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Chapter 17: Problem 26

(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 \(27.5^{\circ} \mathrm{C} ?\)

Short Answer

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

Step by step solution

01

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.
02

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.
03

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 \( A_0 \).
04

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.
05

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.
06

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 \( \Delta A \).

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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, 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.

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Most popular questions from this chapter

One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 \(\mathrm{W}\) is required to keep the interior surface of the box 65.0 \(\mathrm{C}^{\circ}\) (about 120 \(\mathrm{F}^{\circ}\) ) above the temperature of the outer surface. The total area of the box is 2.18 \(\mathrm{m}^{2}\) , and the wall thickness is 3.90 \(\mathrm{cm}\) . Find the thermal conductivity of the material in SI units.

The Sizes of Stars. The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume \(e=1\) for these surfaces. Find the radii of the following stars (assumed to be spherical): (a) Rigel, the bright blue star in the constellation Orion, which radiates energy at a rate of \(2.7 \times 10^{32} \mathrm{W}\) and has surface temperature \(11,000 \mathrm{K} ;\) (b) Procyon \(\mathrm{B}\) (visible only using a telescope), which radiates energy at a rate of \(2.1 \times 10^{23} \mathrm{W}\) and has surface temperature \(10,000 \mathrm{K}\) . (c) Compare your answers to the radius of the earth, the radius of the sun, and the distance between the earth and the sun. (Rigel is an example of a supergiant star, and Procyon \(\mathrm{B}\) is an example of a white dwarf star.)

A technician measures the specific heat of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat transferred to the liquid for 120 \(\mathrm{s}\) at a constant rate of 65.0 \(\mathrm{W}\) . The mass of the liquid is 0.780 \(\mathrm{kg}\) , and its temperature increases from \(18.55^{\circ} \mathrm{C}\) to \(22.54^{\circ} \mathrm{C}\) (a) Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat? Explain.

What is the rate of energy radiation per unit area of a black-body at a temperature of (a) 273 \(\mathrm{K}\) and \((\mathrm{b}) 2730 \mathrm{K} ?\)

A Walk in the Sun. Consider a poor lost soul waking at 5 \(\mathrm{km} / \mathrm{h}\) on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of 280 \(\mathrm{W}\) , and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k^{\prime} A_{\mathrm{skin}}\) \(\left(T_{\text { air }}-T_{\text { skin }}\right),\) where \(k^{\prime}\) is 54 \(\mathrm{J} / \mathrm{h} \cdot \mathrm{C}^{\circ} \cdot \mathrm{m}^{2}\) the exposed skin area \(A_{\mathrm{skin}}\) is \(1.5 \mathrm{m}^{2},\) the air temperature \(T_{\text { air }}\) is \(47^{\circ} \mathrm{C}\) the exposed skin area \(A_{\mathrm{skin}}\) is \(36^{\circ} \mathrm{C}\) (iii) the skin absorbs radiant energy from the sun at a rate of 1400 \(\mathrm{W} / \mathrm{m}^{2}\) , (iv) the skin absorbs radiant energy from the environment, which has temperature \(47^{\circ} \mathrm{C} .\) (a) Calculate the net rate (in wats) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e=1\) and that the skin temperature is initially \(36^{\circ} \mathrm{C} .\) Which mechanism is the most important? (b) At what rate (in \(L / h )\) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at \(36^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} . )(\mathrm{c})\) Suppose instead the person is protected by light-colored clothing \((e \approx 0)\) so that the exposed skin area is only \(0.45 \mathrm{m}^{2} .\) What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.
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