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Chapter 12: Problem 3

On the moon the surface temperature ranges from \(375 \mathrm{~K}\) during the day to \(1.00 \times 10^{2} \mathrm{~K}\) at night. What are these temperatures on the (a) Celsius and (b) Fahrenheit scales?

Short Answer

Expert verified
(a) Day: 101.85 掳C, 215.33 掳F; (b) Night: -173.15 掳C, -279.67 掳F.

Step by step solution

01

Understanding the Temperature Conversion

The problem requires converting temperatures from Kelvin (K) to both Celsius (掳C) and Fahrenheit (掳F). The conversion between these scales involves specific formulas. To convert from Kelvin to Celsius, use the formula: \[ 掳C = K - 273.15 \]
02

Convert Day Temperature from Kelvin to Celsius

Convert the day temperature of 375 K to Celsius using the formula from Step 1.Substitute 375 for K: \[ 掳C = 375 - 273.15 = 101.85 掳C \]
03

Convert Night Temperature from Kelvin to Celsius

Convert the night temperature of \(1.00 \times 10^2\) K to Celsius. This is 100 K.Substitute 100 for K:\[ 掳C = 100 - 273.15 = -173.15 掳C \]
04

Convert Day Temperature from Kelvin to Fahrenheit

Convert 375 K to Fahrenheit using the formula: \[ 掳F = (K - 273.15) \times \frac{9}{5} + 32 \]Substitute 375 for K:\[ 掳F = (375 - 273.15) \times \frac{9}{5} + 32 \]\[ 掳F = 101.85 \times \frac{9}{5} + 32 \]\[ 掳F = 183.33 + 32 = 215.33 掳F \]
05

Convert Night Temperature from Kelvin to Fahrenheit

Convert 100 K to Fahrenheit using the formula:\[ 掳F = (100 - 273.15) \times \frac{9}{5} + 32 \]\[ 掳F = -173.15 \times \frac{9}{5} + 32 \]\[ 掳F = -311.67 + 32 = -279.67 掳F \]
06

Conclusion

The temperatures on the moon are: (a) Day: 101.85 掳C (or 215.33 掳F) (b) Night: -173.15 掳C (or -279.67 掳F)

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Key Concepts

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

Kelvin to Celsius Conversion
To understand the conversion from Kelvin to Celsius, it's important to know what each scale represents. The Kelvin scale is often used in scientific settings because it is an absolute scale. It starts at absolute zero, or 0 K, where all molecular motion stops. The Celsius scale, which is more common in everyday use, is based on the freezing and boiling points of water at sea level, set at 0 掳C and 100 掳C respectively.
  • The conversion from Kelvin to Celsius is straightforward: you simply subtract 273.15 from the Kelvin temperature.
  • The formula for conversion is: \[ 掳C = K - 273.15 \]
    For example, a Kelvin temperature of 375, when converted to Celsius is: \[ 375 - 273.15 = 101.85 掳C \]
  • Similarly, if the temperature is at 100 K: \[ 100 - 273.15 = -173.15 掳C \]
Understanding this conversion helps in interpreting temperatures in scientific and engineering contexts, where Kelvin is predominantly used.
Kelvin to Fahrenheit Conversion
The Fahrenheit scale, commonly used in the United States, is known for its slightly more complex relationship with other temperature scales. The conversion from Kelvin to Fahrenheit needs two steps. You first convert Kelvin to Celsius, and then Celsius to Fahrenheit.
The formula for converting from Kelvin directly to Fahrenheit spells out these steps as:\[ 掳F = (K - 273.15) \times \frac{9}{5} + 32 \] Here's how you can apply this formula:
  • For a day temperature on the moon of 375 K: \[ 掳F = (375 - 273.15) \times \frac{9}{5} + 32 = 215.33 掳F \]
  • For a colder night temperature of 100 K: \[ 掳F = (100 - 273.15) \times \frac{9}{5} + 32 = -279.67 掳F \]
This conversion is useful when working with data that integrates both scientific measurements (Kelvin) with more localized usage (Fahrenheit).
Temperature Scales
Before diving into conversions, it's beneficial to understand the temperature scales commonly encountered in science and daily life. Each scale has its unique reference points and usage.
  • Kelvin: Often used in scientific contexts, Kelvin is the SI unit for temperature. Absolute zero, the point at which there is theoretically no molecular motion, is at 0 K, making Kelvin very useful in physical sciences.
  • Celsius: Widely applied across most of the world, Celsius is based on the properties of water. The scale sets the freezing point at 0 掳C and boiling point at 100 掳C under standard atmospheric conditions.
  • Fahrenheit: Used primarily in the United States, this scale is based on older, less precise references. It sets water's freezing and boiling points at 32 掳F and 212 掳F respectively.
Understanding these scales' applications and points of reference is crucial for effectively dealing with temperature-related problems in various fields. Recognizing these differences helps in interpreting temperatures correctly in both everyday and scientific settings.

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Most popular questions from this chapter

At a temperature of \(0{ }^{\circ} \mathrm{C}\), the mass and volume of a fluid are \(825 \mathrm{~kg}\) and \(1.17 \mathrm{~m}^{3}\). The coefficient of volume expansion is \(1.26 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1}\). (a) What is the density of the fluid at this temperature? (b) What is the density of the fluid when the temperature has risen to \(20.0^{\circ} \mathrm{C}\) ?

A steel ruler is calibrated to read true at \(20.0^{\circ} \mathrm{C}\). A draftsman uses the ruler at \(40.0^{\circ} \mathrm{C}\) to draw a line on a \(40.0^{\circ} \mathrm{C}\) copper plate. As indicated on the warm ruler, the length of the line is \(0.50 \mathrm{~m}\). To what temperature should the plate be cooled, such that the length of the line truly becomes \(0.50 \mathrm{~m}\) ?

A steel bicycle wheel (without the rubber tire) is rotating freely with an angular speed of \(18.00 \mathrm{rad} / \mathrm{s}\). The temperature of the wheel changes from -100.0 to \(+300.0{ }^{\circ} \mathrm{C}\). No net external torque acts on the wheel, and the mass of the spokes is negligible. (a) Does the angular speed increase or decrease as the wheel heats up? Why? (b) What is the angular speed at the higher temperature?

Multiple-Concept Example 4 reviews the concepts that are involved in this problem. A ruler is accurate when the temperature is \(25^{\circ} \mathrm{C}\). When the temperature drops to \(-14^{\circ} \mathrm{C}\), the ruler shrinks and no longer measures distances accurately. However, the ruler can be made to read correctly if a force of magnitude \(1.2 \times 10^{3} \mathrm{~N}\) is applied to each end so as to stretch it back to its original length. The ruler has a cross-sectional area of \(1.6 \times 10^{-5} \mathrm{~m}^{2},\) and it is made from a material whose coefficient of linear expansion is \(2.5 \times 10^{-5}\left(\mathrm{C}^{0}\right)^{-1}\). What is Young's modulus for the material from which the ruler is made?

At a temperature of \(0^{\circ} \mathrm{C}\), the mass and volume of a fluid are \(825 \mathrm{~kg}\) and \(1.17 \mathrm{~m}^{3}\). The coefficient of volume expansion is \(1.26 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1}\) (a) What is the density of the fluid at this temperature? (b) What is the density of the fluid when the temperature has risen to \(20.0^{\circ} \mathrm{C} ?\)
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