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Chapter 12: Problem 16

One January morning in 1943, a warm chinook wind rapidly raised the temperature in Spearfish, South Dakota, from below freezing to \(12.0^{\circ} {C}\) . As the chinook died away, the temperature fell to \(-20.0^{\circ} {C}\) in 27.0 minutes. Suppose that a 19-m aluminum flagpole were subjected to this temperature change. Find the average speed at which its height would decrease, assuming the flagpole responded instantaneously to the changing temperature.

Short Answer

Expert verified
The average speed of the height decrease is \(8.7 \times 10^{-6} \text{ m/s}\).

Step by step solution

01

Understand the Temperature Change

The temperature change is from \(12.0^{\circ} \text{C}\) to \(-20.0^{\circ} \text{C}\), which is a drop of \(32.0^{\circ} \text{C}\). We need to calculate the contraction of the aluminum flagpole due to this temperature change.
02

Find the Coefficient of Linear Expansion

For aluminum, the coefficient of linear expansion \(\alpha\) is \(2.31 \times 10^{-5} \text{ C}^{-1}\). We will use this for further calculations.
03

Calculate the Change in Length

The change in length \(\Delta L\) can be found using the formula: \[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]where \(L_0 = 19 \text{ m}\) is the original length of the flagpole, \(\alpha = 2.31 \times 10^{-5} \text{ C}^{-1}\), and \(\Delta T = -32.0^{\circ} \text{C}\). Substituting these values, we get:\[ \Delta L = 19 \times 2.31 \times 10^{-5} \times (-32.0) = -0.0140544 \text{ m} \approx -0.0141 \text{ m} \]
04

Convert Time from Minutes to Seconds

The time over which this change occurs is 27.0 minutes. Converting this to seconds for consistency with SI units gives:\[ 27.0 \text{ minutes} \times 60 \frac{\text{seconds}}{\text{minute}} = 1620 \text{ seconds} \]
05

Calculate the Average Speed of Contraction

The average speed \(v\) of the change in length is given by:\[ v = \frac{\Delta L}{\Delta t} \]where \(\Delta L = -0.0141 \text{ m}\) and \(\Delta t = 1620 \text{ seconds}\). Therefore,\[ v = \frac{-0.0141}{1620} \approx -8.7 \times 10^{-6} \text{ m/s} \]
06

Describe the Average Speed

The negative sign indicates a decrease in length. Thus, the average speed at which the height of the flagpole decreases is approximately \(8.7 \times 10^{-6} \text{ m/s}\).

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Key Concepts

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

Coefficient of Linear Expansion
The coefficient of linear expansion is a crucial concept in understanding how materials respond to temperature changes. It represents the fractional change in length per degree Celsius change in temperature. Metals, like aluminum, expand when heated and contract when cooled.
For our aluminum flagpole, this coefficient is given as \( \alpha = 2.31 \times 10^{-5} \, \text{C}^{-1} \). This means for every degree Celsius the temperature changes, each meter of the flagpole changes its length by \( 2.31 \times 10^{-5} \) times its original length.
  • *Linear Expansion Formula*: The change in length \( \Delta L \) is calculated as \( \Delta L = L_0 \cdot \alpha \cdot \Delta T \), where \( L_0 \) is initial length, \( \alpha \) is the coefficient, and \( \Delta T \) is the temperature change.
  • This coefficient allows predictions about how much a material will expand or contract with temperature changes.
Temperature Change
Temperature change is the difference in temperatures experienced by an object. In our example, the temperature decreased from \(12.0^\circ \text{C} \) to \(-20.0^\circ \text{C} \).
To quantify this, we calculate the temperature change \( \Delta T \) which is \( -20.0 - 12.0 = -32.0^\circ \text{C} \). This steep drop affects the physical properties of materials like the aluminum in the flagpole, causing contraction.
  • The direction of temperature change (positive for warming, negative for cooling) directly impacts whether a material expands or contracts.
  • Understanding temperature change helps us manage materials in environments with fluctuating thermal conditions.
Cracking or distortion in structures often arises if such changes aren鈥檛 considered in their design.
Average Speed of Contraction
Average speed of contraction describes how quickly an object's length shortens during cooling. It's defined as the change in length divided by the time period over which this change occurs.
For the aluminum rod, the length decreased by \(-0.0141\text{ m}\) over 27 minutes, which equals 1620 seconds. This gives an average speed of contraction \( v \) as: \( v = \frac{-0.0141}{1620} \approx -8.7 \times 10^{-6} \, \text{m/s} \).
  • The negative sign indicates a decrease, essential for showing contraction instead of expansion.
  • Knowing this speed helps engineers design structures that can withstand rapid cooling without damage.
Such calculations are vital for materials exposed to sudden temperature changes like those in industrial settings or harsh climates.

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Most popular questions from this chapter

Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. While a person is exercising, 0.6 kg of blood flows to the body鈥檚 surface and releases 2000 J of energy. The blood arriving at the surface has the temperature of the body鈥檚 interior, \(37.0^{\circ} {C}\) . Assuming that blood has the same specific heat capacity as water, determine the temperature of the blood that leaves the surface and returns to the interior.

Two bars of identical mass are at \(25^{\circ} {C}\) . One is made from glass and the other from another substance. The specific heat capacity of glass is 840 \({J} /({kg} \cdot {C}^{\circ})\) . When identical amounts of heat are supplied to each, the glass bar reaches a temperature of \(88^{\circ} {C}\) , while the other bar reaches 250.0 'C. What is the specific heat capacity of the other substance?

SSM Dermatologists often remove small precancerous skin lesions by freezing them quickly with liquid nitrogen, which has a temperature of 77 K. What is this temperature on the (a) Celsius and (b) Fahrenheit scales?

Three portions of the same liquid are mixed in a container that prevents the exchange of heat with the environment. Portion A has a mass m and a temperature of \(94.0^{\circ} {C},\) portion \({B}\) also has a mass \(m\) but a temperature of \(78.0^{\circ} {C},\) and portion C has a mass \(m_{{C}}\) and a temperature of \(34.0^{\circ} {C}\) . What must be the mass of portion \({C}\) so that the final temperature \(T_{{f}}\) of the three-portion mixture is \(T_{t}=50.0^{\circ} {C} ?\) Express your answer in terms of \(m ;\) for example, \(m_{{C}}=2.20 {m} .\)

The vapor pressure of water at \(10^{\circ} {C}\) is 1300 \({Pa}\) . (a) What percentage of atmospheric pressure is this? Take atmospheric pressure to be \(1.013 \times 10^{5} {Pa}\) . (b) What percentage of the total air pressure at \(10^{\circ} {C}\) is due to water vapor when the relative humidity is 100\(\% ?\) (c) The vapor pressure of water at \(35^{\circ} {C}\) is 5500 Pa. What is the relative humidity at this temperature if the partial pressure of water in the air has not changed from what it was at \(10^{\circ} {C}\) when the relative humidity was 100\(\% ?\)
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