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\(17.76 \bullet\) A surveyor's \(30.0 \mathrm{~m}\) steel tape is correct at \(20.0^{\circ} \mathrm{C}\). The distance between two points, as measured by this tape on a day when its temperature is \(5.00^{\circ} \mathrm{C},\) is \(25.970 \mathrm{~m} .\) What is the true distance between the points?

Short Answer

Expert verified
The true distance between the two points is approximately \(25.9654 \mathrm{~m}\).

Step by step solution

01

Calculate the Change in Length

First, find the change in temperature by subtracting the original temperature from the current temperature: \(\Delta T = 5.0°C - 20.0°C = -15.0°C\). Then, substitute the defined values into the thermal expansion formula to find the change in length: \(\Delta L = L_0 \alpha \Delta T = 25.970m * 12 \times 10^{-6} \degree C^{-1} * -15.0\degree C = -0.0046 m\).
02

Find the True Length

Substitute the known values into the equation for true length to solve for it: \(L = L_0 + \Delta L = 25.970m - 0.0046m = 25.9654m\). The negative sign signifies that the tape contracted due to the drop in temperature.

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Key Concepts

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

Temperature Change
Understanding temperature change is crucial when it comes to measuring precise distances, especially with materials that are sensitive to thermal fluctuations, like steel. Temperature can affect the length of materials; as the temperature drops, materials generally contract and as it rises, they expand. This behavior can lead to inaccuracies if not properly accounted for. For instance, a steel tape that is calibrated to be accurate at a specific temperature, say 20 degrees Celsius, will slightly alter in length at temperatures above or below that point. Therefore, when measuring distances in varying temperatures, it is important to understand how much a material expands or contracts with temperature changes to ensure accurate measurements.

In our exercise, the surveyor faces a lower temperature scenario, which causes the steel tape to contract, leading to a measurement that if not corrected, would indicate a slightly shorter distance than the true distance between two points.
Linear Expansion
Linear expansion is the increase or decrease in the length of an object due to a temperature change. The extent of linear expansion is proportional to both the original length of the material and the temperature change it experiences. Different materials have different coefficients of linear expansion, usually denoted as \( \alpha \). These coefficients represent the extent to which a material will expand or contract per degree change in temperature.

In our exercise, the linear expansion coefficient for steel is given as \( 12 \times 10^{-6} \degree C^{-1} \) which indicates how much a 1 meter length of steel will expand or contract for a 1 degree Celsius change in temperature. Understanding this concept is vital for engineers, surveyors, and anyone requiring precise measurements, as they must correct for thermal expansion or contraction to ascertain the true size or distance after a temperature fluctuation.
True Length Calculation
The true length calculation is the process of determining the actual distance between two points, factoring in any changes in length due to temperature variations. To find the true distance, we adjust the measured length by the amount of contraction or expansion.

This process often involves first calculating the change in length \( \Delta L \), using the formula \( \Delta L = L_0 \alpha \Delta T \) where \( L_0 \) is the original length, \( \alpha \) is the coefficient of thermal expansion, and \( \Delta T \) is the temperature change. Then, simply add (in the case of expansion) or subtract (in the case of contraction) \( \Delta L \) from \( L_0 \) to find the true length \( L \) of the object.

For the given exercise, where the temperature dropped by 15 degrees Celsius, the steel tape used for measurement contracts. By calculating the contraction and adjusting the measured length, we obtain the true length between the two points, ensuring accurate and reliable results regardless of the environmental conditions.

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Most popular questions from this chapter

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about \(1.50 \mathrm{~kW} / \mathrm{m}^{2} .\) The distance from the earth to the sun is \(1.50 \times 10^{11} \mathrm{~m},\) and the radius of the sun is \(6.96 \times 10^{8} \mathrm{~m} .\) (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?

Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales: (a) the midday temperature at the surface of the moon \((400 \mathrm{~K}) ;\) (b) the temperature at the tops of the clouds in the atmosphere of Saturn \((95 \mathrm{~K}) ;\) (c) the temperature at the center of the sun \(\left(1.55 \times 10^{7} \mathrm{~K}\right)\)

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 \(75^{\circ} \mathrm{C}\) ?

The emissivity of tungsten is 0.350 . A tungsten sphere with radius \(1.50 \mathrm{~cm}\) is suspended within a large evacuated enclosure whose walls are at \(290.0 \mathrm{~K}\). What power input is required to maintain the sphere at \(3000.0 \mathrm{~K}\) if heat conduction along the supports is ignored?

The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A \(75 \mathrm{~kg}(165 \mathrm{lb})\) person of height \(1.83 \mathrm{~m}(6 \mathrm{ft})\) has a body surface area of approximately \(2.0 \mathrm{~m}^{2}\). (a) What is the net amount of heat this person could radiate per second into a room at \(18^{\circ} \mathrm{C}\) (about \(65^{\circ} \mathrm{F}\) ) if his skin's surface temperature is \(30^{\circ} \mathrm{C}\) ? (At such temperatures, nearly all the heat is infrared radiation, for which the body's emissivity is \(1.0,\) regardless of the amount of pigment.) (b) Normally, \(80 \%\) of the energy produced by metabolism goes into heat, while the rest goes into things like pumping blood and repairing cells. Also normally, a person at rest can get rid of this excess heat just through radiation. Use your answer to part (a) to find this person's basal metabolic rate.

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