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Chapter 17: Problem 6

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

Short Answer

Expert verified
(a) 126.85掳C, 260.33掳F; (b) -178.15掳C, -288.67掳F; (c) 1.549972685脳10鈦仿癈, 2.78995083脳10鈦仿癋.

Step by step solution

01

Convert Kelvin to Celsius

To convert Kelvin to Celsius, use the formula \( C = K - 273.15 \). We have three temperatures to convert: (a) For the moon, \( K = 400 \). Thus, \[ C = 400 - 273.15 = 126.85 \](b) For Saturn's clouds, \( K = 95 \). Thus, \[ C = 95 - 273.15 = -178.15 \](c) For the Sun's center, \( K = 1.55 \times 10^7 \). Thus, \[ C = 1.55 \times 10^7 - 273.15 = 1.549972685 \times 10^7 \].
02

Convert Celsius to Fahrenheit

To convert Celsius to Fahrenheit, use the formula \( F = \frac{9}{5}C + 32 \). Calculate for each Celsius temperature from Step 1:(a) For the moon, \( C = 126.85 \): \[ F = \frac{9}{5} \times 126.85 + 32 = 260.33 \](b) For Saturn's clouds, \( C = -178.15 \): \[ F = \frac{9}{5} \times (-178.15) + 32 = -288.67 \](c) For the Sun's center, \( C = 1.549972685 \times 10^7 \): \[ F = \frac{9}{5} \times 1.549972685 \times 10^7 + 32 \approx 2.78995083 \times 10^7 \].

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Key Concepts

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

Kelvin to Celsius
Understanding how to convert temperatures from Kelvin to Celsius is essential for various scientific disciplines. The Kelvin scale is an absolute temperature scale, frequently used in scientific contexts, where 0 K is the absolute zero鈥攖he point at which molecular motion ceases. To convert Kelvin to Celsius, you utilize a simple subtraction formula:
  • Formula: \(C = K - 273.15\)
  • Example: For a temperature of 400 K, the conversion would be \(400 - 273.15 = 126.85 \degree C\).
The Kelvin to Celsius conversion is straightforward because both scales increment by the same amount for each degree. The only difference is the starting point, with Celsius beginning at the freezing point of water and Kelvin starting at absolute zero. Remember this simple conversion formula, and you'll easily navigate between these two scales in any scientific context.
Celsius to Fahrenheit
The Celsius to Fahrenheit conversion is crucial for day-to-day weather understanding, as Fahrenheit is primarily used in the United States. Converting Celsius to Fahrenheit involves a multiplication and then an addition:
  • Formula: \(F = \frac{9}{5}C + 32\)
  • Example: For a temperature of 126.85 掳C, the calculation is \(F = \frac{9}{5} \times 126.85 + 32 = 260.33 \degree F\).
This method transforms Celsius into Fahrenheit by expanding the Celsius scale to match Fahrenheit's larger scale units, and then shifting by 32 degrees to align with the freezing point of water in Fahrenheit. This conversion is vital for accurate communication of temperatures across regions where different units are used.
Temperature Scales
Understanding temperature scales is fundamental to measuring and comparing temperatures in various contexts. Three primary scales are Kelvin, Celsius, and Fahrenheit, each defined by different zero points and increments:
  • Kelvin Scale: Starts at absolute zero (0 K), where no thermal energy remains. It is used in scientific calculations for its direct relation to thermodynamic laws.
  • Celsius Scale: Sets zero at the freezing point of water and 100 at the boiling point, making it ideal for everyday use and scientific research involving water-based processes.
  • Fahrenheit Scale: Uses 32 掳F as the freezing point of water and 212 掳F as the boiling point, typically used in the United States for meteorological purposes.
Each scale has its applications, and knowing how and when to use these scales allows for effective communication and understanding in both scientific and everyday contexts.

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Most popular questions from this chapter

Debye's \(T^{3}\) Law. At very low temperatures the molar heat capacity of rock salt varies with temperature according to Debye's \(T^{3}\) law: $$C=k \frac{T^{3}}{\Theta^{3}}$$ where \(k=1940 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) and \(\Theta=281 \mathrm{K}\) . (a) How much heat is required to raise the temperature of 1.50 \(\mathrm{mol}\) of rock salt from 10.0 \(\mathrm{K}\) to 40.0 \(\mathrm{K} ?(\) Hint: Use Eq. \((17.18)\) in the form \(d Q=n C d T\) and integrate.) (b) What is the average molar heat capacity in this range? (c) What is the true molar heat capacity at 40.0 \(\mathrm{K} ?\)

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

A U.S. penny has a diameter of 1.9000 \(\mathrm{cm}\) at \(20.0^{\circ} \mathrm{C} .\) The coin is made of a metal alloy (mostly zinc) for which the coefficient of linear expansion is \(2.6 \times 10^{-5} \mathrm{K}^{-1} .\) What would its diameter be on a hot day in Death Valley \(\left(48.0^{\circ} \mathrm{C}\right) ?\) On a cold night in the mountains of Greenland \(\left(-53^{\circ} \mathrm{C}\right) ?\)

A \(25,000-\mathrm{kg}\) subway train initially traveling at 15.5 \(\mathrm{m} / \mathrm{s}\) slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are 65.0 \(\mathrm{m}\) long by 20.0 \(\mathrm{m}\) wide by 12.0 \(\mathrm{m}\) high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its specific heat to be 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\).

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