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Like the Kelvin scale, the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine (0\(^\circ\)R). However, the units of this scale are the same size as those of the Fahrenheit scale rather than the Celsius scale. What is the numerical value of the triple-point temperature of water on the Rankine scale?

Short Answer

Expert verified
The triple-point temperature of water on the Rankine scale is 491.688°R.

Step by step solution

01

Understand the Triple-Point and Conversion

The triple-point temperature of water is a fundamental fixed point which is defined as 273.16 Kelvin (K). Our task is to express this temperature in the Rankine scale. We need the conversion relationship between Kelvin and Rankine.
02

Conversion Formula from Kelvin to Rankine

The conversion from Kelvin to Rankine is done using the formula: \[ T_{R} = T_{K} \times \frac{9}{5} \]where \( T_{R} \) is the temperature in Rankine and \( T_{K} \) is the temperature in Kelvin. It is important to understand that 0 Rankine is absolute zero, just like 0 Kelvin.
03

Plug in the Triple-Point Temperature of Water

Now we substitute 273.16 K into the conversion formula:\[ T_{R} = 273.16 \times \frac{9}{5} \]
04

Perform the Calculation

Let's calculate:\[ T_{R} = 273.16 \times \frac{9}{5} = 491.688 \]Thus, the triple-point temperature of water expressed on the Rankine scale is 491.688°R.

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Key Concepts

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

Absolute Temperature Scale
The Rankine and Kelvin scales are both examples of absolute temperature scales. An absolute temperature scale is one where the lowest possible temperature is absolute zero. Absolute zero represents the point at which the molecules of a substance have minimal possible motion, which physically cannot be surpassed. In essence, absolute zero marks the boundary of thermal energy existence, influencing how these scales are developed.
This type of scale measures temperature with a universal starting point, unlike the Celsius or Fahrenheit scales, which are based on arbitrary points, like the freezing point of water. On absolute scales like Kelvin and Rankine, absolute zero is defined as 0 K and 0 °R, ensuring a consistent baseline for scientific measurements. Using absolute temperature scales is essential in scientific fields such as thermodynamics, where temperature values need to avoid negative numbers that can complicate theoretical calculations or simulations.
Kelvin to Rankine Conversion
Converting between Kelvin and Rankine is a straightforward task, as both scales are absolute and linear but differ in their degree sizes. This difference arises due to their historical origins. Kelvin is based on the metric Celsius scale, while Rankine is based on the imperial Fahrenheit scale.
To convert Kelvin to Rankine, multiply the temperature in Kelvin by the fraction \( \frac{9}{5} \).
  • This ratio reflects the relationship between Celsius and Fahrenheit, as each step in the Celsius scale (used in Kelvin) is \( \frac{5}{9} \) of the step in the Fahrenheit scale (used in Rankine).
  • Because they start at absolute zero, the conversion formula directly scales temperatures without needing to adjust with an additional offset, like adding 273.15 for Celsius to Kelvin or 32 for Celsius to Fahrenheit.

Recognizing this simplicity makes conversions easy and quick, an advantage in scientific processes demanding rapid analysis.
Triple-Point Temperature of Water
One of the critical benchmarks in thermodynamics is the triple-point temperature of water. By definition, this is the specific condition where water coexists in all three states: solid, liquid, and gas, establishing a fundamental temperature context.
In the Kelvin scale, this point is precisely set at 273.16 K, which is used as a fixed reference in both Kelvin and Rankine due to its reliability across different situations. This consistency allows it to serve as a pivotal reference for calibrating thermodynamic experiments or developing scientific instruments.
When calculating this temperature on the Rankine scale, the conversion formula \(T_{R} = T_{K} \times \frac{9}{5}\) is employed, leading to a numerical value of 491.688°R. This conversion not only underscores the unity between the absolute temperature scales but also the importance of having standardized reference points like the triple-point of water, which help to unify scientific research for consistent outcomes.

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Most popular questions from this chapter

A vessel whose walls are thermally insulated contains 2.40 kg of water and 0.450 kg of ice, all at 0.0\(^\circ\)C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.0\(^\circ\)C? You can ignore the heat transferred to the container.

One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 W is required to keep the interior surface of the box 65.0 C\(^\circ\) (about 120 F\(^\circ\)) above the temperature of the outer surface. The total area of the box is 2.18 m\(^2\), and the wall thickness is 3.90 cm. Find the thermal conductivity of the material in SI units.

BIO While running, a 70-kg student generates thermal energy at a rate of 1200 W. For the runner to maintain a constant body temperature of 37\(^\circ\)C, this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the energy could not flow out of the student’s body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to 44\(^\circ\)C or higher. The specific heat of a typical human body is 3480 J / kg \(\cdot\) K, slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

During your mechanical engineering internship, you are given two uniform metal bars \(A\) and \(B\), which are made from different metals, to determine their thermal conductivities. Measuring the bars, you determine that both have length 40.0 cm and uniform cross-sectional area 2.50 cm\(^2\). You place one end of bar \(A\) in thermal contact with a very large vat of boiling water at 100.0\(^\circ\)C and the other end in thermal contact with an ice-water mixture at 0.0\(^\circ\)C. To prevent heat loss along the bar's sides, you wrap insulation around the bar. You weigh the amount of ice initially and find it to be 300 g. After 45.0 min has elapsed, you weigh the ice again and find that 191 g of ice remains. The ice-water mixture is in an insulated container, so the only heat entering or leaving it is the heat conducted by the metal bar. You are confident that your data will allow you to calculate the thermal conductivity \(k_A\) of bar \(A\). But this measurement was tedious-you don't want to repeat it for bar \(B\). Instead, you glue the bars together end to end, with adhesive that has very large thermal conductivity, to make a composite bar 80.0 m long. You place the free end of A in thermal contact with the boiling water and the free end of \(B\) in thermal contact with the ice-water mixture. As in the first measurement, the composite bar is thermally insulated. You go to lunch; when you return, you notice that ice remains in the ice-water mixture. Measuring the temperature at the junction of the two bars, you find that it is 62.4\(^\circ\)C. After 10 minutes you repeat that measurement and get the same temperature, with ice remaining in the ice-water mixture. From your data, calculate the thermal conductivities of bar \(A\) and of bar \(B\).

An electric kitchen range has a total wall area of 1.40 m\(^2\) and is insulated with a layer of fiberglass 4.00 cm thick. The inside surface of the fiberglass has a temperature of 175\(^\circ\)C, and its outside surface is at 35.0\(^\circ\)C. The fiberglass has a thermal conductivity of 0.040 W /m \(\cdot\) K. (a) What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of 1.40 m\(^2\)? (b) What electric-power input to the heating element is required to maintain this temperature?

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