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Chapter 17: Problem 10

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

Short Answer

Expert verified
(a) 126.85掳C, 260.33掳F; (b) -178.15掳C, -288.67掳F; (c) 1.54999728脳10鈦仿癈, 2.789995104脳10鈦仿癋.

Step by step solution

01

Convert from Kelvin to Celsius

To convert a temperature from Kelvin (K) to Celsius (掳C), use the formula: \[ 掳C = K - 273.15 \]. Let's apply this formula to the given temperatures: (a) 400 K, (b) 95 K, (c) 1.55 \times 10^7 K.(a) \[400 \, \mathrm{K} - 273.15 = 126.85 \, ^{\circ} \mathrm{C}\](b) \[95 \, \mathrm{K} - 273.15 = -178.15 \, ^{\circ} \mathrm{C}\](c) \[1.55 \times 10^7 \, \mathrm{K} - 273.15 = 1.54999728 \times 10^7 \, ^{\circ} \mathrm{C}\]
02

Convert from Celsius to Fahrenheit

To convert from Celsius (掳C) to Fahrenheit (掳F), use the formula: \[ 掳F = 掳C \times \frac{9}{5} + 32 \]. Apply this to the Celsius values found in Step 1:(a) \[126.85 \, ^{\circ} \mathrm{C} \times \frac{9}{5} + 32 = 260.33 \, ^{\circ} \mathrm{F} \](b) \[-178.15 \, ^{\circ} \mathrm{C} \times \frac{9}{5} + 32 = -288.67 \, ^{\circ} \mathrm{F} \](c) \[1.54999728 \times 10^7 \, ^{\circ} \mathrm{C} \times \frac{9}{5} + 32 = 2.789995104 \times 10^7 \, ^{\circ} \mathrm{F} \]

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Key Concepts

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

Kelvin to Celsius
The Kelvin and Celsius temperature scales are both widely used in scientific contexts. The Kelvin scale is the absolute temperature scale and has its zero point at absolute zero. To convert a temperature from Kelvin to Celsius, you need to know a simple formula: \[ 掳C = K - 273.15 \]
  • Subtract 273.15 from the Kelvin temperature to get the Celsius equivalent.
  • This conversion is straightforward as both scales use the same size for their degrees.
For example, in our exercise, we converted 400 K to Celsius:
  • 400 K - 273.15 gives us 126.85 掳C, equivalent to the moon's midday temperature.
  • 95 K, representing the temperature at Saturn's cloud tops, becomes -178.15 掳C.
  • Finally, the extremely high temperature at the sun's center, 1.55 \times 10^7 K, converts to approximately 1.55 \times 10^7 掳C.
Remember, this formula is essential for many scientific calculations that require temperature measures in Celsius.
Celsius to Fahrenheit
Once you have temperatures in Celsius, converting them to Fahrenheit involves another simple formula. Fahrenheit is often used in everyday settings, especially in the United States. The conversion formula between Celsius and Fahrenheit is: \[ 掳F = 掳C \times \frac{9}{5} + 32 \]
  • To convert, multiply the Celsius value by \( \frac{9}{5} \) and then add 32 to the result.
  • This formula reflects the different sizes of degrees and zero points in the two scales.
Applying this formula to our converted Celsius values gives us temperatures in Fahrenheit:
  • 126.85 掳C becomes approximately 260.33 掳F for the lunar surface temperature.
  • -178.15 掳C converts to about -288.67 掳F at Saturn's cloud tops.
  • 1.55 \times 10^7 掳C results in roughly 2.79 \times 10^7 掳F for the sun's core.
Understanding these calculations is helpful for translating temperature data across different scales used in various fields.
Temperature Scales
To appreciate the conversions, it's crucial to understand the concept of temperature scales. Different scales are used depending on the context and geographic location:
  • Kelvin: The Kelvin scale is used primarily in science, particularly for thermodynamic calculations. It starts at absolute zero, where all molecular movement stops.
  • Celsius: Widely used in most parts of the world for everyday temperature measurement. It sets the freezing and boiling points of water at 0 掳C and 100 掳C, respectively.
  • Fahrenheit: Used primarily in the United States for weather forecasts and other applications. It places the freezing point of water at 32 掳F and the boiling point at 212 掳F.
Each scale has its advantages and common usage areas. Kelvin's advantage is the directness it offers in scientific research, which is often necessary when dealing with very low or very high temperatures. Meanwhile, Celsius and Fahrenheit provide practical scalability for daily weather and environmental observations. Understanding these differences is key to properly converting and using temperatures across these scales.

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Most popular questions from this chapter

A tube leads from a \(0.150-\mathrm{kg}\) calorimeter to a flask in which water is boiling under atmospheric pressure. The calorimeter has specific heat capacity 420 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , and it originally contains 0.340 \(\mathrm{kg}\) of water at \(15.0^{\circ} \mathrm{C}\) . Steam is allowed to condense in the calorimeter at atmospheric pressure until the temperature of the calorimeter and contents reaches \(71.0^{\circ} \mathrm{C}\) , at which point the total mass of the calorimeter and its contents is found to be 0.525 \(\mathrm{kg}\) . Compute the heat of vaporization of water from these data.

One end of an insulated metal rod is maintained at \(100.0^{\circ} \mathrm{C}\) and the other end is maintained at \(0.00^{\circ} \mathrm{C}\) by an ice-water mixture. The rod is 60.0 \(\mathrm{cm}\) long and has a cross-sectional area of 1.25 \(\mathrm{cm}^{2}\) . The heat conducted by the rod melts 8.50 \(\mathrm{g}\) of ic. 0 min. Find the thermal conductivity \(k\) of the metal.

(a) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.

A laboratory technician drops a \(0.0850-\mathrm{kg}\) sample of unknown material, at a temperature of \(100.0^{\circ} \mathrm{C}\) , into a calorimeter. The calorimeter can, initially at \(19.0^{\circ} \mathrm{C}\) , is made of 0.150 \(\mathrm{kg}\) of copper and contains 0.200 \(\mathrm{kg}\) of water. The final temperature of the calorimeter can and contents is \(26.1^{\circ} \mathrm{C}\) . Compute the specific heat capacity of the sample.

A copper calorimeter can with mass 0.100 kg contains 0.160 \(\mathrm{kg}\) of water and 0.0180 \(\mathrm{kg}\) of ice in thermal equilibrium at atmospheric pressure. If 0.750 \(\mathrm{kg}\) of lead at a temperature of \(255^{\circ} \mathrm{C}\) is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.
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