/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 The tallest building in the worl... [FREE SOLUTION] | 91Ó°ÊÓ

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The tallest building in the world, according to some architectural standards, is the Taipei 101 in Taiwan, at a height of 1671 feet. Assume that this height was measured on a cool spring day when the temperature was \(15.5^{\circ} \mathrm{C}\) . You could use the building as a sort of giant thermometer on a hot summer day by carefully measuring its height. Suppose you do this and discover that the Taipei 101 is 0.471 foot taller than its official height. What is the temperature, assuming that the building is in thermal equilibrium with the air and that its entire frame is made of steel?

Short Answer

Expert verified
The temperature is approximately 38.9 °C.

Step by step solution

01

Understand the Physics Concept

Tall objects like Taipei 101 expand and contract with temperature changes due to thermal expansion. We use the linear expansion formula to find the temperature change when given the change in height.
02

Thermal Expansion Formula

The formula for linear thermal expansion is given by: \[\Delta L = \alpha L_0 \Delta T\]where \(\Delta L\) is the change in length (height), \(\alpha\) is the coefficient of linear expansion for the material (steel), \(L_0\) is the original length (1671 feet), and \(\Delta T\) is the change in temperature.
03

Coefficient of Expansion for Steel

Look up the coefficient of linear expansion for steel in tables. The common value is \( \alpha = 12 \times 10^{-6} \, \text{per degree Celsius} \).
04

Solve for Temperature Change

Rearrange the linear expansion formula to solve for \(\Delta T\):\[\Delta T = \frac{\Delta L}{\alpha L_0}\]Substitute \(\Delta L = 0.471\) feet, \(\alpha = 12 \times 10^{-6}\), and \(L_0 = 1671\) feet into the formula:\[\Delta T = \frac{0.471}{12 \times 10^{-6} \times 1671}\]
05

Calculate Temperature Change

Perform the calculation:\[\Delta T = \frac{0.471}{12 \times 10^{-6} \times 1671} \approx 23.4 \, \text{°C}\]
06

Determine Final Temperature

Add the temperature change to the original temperature:\[\text{Final Temperature} = 15.5 \, \text{°C} + 23.4 \, \text{°C} = 38.9 \, \text{°C}\]

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Key Concepts

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

Coefficient of Linear Expansion
When materials are exposed to temperature changes, they typically expand or contract. This natural phenomenon is known as thermal expansion. A pivotal concept in understanding this is the Coefficient of Linear Expansion. It essentially determines how much a certain material will expand or contract per unit of original length for each degree of temperature change. This coefficient is a material-specific constant and varies for different substances. For steel, which is the material of the Taipei 101 building, this coefficient is measured at approximately \(12 \times 10^{-6}\) per degree Celsius.
  • This value suggests that for every degree Celsius change, each meter of steel expands or contracts by a very small amount of 12 millionths of its length.
  • Knowing the coefficient is crucial in the field of engineering, as it aids in anticipating how structures will react to temperature variations.
Understanding this concept allows engineers to build structures that can withstand the natural expansions and contractions brought on by temperature fluctuations, ensuring stability and safety over a variety of environmental conditions.
Temperature Change Calculation
Calculating the change in temperature when a structure like Taipei 101 elongates is a straightforward application of the linear thermal expansion formula. The formula \( \Delta L = \alpha L_0 \Delta T \) connects the change in length \( \Delta L \), the initial length \( L_0 \), the coefficient of linear expansion \( \alpha \), and the temperature change \( \Delta T \). By rearranging the formula, you can solve for the temperature change: \[\Delta T = \frac{\Delta L}{\alpha L_0}\]
  • This equation asserts that the temperature change is proportional to the change in length and inversely proportional to both the original length and the coefficient of linear expansion.
  • To perform this calculation, substitute\(\Delta L = 0.471\,\text{feet}\), \(\alpha = 12 \times 10^{-6}\), and \(L_0 = 1671\,\text{feet}\) into the formula.
Upon solving, the calculation returns a temperature rise of approximately 23.4 °C, indicating the difference between the initial measurement temperature and the current condition after expansion.
Physics Problem Solving
Physics problems like determining the temperature change due to thermal expansion help solidify understanding of basic principles by applying them to real-world scenarios. Approach these problems step-by-step, beginning with a clear understanding of the concepts involved.
  • First, identify the known quantities: the original and expanded lengths, material properties, and initial temperature.
  • Next, apply the relevant formula, which for thermal expansion is \( \Delta L = \alpha L_0 \Delta T \). Rearrange it as necessary to solve for the unknown, such as \(\Delta T\) in this instance.
  • Finally, calculate the solution accurately, and interpret it in the context of the initial problem.
Understanding each step not only aids in solving the problem at hand but also consolidates your grasp on fundamental physics principles. This approach empowers you to tackle various other physics challenges with confidence, knowing that you can systematically break complex problems into manageable parts.

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Most popular questions from this chapter

A metal rod is 40.125 \(\mathrm{cm}\) long at \(20.0^{\circ} \mathrm{C}\) and 40.148 \(\mathrm{cm}\) long at \(45.0^{\circ} \mathrm{C}\) . Calculate the average coefficient of linear expansion of the rod for this temperature range.

In a container of negligible mass, 0.0400 \(\mathrm{kg}\) of steam at \(100^{\circ} \mathrm{C}\) and atmospheric pressure is added to 0.200 \(\mathrm{kg}\) of water at \(50.0^{\circ} \mathrm{C} .\) (a) If no heat is lost to the surroundings, what is the final temperature of the system?(b) At the final temperature, how many kilograms are there of steam and how many of liquid water?

(a) A wire that is 1.50 \(\mathrm{m}\) long at \(20.0^{\circ} \mathrm{C}\) is found to increase in length by 1.90 \(\mathrm{cm}\) when warmed to \(420.0^{\circ} \mathrm{C}\) . Compute its average coefficient of linear expansion for this temperature range. (b) The wire is stretched just (zero tension) at \(420.0^{\circ} \mathrm{C}\) . Find the stress in the wire if it is cooled to \(20.0^{\circ} \mathrm{C}\) without being allowed to contract. Young's modulus for the wire is \(20 \times 10^{11} \mathrm{Pa}\) .

A \(25,000-\mathrm{kg}\) subway train initially traveling at 15.5 \(\mathrm{m} / \mathrm{s}\) slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are 65.0 \(\mathrm{m}\) long by 20.0 \(\mathrm{m}\) wide by 12.0 \(\mathrm{m}\) high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its specific heat to be 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) .

"The Ship of the Desert" Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400 \(\mathrm{kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day \(\left(12 \text { hours) instead of letting it rise to } 40.0^{\circ} \mathrm{C} \text { . (Note: The }\right.\) specific heat of a camel or other mammal is about the same as that of a typical human, 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) )

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