Chapter 17: Problem 76
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.
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.
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 \(-15.0\, \text{°C}\).
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.
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.
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: \( \text{True distance} = \frac{L_{measured}}{L_{actual}/L_0} \).