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Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales: (a) the midday temperature at the surface of the moon (400 K); (b) the temperature at the tops of the clouds in the atmosphere of Saturn (95 K); (c) the temperature at the center of the sun \((1.55 \times 10{^7} K)\).

Short Answer

Expert verified
400 K: 126.85 °C, 260.33 °F; 95 K: -178.15 °C, -288.67 °F; Sun's core: 1.5499973 x 10⁷ °C, 2.789995 x 10⁷ °F.

Step by step solution

01

Understanding Conversion Formulas

To convert Kelvin (K) to Celsius (°C), use the formula: \( °C = K - 273.15 \). To convert Kelvin (K) to Fahrenheit (°F), use the formula: \( °F = (K - 273.15) \times \frac{9}{5} + 32 \).
02

Convert 400 K to Celsius

Using the formula \( °C = K - 273.15 \), calculate: \( 400 - 273.15 = 126.85 \ °C \).
03

Convert 400 K to Fahrenheit

Using the formula \( °F = (400 - 273.15) \times \frac{9}{5} + 32 \), calculate: \( 126.85 \times \frac{9}{5} + 32 = 260.33 \ °F \).
04

Convert 95 K to Celsius

Using the formula \( °C = K - 273.15 \), calculate: \( 95 - 273.15 = -178.15 \ °C \).
05

Convert 95 K to Fahrenheit

Using the formula \( °F = (95 - 273.15) \times \frac{9}{5} + 32 \), calculate: \( -178.15 \times \frac{9}{5} + 32 = -288.67 \ °F \).
06

Convert \(1.55 \times 10^7\) K to Celsius

Using the formula \( °C = K - 273.15 \), calculate: \( 1.55 \times 10^7 - 273.15 = 1.5499973 \times 10^7 \ °C \).
07

Convert \(1.55 \times 10^7\) K to Fahrenheit

Using the formula \( °F = (1.55 \times 10^7 - 273.15) \times \frac{9}{5} + 32 \), calculate: \( 1.5499973 \times 10^7 \times \frac{9}{5} + 32 = 2.789995 \times 10^7 \ °F \).

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Key Concepts

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

Kelvin to Celsius
Converting temperatures from Kelvin to Celsius is quite straightforward and follows a simple formula. The Kelvin and Celsius scales are directly related with a linear offset.
To convert a temperature from Kelvin to Celsius, just subtract 273.15 from the Kelvin temperature:
  • Formula: \( °C = K - 273.15 \)
  • This formula arises because 0 °C is equivalent to 273.15 K. The 273.15 is the freezing point of water in Kelvin, serving as the reference point between these scales.
This simplicity is due to Kelvin being simply an absolute scale of Celsius. Therefore, an increase or decrease in one degree Kelvin is exactly the same change in degrees as on the Celsius scale. For everyday applications, Kelvin is typically used in scientific contexts where absolute temperatures are needed, while Celsius is more common in everyday life.
For example, if you're converting the moon's midday temperature at 400 K, the calculation is:
  • \( 400 - 273.15 = 126.85 \ \degree C \)
Always remember, when dealing with temperature conversions, the precise subtraction of 273.15 is crucial.
Kelvin to Fahrenheit
While converting from Kelvin to Fahrenheit involves a bit more math than converting to Celsius, it's still quite manageable with a methodical approach. Fahrenheit, unlike Celsius, does not directly align with Kelvin through a straightforward addition or subtraction due to its differing scale intervals.
Here's the step-by-step formula:
  • First, convert Kelvin to Celsius: \( K - 273.15 \).
  • Next, convert the Celsius temperature to Fahrenheit using: \( °F = (°C \times \frac{9}{5}) + 32 \).
  • Combine these into a single formula: \( °F = (K - 273.15) \times \frac{9}{5} + 32 \).
Fahrenheit's different starting and end points (freezing point of water at 32 °F and boiling at 212 °F) make these conversions slightly more involved. For example, converting the midday temperature on the moon (400 K):
  • \( °F = (400 - 273.15) \times \frac{9}{5} + 32 = 260.33 \ \degree F \)
Handling each step carefully ensures accurate conversion outcomes.
Temperature scales
Understanding temperature scales is essential for interpreting different temperature readings across scientific fields and everyday contexts. Primarily, three temperature scales are widely used: Kelvin, Celsius, and Fahrenheit.
  • Kelvin: Absolute thermodynamic scale. It starts from absolute zero, the point where molecular motion stops, set at 0 K. Kelvin numbers share the same step increment as Celsius, making direct conversions straightforward.
  • Celsius: Based on the freezing and boiling points of water, 0 °C and 100 °C respectively. Predominantly used globally for weather and most routine temperatures.
  • Fahrenheit: Commonplace in the United States. It defines 32 °F as the freezing point of water and 212 °F as its boiling point. As such, smaller temperature intervals make changes appear more significant compared to Celsius or Kelvin.
Understanding these scales involves recognizing their origins and practical applications in differing contexts. Scientists often use Kelvin for precise measurements, essential in fields like astronomy and physics. Celsius is well-suited for daily weather reporting and various regional applications, whereas Fahrenheit, mainly used in the U.S., reflects a cultural preference rather than scientific necessity. Through mastering these scales, you can accurately interpret and convert temperatures across disciplines.

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Most popular questions from this chapter

A copper calorimeter can with mass 0.446 kg contains 0.0950 kg of ice. The system is initially at 0.0\(^\circ\)C. (a) If 0.0350 kg of steam at 100.0\(^\circ\)C and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

A metal sphere with radius 3.20 cm is suspended in a large metal box with interior walls that are maintained at 30.0\(^\circ\)C. A small electric heater is embedded in the sphere. Heat energy must be supplied to the sphere at the rate of 0.660 J/s to maintain the sphere at a constant temperature of 41.0\(^\circ\)C. (a) What is the emissivity of the metal sphere? (b) What power input to the sphere is required to maintain it at 82.0\(^\circ\)C? What is the ratio of the power required for 82.0\(^\circ\)C to the power required for 41.0\(^\circ\)C? How does this ratio compare with 2\(^4\)? Explain.

Convert the following Celsius temperatures to Fahrenheit: (a) -62.8\(^\circ\)C, the lowest temperature ever recorded in North America (February 3, 1947, Snag, Yukon); (b) 56.7\(^\circ\)C, the highest temperature ever recorded in the United States (July 10, 1913, Death Valley, California); (c) 31.1\(^\circ\)C, the world’s highest average annual temperature (Lugh Ferrandi, Somalia).

The molar heat capacity of a certain substance varies with temperature according to the empirical equation $$C = 29.5 J/mol \cdot K + (8.20 \times 10{^-}{^3} J/mol \cdot K{^2})T$$ How much heat is necessary to change the temperature of 3.00 mol of this substance from 27\(^\circ\)C to 227\(^\circ\)C? (Hint: Use Eq. (17.18) in the form d\(Q\) = n\(C\) d\(T\) and integrate.)

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume e = 1 for these surfaces. Find the radii of the following stars (assumed to be spherical): (a) Rigel, the bright blue star in the constellation Orion, which radiates energy at a rate of \(2.7 \times 10{^3}{^2} W\) and has surface temperature 11,000 K; (b) Procyon B (visible only using a telescope), which radiates energy at a rate of \(2.1 \times 10{^2}{^3} W\) and has surface temperature 10,000 K. (c) Compare your answers to the radius of the earth, the radius of the sun, and the distance between the earth and the sun. (Rigel is an example of a supergiant star, and Procyon B is an example of a white dwarf star.)

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