Ch 18: A Macroscopic Description of Matter

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 18: A Macroscopic Description of Matter

Problem 17

Chapter 18, Problem 17

A surveyor has a steel measuring tape that is calibrated to be 100.000 m long (i.e., accurate to ±1 mm) at 20°C. If she measures the distance between two stakes to be 65.175 m on a 3°C day, does she need to add or subtract a correction factor to get the true distance? How large, in mm, is the correction factor?

Verified step by step guidance

Step 1: Understand the problem. The steel measuring tape expands or contracts with temperature changes due to thermal expansion. The tape is calibrated at 20°C, but the measurement is taken at 3°C. We need to determine the correction factor to account for this temperature difference.

Step 2: Recall the formula for linear thermal expansion:

ΔL = L \cdot α \cdot ΔT

, where

ΔL

is the change in length,

L

is the original length,

α

is the coefficient of linear expansion for steel, and

ΔT

is the temperature change.

Step 3: Calculate the temperature change:

ΔT = 20 - 3 = 17 ° C

. The tape is colder than its calibration temperature, so it contracts.

Step 4: Substitute the values into the formula. Use the coefficient of linear expansion for steel, which is approximately

11 \times 10 ⁻ 6 / ° C

. The original length of the tape is

65.175

m. The formula becomes:

ΔL = 65.175 \cdot 11 \times 10 ⁻ 6 \cdot 17

Step 5: Interpret the result. The value of

ΔL

will be negative, indicating contraction. Convert the result to millimeters by multiplying by 1000 (since 1 m = 1000 mm). The correction factor should be subtracted from the measured distance to get the true distance.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Thermal Expansion

Thermal expansion refers to the tendency of materials to change in size or volume in response to changes in temperature. For metals like steel, this means that as temperature decreases, the material contracts, leading to a shorter length. Understanding this concept is crucial for determining how much the measuring tape's length will change when the temperature drops from 20°C to 3°C.

Coefficient of Linear Expansion

The coefficient of linear expansion is a material-specific constant that quantifies how much a unit length of a material expands or contracts per degree change in temperature. For steel, this coefficient is typically around 11 x 10^-6 /°C. This value is essential for calculating the correction factor needed to adjust the measured length of the tape based on the temperature difference.

Correction Factor

A correction factor is an adjustment made to account for systematic errors in measurements due to external conditions, such as temperature. In this scenario, the correction factor will be calculated using the change in temperature and the coefficient of linear expansion to determine how much the tape's length has changed, ensuring that the final measurement reflects the true distance between the stakes.

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