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On the Rankine temperature scale, which is sometimes used in engineering applications, the ice point is at \(491.67^{\circ} \mathrm{R}\) and the steam point is at \(671.67^{\circ} \mathrm{R}\). Determine a relationship (analogous to Equation 12.1 ) between the Rankine and Fahrenheit temperature scales.

Short Answer

Expert verified
\(T_F = T_R - 459.67\) and \(T_R = T_F + 459.67\).

Step by step solution

01

Understand Given Points on Rankine Scale

The ice point refers to the freezing point of water, which is given as \(491.67^{\circ} \mathrm{R}\) on the Rankine scale. The steam point, which refers to the boiling point of water, is given as \(671.67^{\circ} \mathrm{R}\).
02

Understand Corresponding Points on Fahrenheit Scale

In Fahrenheit, the ice point is at \(32^{\circ} \mathrm{F}\) and the steam point is at \(212^{\circ} \mathrm{F}\).
03

Derive Basic Relationship Formula

Use the two-point form, similar to deriving a linear relationship: \[\frac{T_R - 491.67}{671.67 - 491.67} = \frac{T_F - 32}{212 - 32}\]where \(T_R\) is the temperature in Rankine and \(T_F\) is the temperature in Fahrenheit.
04

Simplify the Formula

Solving the equation: 1. Simplify the differences: \(671.67 - 491.67 = 180\) and \(212 - 32 = 180\).2. The equation becomes: \[\frac{T_R - 491.67}{180} = \frac{T_F - 32}{180}\] 3. Cross multiply to get:\[T_R - 491.67 = T_F - 32\] 4. Rearrange the formula to express \(T_F\) in terms of \(T_R\):\[T_F = T_R - 491.67 + 32\] 5. Simplify:\[T_F = T_R - 459.67\]
05

Express the Inverse Relationship

Rearrange to express Rankine in terms of Fahrenheit if needed for understanding:\[T_R = T_F + 459.67\]

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Key Concepts

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

Rankine Temperature Scale
The Rankine temperature scale is an absolute scale, much like the Kelvin scale in the metric system. It is primarily used in engineering fields, particularly in thermodynamics for temperature measurements where heat energy calculations are essential. The Rankine scale sets its zero point at absolute zero, equating to -459.67°F. This zero point is where molecular motion stops. The unique feature about Rankine is that its degree intervals are the same size as those of Fahrenheit. This means a change of 1° Rankine is equal to a change of 1° Fahrenheit.

The ice point on the Rankine scale is at 491.67°R, whereas the steam point is at 671.67°R. These values are crucial when converting between Rankine and Fahrenheit, especially in engineering calculations where precision is needed.
Fahrenheit Temperature Scale
The Fahrenheit temperature scale is popular in the United States for non-scientific applications, such as weather reports and household thermostats. It is named after the German physicist Daniel Gabriel Fahrenheit who developed it early in the 18th century. On this scale, the freezing point of water is 32°F, and the boiling point is 212°F, establishing 180 equal divisions, or degrees, between these two benchmarks. Because the degree size in Fahrenheit mirrors that of Rankine, it simplifies conversions between these two units.

One interesting note is how the Fahrenheit scale was originally conceptualized using a salt-water mixture for the zero point, which was defined as the freezing point of an ice-water-salt concoction.
Temperature Conversion Formula
Temperature conversion between different scales is important for many practical applications, especially in scientific and engineering contexts. For converting Rankine to Fahrenheit, you can use the linear conversion formula: \[ T_F = T_R - 459.67 \]where \( T_F \) is the temperature in Fahrenheit and \( T_R \) is the temperature in Rankine. This equation highlights that the two scales are directly proportional with an offset, which accounts for their alignment in degree intervals.

This conversion illustrates that the only difference between the two is essentially an additive constant which aligns the absolute zero of Rankine with -459.67°F in Fahrenheit. Conversely, to convert Fahrenheit to Rankine, simply reverse the process:\[ T_R = T_F + 459.67 \] These formulas are essential for calculations involving heat and energy transfer.
Thermodynamics
Thermodynamics is the study of heat, energy, and the work it can produce. It's a core discipline in physics and engineering, dealing with the principles governing temperature, energy transfer, and the behavior of systems under heat action. Temperature scales like Rankine and Fahrenheit play a vital role in understanding and applying thermodynamic principles, especially since the Rankine scale is absolute, simplifying calculations involving absolute temperatures.

Key concepts in thermodynamics include:
  • First Law of Thermodynamics, which deals with energy conservation.
  • Second Law of Thermodynamics, which introduces the concept of entropy.
  • Third Law of Thermodynamics, which states that the entropy of a system approaches a constant value as the temperature approaches absolute zero.
Using these concepts, engineers can design systems such as engines, refrigerators, and even power plants, making temperature conversion integral to precise and accurate thermodynamic problem-solving.

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Most popular questions from this chapter

On a cool autumn morning, the relative humidity is \(100 \%\). (a) Does this mean that the partial pressure of water in the air equals atmospheric pressure? Why or why not? (b) Suppose the air warms up in the afternoon, but the partial pressure of water in the air does not change. Is the humidity still \(100 \%\) ? Provide a reason for your answer. The vapor pressure of water at \(20{ }^{\circ} \mathrm{C}\) is \(2500 \mathrm{~Pa}\). (a) What percentage of atmospheric pressure is this? (b) What percentage of the total air pressure at \(20{ }^{\circ} \mathrm{C}\) is due to water vapor if the relative humidity is \(100 \%\) ? (c) The vapor pressure of water at \(35^{\circ} \mathrm{C}\) is \(5500 \mathrm{~Pa}\). What is the relative humidity at this temperature if the partial pressure of water in the air has not changed from that at \(20{ }^{\circ} \mathrm{C}\) ?

(a) Two portions of the same liquid have the same mass, but different temperatures. They are mixed in a container that prevents the exchange of heat with the environment. Portion \(\mathrm{A}\) has the higher initial temperature of \(T_{0 \mathrm{~A}}\), while portion B has an initial temperature of \(T_{0 \mathrm{~B}}\). Relative to these two temperatures, where on the temperature scale will the final equilibrium temperature of the mixture be located? Explain. (b) Three portions of the same liquid are mixed in a container that prevents the exchange of heat with the environment. Portion A has a mass \(m\) and a temperature of \(94.0^{\circ} \mathrm{C}\), portion \(\mathrm{B}\) has a mass \(m\) and a temperature of \(78.0^{\circ} \mathrm{C}\), and portion \(\mathrm{C}\) has a mass \(2 \mathrm{~m}\) and a temperature of \(34.0{ }^{\circ} \mathrm{C}\). Considering your answer to Concept Question (a) and without doing a heat-lost-equals-heat-gained calculation, deduce the final temperature of the mixture at equilibrium. Justify your answer. Three portions of the same liquid are mixed in a container that prevents the exchange of heat with the environment. Portion A has a mass \(m\) and a temperature of \(94.0^{\circ} \mathrm{C}\), portion \(\mathrm{B}\) has a mass \(m\) and a temperature of \(78.0{ }^{\circ} \mathrm{C}\), and portion \(\mathrm{C}\) has a mass \(m_{\mathrm{C}}\) and a temperature of \(34.0{ }^{\circ} \mathrm{C}\). What must be the mass of portion \(\mathrm{C}\) such that the final temperature \(T_{\mathrm{f}}\) of the three-portion mixture is \(T_{\mathrm{f}}=50.0{ }^{\circ} \mathrm{C}\) ? Express your answer in terms of \(m ;\) e.g., \(m_{\mathrm{C}}=2.20 \mathrm{~m}\).

A temperature of absolute zero occurs at \(-273.15^{\circ} \mathrm{C}\). What is this temperature on the Fahrenheit scale?

Interactive Solution \(\underline{12.61} 12.61\) at provides a model for solving problems such as this. A \(42-\mathrm{kg}\) block of ice at \(0{ }^{\circ} \mathrm{C}\) is sliding on a horizontal surface. The initial speed of the ice is \(7.3 \mathrm{~m} / \mathrm{s}\) and the final speed is \(3.5 \mathrm{~m} / \mathrm{s}\). Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice, and determine the mass of ice that melts into water at \(0^{\circ} \mathrm{C}\).

Concept Questions (a) When heat is added to an object, what factors determine its change in temperature, assuming there is no phase change? (b) Suppose the same amount of heat is applied to two bars without their changing phase. They have the same mass, but experience different changes in temperature. Are the specific heat capacities the same for each bar? Why or why not? Problem Two bars of identical mass are at \(25^{\circ} \mathrm{C}\). One is made from glass and the other from another substance. When identical amounts of heat are supplied to each, the glass bar reaches a temperature of \(88^{\circ} \mathrm{C}\), while the other reaches \(250.0{ }^{\circ} \mathrm{C}\). What is the specific heat capacity of the other substance?

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