/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 On the moon the surface temperat... [FREE SOLUTION] | 91影视

91影视

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

Short Answer

Expert verified
375 K is 101.85 掳C and 215.33 掳F; 100 K is -173.15 掳C and -279.67 掳F.

Step by step solution

01

Converting Kelvin to Celsius

To convert Kelvin to Celsius, subtract 273.15 from the Kelvin temperature. For 375 K, the Celsius temperature is \(375 - 273.15 = 101.85 \ C\). For \(1.00 \times 10^{2}\ K\), it鈥檚 \(100 - 273.15 = -173.15 \ C\).
02

Converting Celsius to Fahrenheit

To convert Celsius to Fahrenheit, use the formula \(F = C \times \frac{9}{5} + 32\). For 101.85 掳C, degree F is \(101.85 \times \frac{9}{5} + 32 = 215.33 \ F\), and for -173.15 掳C, it's \(-173.15 \times \frac{9}{5} + 32 = -279.67 \ F\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Kelvin to Celsius conversion
The Kelvin and Celsius scales are two temperature measurement systems widely used in scientific contexts. To convert a temperature from Kelvin to Celsius, you need to remember a simple rule: subtract 273.15 from the Kelvin value. For example, if the temperature is 375 K, then the equivalent temperature in Celsius is calculated as follows:\[ \text{Celsius} = 375 - 273.15 = 101.85 \ 掳C \]Similarly, for a cooler temperature of 100 K (expressed as \(1.00 \times 10^{2} \ K\)), the conversion to Celsius would be:\[ \text{Celsius} = 100 - 273.15 = -173.15 \ 掳C \]This easy subtraction makes it straightforward to convert temperatures between the Kelvin and Celsius scales. The Kelvin scale starts at absolute zero, the point where theoretically, molecular motion ceases, making it significantly useful for scientific calculations.
Celsius to Fahrenheit conversion
To change temperatures from Celsius to Fahrenheit, we use a common formula that incorporates a simple multiplication and addition.The formula is: \[ F = C \times \frac{9}{5} + 32 \]Let's apply this formula to convert the temperatures from Celsius to Fahrenheit. If you have a temperature of 101.85 掳C, the conversion would be:\[ F = 101.85 \times \frac{9}{5} + 32 = 215.33 \ 掳F \]For a much colder temperature like -173.15 掳C, it becomes:\[ F = -173.15 \times \frac{9}{5} + 32 = -279.67 \ 掳F \]This formula works because it takes into account the different size units of these two scales - with 180 degrees Fahrenheit covering the same range as 100 degrees Celsius between water's freezing and boiling points.
temperature scales
There are three main temperature scales used today: Kelvin, Celsius, and Fahrenheit. Each scale has its own point of origin and application. - **Kelvin Scale:** - It starts at absolute zero, the theoretical point where all molecular motion stops. - It is primarily used in science due to its relation to absolute temperature, where 0 K signifies the point of having no thermal energy. - No negative numbers, simplifies a lot of thermodynamic calculations. - **Celsius Scale:** - Based on the freezing and boiling points of water, 0 掳C and 100 掳C respectively. - It is widely used around the world for everyday temperature measurements. - **Fahrenheit Scale:** - Common in the United States for day-to-day temperature measurement. - Water freezes at 32 掳F and boils at 212 掳F. Understanding these scales is crucial for converting temperatures, understanding weather reports, and conducting scientific analysis.
moon surface temperature
The moon, unlike Earth, doesn't have an atmosphere to regulate temperatures. Because of this, the surface temperatures vary dramatically between day and night. - During the lunar day, temperatures can soar to around 375 K. Converting this to other scales, it equates to: - 101.85 掳C in Celsius - 215.33 掳F in Fahrenheit - At night, without heat retention from an atmosphere, temperatures can plunge to as low as 100 K. In other units, this is: - -173.15 掳C in Celsius - -279.67 掳F in Fahrenheit These extreme temperature swings make the moon's environment harsh for both human and robotic exploration missions. Understanding how temperatures convert between scales helps scientists plan for these conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An aluminum can is filled to the brim with a liquid. The can and the liquid are heated so their temperatures change by the same amount. The can's initial volume at \(5^{\circ} {C}\) is \(3.5 \times 10^{-4} {m}^{3} .\) The coefficient of volume expansion for aluminum is \(69 \times 10^{-6}({C}^{9})^{-1}\) When the can and the liquid are heated to \(78^{\circ} {C}, 3.6 \times 10^{-6} {m}^{3}\) of liquid spills over. What is the coefficient of volume expansion of the liquid?

A snow maker at a resort pumps 130 kg of lake water per minute and sprays it into the air above a ski run. The water droplets freeze in the air and fall to the ground, forming a layer of snow. If all the water pumped into the air turns to snow, and the snow cools to the ambient air temperature of \(-7.0^{\circ} {C},\) how much heat does the snow-making process release each minute? Assume that the temperature of the lake water is \(12.0^{\circ} \mathrm{C},\) and use \(2.00 \times 10^{3} {J} /({kg} \cdot {C}^{\circ})\) for the specific heat capacity of snow.

A 0.200 -kg piece of aluminum that has a temperature of \(-155^{\circ} {C}\) is added to 1.5 \({kg}\) of water that has a temperature of \(3.0^{\circ} {C}\) . At equilibrium the temperature is \(0.0^{\circ} {C}\) . Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.

An insulated container is partly filled with oil. The lid of the container is removed, 0.125 kg of water heated to \(90.0^{\circ} {C}\) is poured in, and the lid is replaced. As the water and the oil reach equilibrium, the volume of the oil increases by \(1.20 \times 10^{-5} {m}^{3}\) . The density of the oil is 924 \({kg} / {m}^{3}\) , its specific heat capacity is \(1970{J} /({kg} \cdot {C}^{\circ}),\) and its coefficient of volume expansion is \(721 \times 10^{-6}({C}^{0})^{-1}.\) What is the temperature when the oil and the water reach equilibrium?

When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature \((37^{\circ} {C})\) by warming up the water in your stomach. Could drinking ice water, then, substitute for exercise as a way to 鈥渂urn calories?鈥 Suppose you expend 430 kilocalories during a brisk hour-long walk. How many liters of ice water \((0^{\circ} {C})\)) would you have to drink in order to use up 430 kilocalories of metabolic energy? For comparison, the stomach can hold about 1 liter.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.