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

Short Answer

Expert verified
The true distance is approximately 25.974 m.

Step by step solution

01

Understand the problem

We need to find the true distance between two points measured using a steel tape at a different temperature. The steel tape is calibrated at 20.0\(^\circ\)C, but the measurement is done at 5.00\(^\circ\)C.
02

Know the formula

The linear expansion of steel can be calculated using:\[ L = L_0 (1 + \alpha \Delta T) \]where \(L\) is the new length, \(L_0\) is the original length, \(\alpha\) is the coefficient of linear expansion for steel, and \(\Delta T\) is the change in temperature.
03

Identify given information

We have the original length of the tape, \(L_0 = 30.0\,m\). The measured length, \(L_{measured} = 25.970\,m\), and the temperature change, \(\Delta T = 5.00 - 20.0 = -15.0\,\text{°C}\). The coefficient of linear expansion for steel, \(\alpha\), is approximately \(11 \times 10^{-6} \, \text{°C}^{-1}\).
04

Calculate the actual length of the tape at the new temperature

Substitute the values into the formula to find \(L\) at 5.00\(^\circ\)C:\[ L = 30.0 \times (1 + 11 \times 10^{-6} \times (-15)) \]\[ L \approx 30.0 \times (1 - 0.000165) = 30.0 \times 0.999835 = 29.99505\,m \]
05

Determine the true distance between the points

Since the tape shrinks when cooled, the measured length is shorter. Therefore, to find the true length:\[ \text{True distance} = \frac{25.970}{\frac{29.99505}{30.0}} \]\[ \text{True distance} \approx \frac{25.970}{0.999835} \approx 25.974\,m \]

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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, denoted by \( \alpha \), is a material-specific value that quantifies how much a material will expand or contract when subjected to a temperature change. It is essential in assessing the thermal behavior of materials, especially metals like steel.
The coefficient of linear expansion is typically small, but crucial in engineering and physics calculations. For example, the coefficient for steel is approximately \( 11 \times 10^{-6} \, \text{°C}^{-1} \). This means that for every degree Celsius change in temperature, a 1-meter length of steel will change its length by \( 11 \times 10^{-6} \) meters.
  • To calculate length changes: \( L = L_0 (1 + \alpha \Delta T) \) where \( L_0 \) is the original length, and \( \Delta T \) is the temperature change.
  • This coefficient helps ensure accurate measurements in varying temperature conditions, which is especially important for tasks like surveying or engineering projects.
Understanding this concept allows us to predict and compensate for the expansion or contraction of materials, ensuring precision in measurements.
Temperature Change
Temperature change, denoted as \( \Delta T \), is the difference between the initial and final temperatures experienced by an object. In the context of linear thermal expansion, \( \Delta T \) affects how much a material will expand or contract.
In our exercise, the temperature change is from 20.0\( ^\circ \)C, the original calibration temperature of the tape, to 5.00\( ^\circ \)C, the temperature on the day of measurement. This results in a \( \Delta T \) of \(-15.0\, \text{°C}\).
  • Temperature change can be calculated as: \( \Delta T = T_{final} - T_{initial} \).
  • A negative temperature change indicates contraction, while a positive indicates expansion.
Tracking temperature variations is vital as it directly impacts the dimensions of materials sensitive to thermal changes, emphasizing the importance of considering \( \Delta T \) in accurate measurements and calculations.
True Distance Calculation
When measuring distances with materials susceptible to thermal expansion, like a steel tape, determining the true distance requires correcting for temperature-induced length changes. Measuring at a temperature different from the calibration point causes inaccuracies due to the expansion or contraction of the tape.
To find the true distance, a two-step process involves recalculating the tape's actual length at the current temperature and then adjusting the measured distance to get the true measurement.
  • The actual length of the tape at the new temperature is found with \( L = L_0 (1 + \alpha \Delta T) \).
  • Adjust the measured distance to find the true distance: \( \text{True distance} = \frac{L_{measured}}{L_{actual}/L_0} \).
This correction process ensures that the final measurement reflects the true distance between points, accounting for thermal effects on the measuring tool. It highlights the precision needed in surveying and similar endeavors.

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Most popular questions from this chapter

A thirsty nurse cools a 2.00-L bottle of a soft drink (mostly water) by pouring it into a large aluminum mug of mass 0.257 kg and adding 0.120 kg of ice initially at -15.0\(^\circ\)C. If the soft drink and mug are initially at 20.0\(^\circ\)C, what is the final temperature of the system, assuming that no heat is lost?

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 190.0 J/s from a furnace at 400.0\(^\circ\)C to a container of boiling water under 1 atmosphere. What must the rod’s diameter be?

Rods of copper, brass, and steel-each with crosssectional area of 2.00 cm\(^2\)-are welded together to form a Y-shaped figure. The free end of the copper rod is maintained at 100.0\(^\circ\)C, and the free ends of the brass and steel rods at 0.0\(^\circ\)C. Assume that there is no heat loss from the surfaces of the rods. The lengths of the rods are: copper, 13.0 cm; brass, 18.0 cm; steel, 24.0 cm. What is (a) the temperature of the junction point; (b) the heat current in each of the three rods?

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about 1.50 kW/m\(^2\). The distance from the earth to the sun is \(1.50 \times 10{^1}{^1} m\), and the radius of the sun is \(6.96 \times 10{^8} m\). (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?

A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at 20.0\(^\circ\)C. A 0.250-kg block of iron at 85.0\(^\circ\)C is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

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