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Convert the following temperatures to the Kelvin scale: (a) \(-183^{\circ} \mathrm{C}\) (the melting point of oxygen) (b) \(6000^{\circ} \mathrm{C}\) (temperature at the surface of the sun) (c) \(-269^{\circ} \mathrm{C}\) (the boiling point of helium) (d) \(800^{\circ} \mathrm{C}\) (the melting point of sodium chloride)

Short Answer

Expert verified
(a) 90.15 K, (b) 6273.15 K, (c) 4.15 K, (d) 1073.15 K.

Step by step solution

01

Understanding the Conversion Formula

To convert a temperature from Celsius (^{ ext{°C}} ) to Kelvin (K), use the formula: \[ K = ^{ ext{°C}} + 273.15 \]
02

Convert -183°C to Kelvin

Substitute \(-183\) for \( ^{ ext{°C}} \) in the conversion formula: \[ K = -183 + 273.15 = 90.15 \] So, \(-183^{ ext{°C}}\) equals \(90.15 \kern{1pt} ext{K} \).
03

Convert 6000°C to Kelvin

Substitute \(6000\) for \( ^{ ext{°C}} \) in the conversion formula: \[ K = 6000 + 273.15 = 6273.15 \] Thus, \(6000^{ ext{°C}}\) converts to \(6273.15 \kern{1pt} ext{K} \).
04

Convert -269°C to Kelvin

Insert \(-269\) in place of \( ^{ ext{°C}} \) in the conversion formula: \[ K = -269 + 273.15 = 4.15 \] Hence, \(-269^{ ext{°C}}\) is \(4.15 \kern{1pt} ext{K} \).
05

Convert 800°C to Kelvin

Substitute \(800\) for \( ^{ ext{°C}} \) in the formula: \[ K = 800 + 273.15 = 1073.15 \] Therefore, \(800^{ ext{°C}}\) equals \(1073.15 \kern{1pt} ext{K} \).

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Key Concepts

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

Kelvin Scale
In the world of scientific measurement, the Kelvin scale is a crucial tool for understanding temperature. Named after the renowned physicist Lord Kelvin, it is the primary temperature scale in the scientific community. The Kelvin scale is unique because it starts at absolute zero, the coldest possible temperature, where molecular motion stops. This makes the Kelvin a true zero point, unlike Celsius or Fahrenheit, which are based on arbitrary points related to the freezing or boiling of water.

Importantly, the Kelvin scale is an absolute scale and does not use the term "degree"; temperatures are expressed simply as Kelvin (K). For instance, 273.15 K is the freezing point of water, equivalent to 0°C. By choosing absolute zero as its starting point, the Kelvin scale provides a more straightforward understanding of thermodynamic temperature.
Celsius to Kelvin
Temperature conversions between Celsius and Kelvin are straightforward and widely used in scientific calculations. To convert from Celsius to Kelvin, one simply adds 273.15 to the Celsius temperature. This relationship comes from the fact that both scales are offset by 273.15 due to their respective starting points, with 0°C being equivalent to 273.15 K.

When performing these conversions, remember that:
  • A temperature like \(-183^{\circ} \mathrm{C}\) is simply \(273.15 - 183 = 90.15\ \mathrm{K}\)
  • Likewise, for \(6000^{\circ} \mathrm{C}\), the conversion is \(6000 + 273.15 = 6273.15\ \mathrm{K}\)
  • It is the same straightforward process for any Celsius temperature, making it easy to switch between the two scales.
The ease of this calculation makes it an essential tool for students and scientists alike, allowing them to express temperatures in Kelvin, the standard used in most scientific and engineering applications.
Scientific Measurement
Scientific measurement involves precise and consistent quantification of phenomena, moving beyond everyday observations to more rigorous standards. In the realm of temperature, this means employing scales like the Kelvin scale.

Scientific measurements are crucial for replicability and standardization in research, enabling scientists to communicate findings with accuracy and consistency. This process ensures that results are not only reliable but also comparable across various experiments and studies. The Kelvin scale plays a pivotal role here, as it provides a uniform standard for temperature measurement, making scientific communication clearer and more effective.

Whether it's measuring the surface temperature of the sun (\(6000^{\circ} \mathrm{C} = 6273.15\ \mathrm{K}\)) or the boiling point of helium at \(-269^{\circ} \mathrm{C}\) (\(4.15\ \mathrm{K}\)), employing accurate and standardized measurements is integral to the advancement of science.

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Most popular questions from this chapter

A mixture of neon and argon has a density of \(1.64 \mathrm{~g} \cdot \mathrm{L}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(800.0\) Torr. Compute the ratio of the number of moles of neon to the number of moles of argon in the mixture.

Explosions occur when a substance decomposes rapidly with the production of a large volume of gases. When detonated, TNT (trinitrotoluene), decomposes according to the equation $$ \begin{aligned} 2 \mathrm{C}_{7} \mathrm{H}_{5}\left(\mathrm{NO}_{2}\right)_{3}(s) & \rightarrow \\ 2 \mathrm{C}(s)+12 \mathrm{CO}(g)+5 \mathrm{H}_{2}(g)+3 \mathrm{~N}_{2}(g) \end{aligned} $$ What is the total volume of gases produced from \(1.00\) kilogram of TNT at \(0^{\circ} \mathrm{C}\) and \(1.00\) atm? What pressure is produced if the reaction is confined to a 50 -liter container at \(500^{\circ} \mathrm{C}\) ? Assume that you can use the ideal-gas equation.

About \(50 \%\) of U.S. and most Canadian sulfur is produced by the Claus process, in which sulfur is obtained from the \(\mathrm{H}_{2} \mathrm{~S}(g)\) that occurs in natural gas deposits or is produced when sulfur is removed from petroleum. The reactions are described by the equations (1) \(2 \mathrm{H}_{2} \mathrm{~S}(g)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (2) \(\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightarrow 3 \mathrm{~S}(l)+2 \mathrm{H}_{2} \mathrm{O}(g)\) How many metric tons of sulfur can be produced from 2.00 million liters of \(\mathrm{H}_{2} \mathrm{~S}(g)\) at \(6.00\) bar and \(200.0^{\circ} \mathrm{C}\) ?

Nitroglycerin decomposes according to the equation $$ \begin{array}{r} 4 \mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(s) \rightarrow \\\ 12 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(l)+6 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \end{array} $$ What is the total volume of the gases produced when collected at \(1.00\) bar and \(25^{\circ} \mathrm{C}\) from the decomposition of \(10.0\) grams of nitroglycerin? What pressure is produced if the reaction is confined to a volume of \(0.500\) liters at \(25^{\circ} \mathrm{C} ?\) Assume that you can use the ideal-gas equation. Neglect any pressure due to water vapor.

Sodium peroxide is used in self-contained breathing devices to absorb exhaled carbon dioxide and simultaneously produce oxygen. The equation for the reaction is $$ 2 \mathrm{Na}_{2} \mathrm{O}_{2}(s)+2 \mathrm{CO}_{2}(g) \rightarrow 2 \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{O}_{2}(g) $$ How many liters of \(\mathrm{CO}_{2}(g)\), measured at \(0^{\circ} \mathrm{C}\) and one atm, can be absorbed by one kilogram of \(\mathrm{Na}_{2} \mathrm{O}_{2}(s) ?\) How many liters of \(\mathrm{O}_{2}(g)\) will be produced under the same conditions?

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