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The Rankine temperature scale is an absolute temperature scale that uses Fahrenheit degrees; that is, temperatures are measured in Fahrenheit degrees, starting at absolute zero. Find the relationships between temperature values on the Rankine scale and those on the Fahrenheit, Kelvin, and Celsius scales.

Short Answer

Expert verified
Answer: 1. Rankine to Fahrenheit: F = R - 459.67 2. Rankine to Kelvin: K = R * (5/9) 3. Rankine to Celsius: C = (R - 491.67) * (5/9)

Step by step solution

01

Absolute Zero

Absolute zero is the lowest possible temperature, where particles have minimal motion. In the Kelvin scale, absolute zero is defined as 0K. The Celsius equivalent of this temperature is -273.15°C, and its Fahrenheit equivalent is -459.67°F. The Rankine scale uses Fahrenheit degrees and starts at absolute zero, which means absolute zero on the Rankine scale is 0R.
02

Fahrenheit, Celsius, and Kelvin Relationships

First, let's review the relationships between Fahrenheit (F), Celsius (C), and Kelvin (K) temperature scales: 1. Fahrenheit to Celsius conversion: C = (F - 32) * (5/9) 2. Celsius to Kelvin conversion: K = C + 273.15 3. Fahrenheit to Kelvin conversion: K = (F + 459.67) * (5/9) We will use these relationships to find the relationships between temperature values on the Rankine scale (R) and those on the Fahrenheit, Kelvin, and Celsius scales.
03

Rankine to Fahrenheit

Since Rankine starts at absolute zero and uses Fahrenheit degrees, each increment of 1R corresponds to an increment of 1°F. Therefore, Rankine and Fahrenheit scales have the same difference between their degrees (units). To convert Rankine (R) to Fahrenheit (F), we simply subtract the absolute zero value in Fahrenheit (-459.67°F) to get: F = R - 459.67
04

Rankine to Kelvin

To convert Rankine (R) to Kelvin (K), we first convert Rankine to Fahrenheit using the relationship derived in the previous step, and then convert Fahrenheit to Kelvin using the known relationship: F = R - 459.67 K = (F + 459.67) * (5/9) Now, we substitute the first equation into the second equation to get: K = ((R - 459.67) + 459.67) * (5/9) Simplifying the equation, we get: K = R * (5/9)
05

Rankine to Celsius

To convert Rankine (R) to Celsius (C), we first convert Rankine to Fahrenheit using the relationship derived earlier, and then convert Fahrenheit to Celsius using the known relationship: F = R - 459.67 C = (F - 32) * (5/9) Now, we substitute the first equation into the second equation to get: C = ((R - 459.67) - 32) * (5/9) Simplifying the equation, we get: C = (R - 491.67) * (5/9) In summary, the relationships between temperature values on the Rankine scale and those on the Fahrenheit, Kelvin, and Celsius scales are: 1. Rankine to Fahrenheit: F = R - 459.67 2. Rankine to Kelvin: K = R * (5/9) 3. Rankine to Celsius: C = (R - 491.67) * (5/9)

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Key Concepts

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

Absolute Zero
Imagine a space where particles stand completely still, this theoretical point is known as absolute zero. It represents the lowest limit of the thermodynamic temperature scale, where molecular motion ceases. On the Kelvin scale, this is marked as 0K. In terms of Celsius and Fahrenheit, it corresponds to -273.15°C and -459.67°F respectively. The Rankine temperature scale, being an absolute scale like Kelvin, also starts at 0R. Understanding the concept of absolute zero is crucial for temperature scale conversions and in appreciating the fundamental limits of nature. It's fascinating that such an extreme temperature is used as a starting point for two of our temperature measurement scales!
Temperature Scale Conversions
Dealing with temperature measurements often requires converting from one scale to another. For students, mastering these conversions is important for solving various scientific problems. The Rankine scale (R) is directly linked to the Fahrenheit (F) scale, as they both start at absolute zero and move in the same degree increments. To convert from Rankine to Fahrenheit, simply subtract absolute zero in Fahrenheit:

From Rankine to Fahrenheit:
F = R - 459.67

For Kelvin (K) and Celsius (C), the Rankine scale requires different equations. Converting from Rankine to Kelvin omits the need to subtract or add Fahrenheit's absolute zero value:

From Rankine to Kelvin:
K = R * (5/9)

While Celsius, being off by 32 degrees and scaled differently, involves adjusting for these discrepancies:

From Rankine to Celsius:
C = (R - 491.67) * (5/9)

These conversions are not only a practice in algebra but also teach us about the relationship between different scales and how they were designed around various reference points.
Fahrenheit, Celsius, and Kelvin Relationships
Understanding the relationships between Fahrenheit, Celsius, and Kelvin is a fundamental pursuit in many scientific fields. These scales connect through simple mathematical relationships, allowing for temperature conversions vital in various scenarios. From everyday weather forecasting to complex scientific research, being adept at these interrelationships is indispensable.

For converting Fahrenheit to Celsius, one must subtract 32 (the freezing point of water on the Fahrenheit scale) and then scale the temperature using the ratio of the temperature intervals between boiling and freezing of water on both scales:

Fahrenheit to Celsius:
C = (F - 32) * (5/9)

Kelvin, on the other hand, is a direct Celsius scale offset by 273.15 (the difference in the absolute zero between the scales):

Celsius to Kelvin:
K = C + 273.15

To bypass Celsius when converting Fahrenheit to Kelvin, the absolute zero adjustments and scaling ratio are applied simultaneously:

Fahrenheit to Kelvin:
K = (F + 459.67) * (5/9)

These relationships elucidate the coherent design of temperature scales which, once understood, can be navigated with ease. Students and professionals aliketap into this knowledge regularly to interpret and communicate temperature information accurately across various disciplines.

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Most popular questions from this chapter

You are outside on a hot day, with the air temperature at \(T_{0}\). Your sports drink is at a temperature \(T_{\mathrm{d}}\) in a sealed plastic bottle. There are a few remaining ice cubes in the sports drink, which are at a temperature \(T_{\mathrm{i}}\), but they are melting fast. a) Write an inequality expressing the relationship among the three temperatures. b) Give reasonable values for the three temperatures in degrees Celsius.

On a hot summer day, a cubical swimming pool is filled to within \(1.0 \mathrm{~cm}\) of the top with water at \(21{ }^{\circ} \mathrm{C} .\) When the water warms to \(37^{\circ} \mathrm{C}\), the pool overflows. What is the depth of the pool?

In order to create a tight fit between two metal parts, machinists sometimes make the interior part larger than the hole into which it will fit and then either cool the interior part or heat the exterior part until they fogether. Suppose an aluminum rod with diameter \(D_{1}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right)\) is to be fit into a hole in a brass plate that has a diameter \(D_{2}=10.000 \mathrm{~mm}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right) .\) The machinists can cool the rod to \(77.0 \mathrm{~K}\) by immersing it in liquid nitrogen. What is the largest possible diameter that the rod can have at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\) and just fit into the hole if the rod is cooled to \(77.0 \mathrm{~K}\) and the brass plate is left at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C} ?\) The linear expansion coefficients for aluminum and brass are \(22 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\) and \(19 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\), respectively.

The lowest air temperature recorded on Earth is \(-129^{\circ} \mathrm{F}\) in Antarctica. Convert this temperature to the Celsius scale.

On a cool morning, with the temperature at \(15.0^{\circ} \mathrm{C}\), a painter fills a 5.00 -gal aluminum container to the brim with turpentine. When the temperature reaches \(27.0^{\circ} \mathrm{C}\), how much fluid spills out of the container? The volume expansion coefficient for this brand of turpentine is \(9.00 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\).

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