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Some textbooks use the unit \(\mathrm{K}^{-1}\) rather than \({ }^{\circ} \mathrm{C}^{-1}\) for values of the linear expansion coefficient; see Table 17.2 How will the numerical values of the coefficient differ if expressed in \(\mathrm{K}^{-1}\) ?

Short Answer

Expert verified
Answer: No, there is no difference between the linear expansion coefficient when expressed in K鈦宦 and 掳C鈦宦. This is because both Kelvin and Celsius scales have the same incremental difference in temperature, which influences the numerical value of the linear expansion coefficient.

Step by step solution

01

Relationship of temperature scales

Both the Celsius (C) and Kelvin (K) scales of temperature have the same incremental difference (1 degree), but their numbering systems start from different points: absolute zero (0 K) for the Kelvin scale and the freezing point of water (0 掳C) for the Celsius scale. The relationship between these two scales can be expressed as: T(K) = T(掳C) + 273.15
02

Analyzing the expression of linear expansion coefficient in both units

The linear expansion coefficient, 伪, describes the tendency of a solid to expand or contract as a function of temperature changes. Mathematically, it is given by: 螖L = L鈧伪螖T, where 螖L is the change in length, L鈧 is the initial length, and 螖T is the change in temperature. Since the incremental difference in temperature for both Kelvin and Celsius scales is the same, we should not observe any significant difference when expressed in K鈦宦 or 掳C鈦宦. The linear expansion coefficient should remain the same.
03

Comparing numerical values of the same coefficient in K鈦宦 and 掳C鈦宦

It can be verified that the numerical values of the coefficients will not differ when expressed in K鈦宦 or 掳C鈦宦. This is because the main effect of the change of temperature is what influences the numerical value of the linear expansion coefficient. Since the increments are the same in both scales, the values of 伪 in K鈦宦 and 掳C鈦宦 will be equal for the same temperature change.

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Key Concepts

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

Temperature Scales
Understanding different temperature scales is essential in thermodynamics. The Celsius and Kelvin scales are two commonly used temperature measurement systems. These scales increment similarly, meaning a change of 1 degree Celsius is equal to a change of 1 Kelvin.
The difference lies in their starting points:
  • Celsius Scale: Begins at the freezing point of water, which is 0 掳C. Used widely in most parts of the world for everyday temperature readings.
  • Kelvin Scale: Starts at absolute zero (0 K), the point at which absolutely no molecular motion occurs. It is primarily used in scientific contexts.
To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature: \[ T(K) = T(掳C) + 273.15 \]
This makes both scales incredibly useful and interchangeable in scientific equations, as they maintain a consistent interval.
Linear Expansion Coefficient
The linear expansion coefficient is a crucial concept in understanding how materials expand or contract with temperature changes. When materials are heated, they tend to expand; conversely, they contract when cooled. This behavior can be quantified using the linear expansion coefficient, usually denoted by \( \alpha \).
Mathematically, the change in length of an object, \( \Delta L \), due to a temperature change, \( \Delta T \), is described by the formula:\[ \Delta L = L_0 \alpha \Delta T \]where \( L_0 \) is the original length of the material.
The linear expansion coefficient is expressed in units of either \( \mathrm{K}^{-1} \) or \( ^{\circ} \mathrm{C}^{-1} \). Because both temperature scales progress with the same incremental change, the numerical value of \( \alpha \) remains unaffected by whether it is expressed per Kelvin or per Celsius degree. This uniformity simplifies calculations and comparisons across various scientific and engineering disciplines.
Kelvin and Celsius Relationship
The connection between Kelvin and Celsius is fundamental in thermal physics and engineering. Since both scales increase at identical rates, measuring temperature changes in degrees Celsius is equivalent to measuring changes in Kelvin. This relationship is beneficial because:
  • It ensures compatibility: Scientists and engineers can choose either scale without affecting the outcomes of their calculations.
  • The conversion factor is straightforward: By adding 273.15, you convert a temperature from Celsius to Kelvin.
This relationship allows the direct application of thermal expansion equations regardless of the unit used. Therefore, whether you use \( \mathrm{K}^{-1} \) or \( ^{\circ} \mathrm{C}^{-1} \) to express the linear expansion coefficient, the scientific principles and results remain consistent. This unified approach is not only practical but also essential for cross-disciplinary studies and global scientific communication.

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Most popular questions from this chapter

At room temperature, an iron horseshoe, when dunked into a cylindrical tank of water (radius of \(10.0 \mathrm{~cm})\) causes the water level to rise \(0.25 \mathrm{~cm}\) above the level without the horseshoe in the tank. When heated in the blacksmith's stove from room temperature to a temperature of \(7.00 \cdot 10^{2} \mathrm{~K}\) worked into its final shape, and then dunked back into the water, how much does the water level rise above the "no horseshoe" level (ignore any water that evaporates as the horseshoe enters the water)? Note: The linear expansion coefficient for iron is roughly that of steel: \(11 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).

Express each of the following temperatures in degrees Celsius and in kelvins. a) \(-19^{\circ} \mathrm{F}\) b) \(98.6^{\circ} \mathrm{F}\) c) \(52^{\circ} \mathrm{F}\)

Two solid objects are made of different materials. Their volumes and volume expansion coefficients are \(V_{1}\) and \(V_{2}\) and \(\beta_{1}\) and \(\beta_{2}\), respectively. It is observed that during a temperature change of \(\Delta T\), the volume of each object changes by the same amount. If \(V_{1}=2 V_{2}\) what is the ratio of the volume expansion coefficients?

The volume of \(1.00 \mathrm{~kg}\) of liquid water over the temperature range from \(0.00^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C}\) fits reasonably well to the polynomial function \(V=1.00016-\left(4.52 \cdot 10^{-5}\right) T+\) \(\left(5.68 \cdot 10^{-6}\right) T^{2}\), where the volume is measured in cubic meters and \(T\) is the temperature in degrees Celsius. a) Use this information to calculate the volume expansion coefficient for liquid water as a function of temperature. b) Evaluate your expression at \(20.0^{\circ} \mathrm{C}\), and compare the value to that listed in Table \(17.3 .\)

A medical device used for handling tissue samples has two metal screws, one \(20.0 \mathrm{~cm}\) long and made from brass \(\left(\alpha_{\mathrm{b}}=18.9 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right)\) and the other \(30.0 \mathrm{~cm}\) long and made from aluminum \(\left(\alpha_{\mathrm{a}}=23.0 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right)\). A gap of \(1.00 \mathrm{~mm}\) exists between the ends of the screws at \(22.0^{\circ} \mathrm{C}\). At what temperature will the two screws touch?

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