \alpha \) is used to calculate how much a side of an object, like a rectangular plate, will expand with a rise in temperature, \( \Delta T \). Remember: the units of \( \alpha \) are usually per degree Celsius \( \text{C}^\circ \). This makes it easy to multiply with the change in temperature, ensuring our final expansion calculation is consistent.
Rectangular Plate
To better understand how materials expand, let's focus on a rectangular plate. A rectangle's area is given by the product of its length and width \( (a \times b) \), making it a straightforward shape for calculating changes when subject to thermal expansion.
Because both dimensions of the rectangle (length and width) can expand due to heat, the resulting change in the area is more complex than linear expansion alone. With the original area of the rectangle known, the change in its area, due to expansion, can be calculated by understanding how each side extends. In practical problems, values like the length of each side or the entire area are given to help have an accurate assessment before factoring in thermal effects.
Temperature Change
The factor that causes materials to expand or contract is temperature change, represented often as \( \Delta T \). As the temperature increases, the atoms in a material vibrate more energetically, causing the material to expand. Conversely, cooling can cause contraction as the vibrations decrease.
In the problem of the rectangular plate, the given temperature change \( \Delta T = 89^\circ \text{C} \) triggers the length increase in both dimensions of the plate. Understanding this concept is key in predicting how the physical properties of materials change in different environments. It’s a prime consideration whether dealing with everyday objects or engineering large constructions.
Area Change Calculation
Calculating the change in the area, \( \Delta A \), of a rectangular plate due to thermal effects involves a few key steps. Firstly, consider how each dimension of the plate grows. Each length expands according to \( \Delta a = \alpha a \Delta T \) and \( \Delta b = \alpha b \Delta T \).
To find out how these expansions change the whole area, use the formula:
  • \( \Delta A = (a + \Delta a)(b + \Delta b) - ab \)
After expanding, you can simplify it by neglecting the small term \( \Delta a \Delta b \), to:get:
  • \( \Delta A = a \Delta b + b \Delta a \)
This expression simplifies accurately calculating the change in area due to the small temperature-driven expansions. Finally, by substituting known values, calculate the new area extent to fully appreciate the scale of thermal expansion in real-world contexts.

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Most popular questions from this chapter

Ethyl alcohol has a boiling point of \(78.0^{\circ} \mathrm{C},\) a freezing point of \(-114^{\circ} \mathrm{C},\) a heat of vaporization of \(879 \mathrm{~kJ} / \mathrm{kg},\) a heat of fusion of \(109 \mathrm{~kJ} / \mathrm{kg},\) and a specific heat of \(2.43 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} .\) How much energy must be removed from \(0.510 \mathrm{~kg}\) of ethyl alcohol that is initially a gas at \(78.0^{\circ} \mathrm{C}\) so that it becomes a solid at \(-114^{\circ} \mathrm{C} ?\)

What mass of butter, which has a usable energy content of \(6.0 \mathrm{Cal} / \mathrm{g}(=6000 \mathrm{cal} / \mathrm{g}),\) would be equivalent to the change in gravitational potential energy of a \(73.0 \mathrm{~kg}\) man who ascends from sea level to the top of Mt. Everest, at elevation \(8.84 \mathrm{~km}\) ? Assume that the average \(g\) for the ascent is \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).

At what temperature is the Fahrenheit scale reading equal to (a) twice that of the Celsius scale and (b) half that of the Celsius scale?

Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of \(3.6 \mathrm{~kg}\) and contains \(14 \mathrm{~kg}\) of water. \(\mathrm{A} 1.8 \mathrm{~kg}\) piece of the metal initially at a temperature of \(180^{\circ} \mathrm{C}\) is dropped into the water. The container and water initially have a temperature of \(16.0^{\circ} \mathrm{C}\), and the final temperature of the entire (insulated) system is \(18.0^{\circ} \mathrm{C}\).

A room is lighted by four \(100 \mathrm{~W}\) incandescent lightbulbs. (The power of \(100 \mathrm{~W}\) is the rate at which a bulb converts electrical energy to heat and the energy of visible light.) Assuming that \(73 \%\) of the energy is converted to heat, how much heat does the room receive in \(6.9 \mathrm{~h} ?\)

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