/*! 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 3 A steel tape is calibrated at \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A steel tape is calibrated at \(20^{\circ} \mathrm{C}\). On a cold day when the temperature is \(-15^{\circ} \mathrm{C}\), what will be the percent error in the tape? \(\alpha_{\text {steel }} 1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) For a temperature change from \(20^{\circ} \mathrm{C}\) to \(-15{ }^{\circ} \mathrm{C}\), we have \(\Delta T-35\) \({ }^{\circ} \mathrm{C}\). Then, $$ \frac{\Delta L}{L_{0}}=\alpha \Delta T=\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(-35^{\circ} \mathrm{C}\right)=-3.9 \times 10^{-4}=-0.039 \% $$

Short Answer

Expert verified
The percent error in the steel tape is -0.039%.

Step by step solution

01

Identify Given Variables

The problem provides us with several key pieces of information:1. The coefficient of linear expansion for steel: \( \alpha = 1.1 \times 10^{-5} \ { }^{\circ} \mathrm{C}^{-1} \).2. The initial calibration temperature of the tape: \( 20^{\circ} \mathrm{C} \).3. The temperature on the cold day: \( -15^{\circ} \mathrm{C} \).
02

Calculate the Temperature Change

Calculate the temperature change \( \Delta T \) by subtracting the cold temperature from the calibration temperature:\[\Delta T = -15^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = -35^{\circ} \mathrm{C}\]
03

Determine Length Change Ratio

Use the formula \( \frac{\Delta L}{L_0} = \alpha \Delta T \) to find the ratio of the change in length to the original length:\[\frac{\Delta L}{L_0} = (1.1 \times 10^{-5} \ { }^{\circ} \mathrm{C}^{-1})(-35^{\circ} \mathrm{C}) = -3.85 \times 10^{-4}\]
04

Calculate Percent Error

Convert the ratio \( \frac{\Delta L}{L_0} \) into a percentage to find the percent error:\[\text{Percent Error} = \left(-3.85 \times 10^{-4}\right) \times 100 = -0.039\%\]
05

Conclusion

This means the steel tape will contract, resulting in a measurement that is 0.039% shorter than expected.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 as \( \alpha \), is a measure that describes how the size of an object changes with a change in temperature. It is crucial in understanding how materials expand or contract when subjected to temperature variations. In the context of linear expansion, this coefficient indicates how much a unit length of a material will expand for a one-degree Celsius change in temperature.
In our example, the steel tape has a coefficient of linear expansion \( \alpha = 1.1 \times 10^{-5} \ { }^{\circ} \mathrm{C}^{-1} \). This value means that for every degree Celsius change, a 1-meter length of the steel tape would expand or contract by \( 1.1 \times 10^{-5} \) meters. This concept is helpful when calibrating measuring instruments in varying environmental conditions.
  • Smaller coefficient values indicate less expansion or contraction with temperature changes.
  • Material properties, like atomic structure and bonding, influence the coefficient.
Understanding the coefficient allows engineers and designers to predict how materials will behave in different temperatures and adjust their designs accordingly.
Temperature change
Temperature change, denoted as \( \Delta T \), refers to the difference in temperature that an object experiences. It plays a key role in determining how much an object made from a given material will expand or contract.
For the steel tape, the initial temperature is \( 20^{\circ} \mathrm{C} \), and it was later exposed to \( -15^{\circ} \mathrm{C} \). To calculate \( \Delta T \), we subtract the final temperature from the initial temperature:
\[ \Delta T = -15^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = -35^{\circ} \mathrm{C} \]
This negative temperature change indicates that the steel tape is experiencing a decrease in temperature, leading it to contract rather than expand.
  • A positive \( \Delta T \) would cause expansion if the material can stretch with heat.
  • Neutral or no \( \Delta T \) implies that there would be no size alteration in the material.
Knowing the temperature change is essential for predicting any dimensional shifts in materials due to environmental changes.
Percent error calculation
Percent error is a way to express how approximate a measured or estimated value is when compared to an expected or true value. In our exercise, we calculated the percent error to determine the impact of temperature changes on the steel tape's measurements.
We use the formula:
\[ \frac{\Delta L}{L_0} = \alpha \cdot \Delta T \]
to find the ratio of change in length \( \Delta L \) to the original length \( L_0 \). The product of the coefficient of linear expansion and the temperature change gives us this ratio, \( -3.85 \times 10^{-4} \) in this situation. Converting this ratio into a percentage helps to visualize the error:
\[ \text{Percent Error} = \left( \frac{\Delta L}{L_0} \right) \times 100 = -0.039\% \]
This percentage shows that, due to contraction, the tape is measuring 0.039% shorter than it should. It's crucial because it allows users to compensate for or correct measurement inaccuracies caused by temperature fluctuations.
  • Negative percent error suggests underestimation.
  • Accuracy of the calculation depends on precision in measuring \( \alpha \) and \( \Delta T \).
Understanding percent error calculation aids in making reliable measurements across different application scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Dry ice freezes at a temperature of \(-109.3^{\circ} \mathrm{F}\). What is that in Celsius?

(a) An aluminum measuring rod, which is correct at \(5.0^{\circ} \mathrm{C}\), measures a certain distance as \(88.42 \mathrm{~cm}\) at \(35.0^{\circ} \mathrm{C}\). Determine the error in measuring the distance due to the expansion of the rod. \((b)\) If this aluminum rod measures a length of steel as \(88.42\) \(\mathrm{cm}\) at \(35.0{ }^{\circ} \mathrm{C}\), what is the correct length of the steel at \(35^{\circ} \mathrm{C}\) ? The coefficient of linear expansion of that sample of aluminum is \(22 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).

Aluminum melts at \(660^{\circ} \mathrm{C}\). How much is that in kelvins?

At \(15.0^{\circ} \mathrm{C}\), a bare wheel has a diameter of \(30.000 \mathrm{~cm}\), and the inside diameter of its steel rim is \(29.930 \mathrm{~cm}\). To what temperature must the rim be heated so as to slip over the wheel? For this type of steel, \(\alpha=1.10 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\).

When a building is constructed at \(-10{ }^{\circ} \mathrm{C}\), a steel beam (crosssectional area \(45 \mathrm{~cm}^{2}\) ) is put in place with its ends cemented in pillars. If the sealed ends cannot move, what will be the compressional force on the beam when the temperature is \(25^{\circ} \mathrm{C}\) ? For this kind of steel, \(\alpha=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) and \(Y=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\). Proceed much as in Problem 15.11: $$ \begin{array}{c} \frac{\Delta L}{L_{0}}=\alpha \Delta T=\left(1.1 \times 10^{-5}{ }^{-5} \mathrm{C}^{-1}\right)\left(35^{\circ} \mathrm{C}\right)=3.85 \times 10^{-4} \\\ F=Y A \frac{\Delta L}{L_{0}}=\left(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\left(45 \times 10^{-4} \mathrm{~m}^{2}\right)\left(3.85 \times 10^{-4}\right)=3.5 \times 10^{5} \mathrm{~N} \end{array} $$

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.