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\(\cdot\) The Eiffel Tower in Paris is 984 ft tall and is made mostly of steel. If this is its height in winter when its temperature is \(-8.00^{\circ} \mathrm{C},\) how much additional vertical distance must you cover if you decide to climb it during a summer heat wave when its temperature is \(40.0^{\circ} \mathrm{C}\) ? (b) Express the coefficient of linear expansion of steel in terms of Fahrenheit degrees.

Short Answer

Expert verified
The additional distance to cover is 0.520 ft; coefficient in Fahrenheit is \(6.11 \times 10^{-6} ^{\circ}\text{F}^{-1}\).

Step by step solution

01

Understand Thermal Expansion

The problem involves the concept of thermal expansion where materials expand or contract with temperature changes. We will be using the formula for linear expansion: \[ \Delta L = L_0 \alpha \Delta T \]where \( \Delta L \) is the change in length, \( L_0 \) is the original length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.
02

Calculate Change in Temperature

First, calculate the temperature change from winter to summer: \[ \Delta T = 40.0^{\circ} \text{C} - (-8.00^{\circ} \text{C}) = 48.0^{\circ} \text{C} \]
03

Find Coefficient of Linear Expansion for Steel

The coefficient of linear expansion for steel is generally \( \alpha = 11 \times 10^{-6} \text{ } ^{\circ}\text{C}^{-1} \). This will be used to calculate the expansion of the Eiffel Tower.
04

Apply the Linear Expansion Formula

Using the formula: \[ \Delta L = 984 \text{ ft} \times 11 \times 10^{-6} \text{ } ^{\circ}\text{C}^{-1} \times 48.0^{\circ} \text{C} \]Calculating gives:\[ \Delta L \approx 0.520 \text{ ft} \]
05

Coefficient of Linear Expansion in Fahrenheit

The relationship between Celsius and Fahrenheit is \( \Delta T_{\text{Fahrenheit}} = \Delta T_{\text{Celsius}} \times \frac{9}{5} \). Thus, the coefficient of linear expansion in terms of Fahrenheit is calculated by adjusting \( \alpha \) to:\[ \alpha_{\text{F}} = \frac{\alpha}{\frac{9}{5}} = 11 \times 10^{-6} ^{\circ}\text{C}^{-1} \times \frac{5}{9} = 6.11 \times 10^{-6} ^{\circ}\text{F}^{-1} \]

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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, often denoted by the Greek letter alpha (\( \alpha \)), is a crucial property of materials that defines how much a material expands per unit length with a change in temperature. For steel, which is the primary material used in the Eiffel Tower, the coefficient is typically \( 11 \times 10^{-6} \, ^{\circ}\text{C}^{-1} \). This means that for each degree Celsius increase in temperature, every meter of steel will expand by 11 millionths of a meter.
It's important to note that this coefficient is specific to materials and defines how reactive they are to temperature changes. Different materials have different coefficients, which is why some materials may expand or contract more than others when exposed to the same temperature changes.
Linear Expansion Formula
When a material's temperature changes, so does its size. This change in size, specifically length, can be calculated using the linear expansion formula:\[ \Delta L = L_0 \alpha \Delta T \]
  • \( \Delta L \) is the change in length.
  • \( L_0 \) is the original length of the material before temperature change.
  • \( \alpha \) is the coefficient of linear expansion of the material.
  • \( \Delta T \) is the temperature change the material undergoes.
Let's consider an example involving the Eiffel Tower. Given its original height in winter (\( L_0 = 984 \, \text{ft} \)), if it experiences a temperature change of \( \Delta T = 48.0^{\circ} \text{C} \) (from \(-8.00^{\circ} \text{C} \) to \( 40.0^{\circ} \text{C} \)), and using steel’s coefficient of linear expansion, you can calculate how much taller the tower becomes. This application of the formula shows how engineering and material science consider thermal effects in design and construction.
Temperature Change
Temperature change, represented as \( \Delta T \), is a straightforward concept but essential in thermal expansion calculations. It is simply the difference between the final temperature and the initial temperature:\[\Delta T = T_{\text{final}} - T_{\text{initial}}\]Using the Eiffel Tower example, the initial winter temperature is \(-8.00^{\circ} \text{C} \)and the final summer temperature is \(40.0^{\circ} \text{C} \). Thus, the temperature change \( \Delta T \) is found by:\[\Delta T = 40.0^{\circ} \text{C} - (-8.00^{\circ} \text{C}) = 48.0^{\circ} \text{C}\]Understanding \( \Delta T \) helps in determining how much a material will expand or contract when subjected to various thermal environments. It is crucial in architectural and engineering designs that account for temperature fluctuations.
Celsius to Fahrenheit Conversion
Sometimes, temperature change calculations require conversions between different units. In particular, converting temperature change from Celsius to Fahrenheit can be crucial when working with global data.The conversion formula is:\[\Delta T_{\text{Fahrenheit}} = \Delta T_{\text{Celsius}} \times \frac{9}{5}\]This formula says that to convert a temperature change from Celsius to Fahrenheit, you multiply by \(\frac{9}{5}\).Using this, we can find that the same \( 48.0^{\circ} \text{C} \)temperature change for the Eiffel Tower becomes:\[\Delta T_{\text{Fahrenheit}} = 48.0^{\circ} \text{C} \times \frac{9}{5} = 86.4^{\circ} \text{F}\]Conversions like this are essential in international projects where materials and temperatures might be measured and reported using different systems.

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Most popular questions from this chapter

\(\cdot\) In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a 200.0 \(\mathrm{W}\) electric immersion heater in 0.320 \(\mathrm{kg}\) of water. (a) How much heat must be added to the water to raise its temperature from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} ?\) (b) How much time is required if all of the heater's power goes into heating the water?

\(\bullet\) A laboratory technician drops an 85.0 g solid sample of unknown material at a temperature of \(100.0^{\circ} \mathrm{C}\) into a calorimeter. The calorimeter can is made of 0.150 \(\mathrm{kg}\) of copper and contains 0.200 \(\mathrm{kg}\) of water, and both the can and water are initially at \(19.0^{\circ} \mathrm{C}\) . The final temperature of the system is measured to be \(26.1^{\circ} \mathrm{C}\) . Compute the specific heat capacity of the sample. (Assume no heat loss to the surroundings.)

\(\bullet\) The effect of urbanization on plant growth. A study published in July 2004 indicated that temperature increases in urban areas in the eastern United States are causing plants to bud up to 7 days early compared with plants in rural areas just a few miles away, thereby disrupting biological cycles. Average temperatures in the urban areas were up to 3.5 \(\mathrm{C}^{\circ}\) higher than in the rural areas. By what percent will the radiated heat per square meter increase due to such a temperature difference if the rural temperature was \(0^{\circ} \mathrm{C}\) the average?

\(\cdot\) Heat loss during breathing. In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is \(-20^{\circ} \mathrm{C},\) what is the amount of heat needed to warm to internal body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the 0.50 \(\mathrm{L}\) of air exchanged with each breath? Assume that the specific heat capacity of 1.3 \(\mathrm{g}\) is 1020 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\) and that 1.0 \(\mathrm{L}\) of air has a mass of 1.3 \(\mathrm{g}\) . (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

\(\cdot\) You have 750 \(\mathrm{g}\) of water at \(10.0^{\circ} \mathrm{C}\) in a large insulated beaker. How much boiling water at \(100.0^{\circ} \mathrm{C}\) must you add to this beaker so that the final temperature of the mixture will be \(75^{\circ} \mathrm{C}\) ?

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