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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 \((400 \mathrm{~K}) ;\) (b) the temperature at the tops of the clouds in the atmosphere of Saturn \((95 \mathrm{~K}) ;\) (c) the temperature at the center of the sun \(\left(1.55 \times 10^{7} \mathrm{~K}\right)\)

Short Answer

Expert verified
400 K = 126.85掳C, 260.33掳F; 95 K = -178.15掳C, -288.67掳F; 1.55 脳 10鈦 K = 1.55 脳 10鈦仿癈, 2.79 脳 10鈦仿癋.

Step by step solution

01

Understand the Kelvin to Celsius Conversion Formula

To convert from Kelvin (K) to Celsius (掳C), you need to subtract 273.15 from the Kelvin temperature: \(掳C = K - 273.15\). We'll use this formula to convert each given Kelvin temperature into Celsius.
02

Convert the Midday Temperature on the Moon (400 K) to Celsius

Subtract 273.15 from 400 K to find the Celsius temperature: \[掳C = 400 - 273.15 = 126.85掳C.\] So, the midday temperature on the moon is 126.85掳C.
03

Convert the Midday Temperature on the Moon (400 K) to Fahrenheit

To convert Celsius to Fahrenheit, use the formula \( 掳F = 掳C \times \frac{9}{5} + 32\). Substitute the Celsius temperature we found:\[掳F = 126.85 \times \frac{9}{5} + 32 = 260.33掳F.\] Thus, the midday temperature on the moon is approximately 260.33掳F.
04

Convert the Temperature at the Tops of the Clouds on Saturn (95 K) to Celsius

Subtract 273.15 from 95 K to find the Celsius temperature: \[掳C = 95 - 273.15 = -178.15掳C.\] Thus, the temperature at the tops of the clouds in Saturn's atmosphere is -178.15掳C.
05

Convert the Temperature at the Tops of the Clouds on Saturn (95 K) to Fahrenheit

Use the formula \( 掳F = 掳C \times \frac{9}{5} + 32\) with the Celsius temperature:\[掳F = -178.15 \times \frac{9}{5} + 32 = -288.67掳F.\] So, the temperature at the tops of Saturn's clouds is approximately -288.67掳F.
06

Convert the Temperature at the Center of the Sun (\(1.55 \times 10^7\ K\)) to Celsius

Subtract 273.15 from \(1.55 \times 10^7\ K\) to find the Celsius temperature: \[掳C = 1.55 \times 10^7 - 273.15 = 1.549972685 \times 10^7掳C.\] Thus, the temperature at the center of the sun is approximately \(1.549972685 \times 10^7掳C\).
07

Convert the Temperature at the Center of the Sun (\(1.55 \times 10^7\ K\)) to Fahrenheit

Use the formula \( 掳F = 掳C \times \frac{9}{5} + 32\):\[掳F = 1.549972685 \times 10^7 \times \frac{9}{5} + 32 = 2.789950833 \times 10^7掳F.\] Therefore, the temperature at the center of the sun is approximately \(2.789950833 \times 10^7掳F\).

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Key Concepts

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

Kelvin to Celsius
Converting temperature from Kelvin to Celsius is quite straightforward. The Kelvin scale is an absolute temperature scale starting at absolute zero, which is the point where all molecular motion theoretically stops. The Celsius scale, on the other hand, is a relative scale that sets the freezing point of water at 0 degrees and the boiling point at 100 degrees. This makes the two scales directly related.

To convert a temperature from Kelvin to Celsius, you simply subtract 273.15 from the Kelvin temperature. This is because the zero point on the Kelvin scale is 273.15 degrees lower than on the Celsius scale.
  • The formula is: \[ 掳C = K - 273.15 \]
For example, if you have a temperature of 400 K, the corresponding Celsius temperature would be \[ 400 - 273.15 = 126.85掳C \].

Understanding this conversion is useful in various scientific fields where absolute temperature measurements are needed.
Kelvin to Fahrenheit
The Kelvin to Fahrenheit conversion involves a two-step process. This is because the Celsius scale is typically used as an intermediary step between Kelvin and Fahrenheit. This helps bridge the difference between the absolute scale of Kelvin and the relative scale of Fahrenheit more smoothly.

First, convert Kelvin to Celsius using the formula:
  • \[ 掳C = K - 273.15 \]
Once you have the Celsius temperature, you can convert it to Fahrenheit using the Celsius to Fahrenheit conversion formula:
  • \[ 掳F = 掳C \times \frac{9}{5} + 32 \]
For example, if you start with a temperature of 95 K, you convert it first to Celsius:
\[ 掳C = 95 - 273.15 = -178.15掳C \].
Then convert Celsius to Fahrenheit: \[ 掳F = -178.15 \times \frac{9}{5} + 32 = -288.67掳F \].

This two-step conversion is essential when working in fields that require precise temperature control or reporting.
Celsius to Fahrenheit
Converting Celsius to Fahrenheit is a common task, especially in countries that use the imperial system for weather forecasts. Celsius is part of the metric system, commonly used in scientific contexts worldwide.

To convert a temperature from Celsius to Fahrenheit, you multiply the Celsius temperature by \( \frac{9}{5} \) and then add 32. This formula accounts for both the difference in scale size and the offset between their zero points:
  • \[ 掳F = 掳C \times \frac{9}{5} + 32 \]
For instance, if the midday temperature on the moon is 126.85掳C, then in Fahrenheit it would be calculated as: \[ 掳F = 126.85 \times \frac{9}{5} + 32 = 260.33掳F \].

This conversion formula can help you interchange Celsius and Fahrenheit temperatures, ensuring you understand and communicate efficiently across different temperature measures.

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Most popular questions from this chapter

A slab of a thermal insulator with a cross-sectional area of \(100 \mathrm{~cm}^{2}\) is \(3.00 \mathrm{~cm}\) thick. Its thermal conductivity is \(0.075 \mathrm{~W} /(\mathrm{m} \cdot \mathrm{K})\). If the temperature difference between opposite faces is \(80 \mathrm{C}^{\circ},\) how much heat flows through the slab in 1 day?

A worker pours \(1.250 \mathrm{~kg}\) of molten lead at a temperature of \(327.3^{\circ} \mathrm{C}\) into \(0.5000 \mathrm{~kg}\) of water at a temperature of \(75.00^{\circ} \mathrm{C}\) in an insulated bucket of negligible mass. Assuming no heat loss to the surroundings, calculate the mass of lead and water remaining in the bucket when the materials have reached thermal equilibrium.

In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is \(-20^{\circ} \mathrm{C},\) what is the amount of heat needed to warm to internal body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the \(0.50 \mathrm{~L}\) of air exchanged with each breath? Assume that the specific heat of air is \(1020 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K})\) and that \(1.0 \mathrm{~L}\) of air has a mass of \(1.3 \mathrm{~g}\). (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

for this temperature? (b) Elevated body temperature. During very vigorous exercise, the body's temperature can go as high as \(40^{\circ} \mathrm{C}\). What would Kelvin and Fahrenheit thermometers read for this temperature? (c) Temperature difference in the body. The surface temperature of the body is normally about \(7 \mathrm{C}^{\circ}\) lower than the internal temperature. Express this temperature difference in kelvins and in Fahrenheit degrees. (d) Blood storage. Blood stored at \(4.0^{\circ} \mathrm{C}\) lasts safely for about 3 weeks, whereas blood stored at \(-160^{\circ} \mathrm{C}\) lasts for 5 years. Express both temperatures on the Fahrenheit and Kelvin scales. (e) Heat stroke. If the body's temperature is above \(105^{\circ} \mathrm{F}\) for a prolonged period, heat stroke can result. Express this temperature on the Celsius and Kelvin scales.

What is the amount of heat entering your skin when it receives the heat released (a) by \(25.0 \mathrm{~g}\) of steam initially at \(100.0^{\circ} \mathrm{C}\) that cools to \(34.0^{\circ} \mathrm{C} ?\) (b) by \(25.0 \mathrm{~g}\) of water initially at \(100.0^{\circ} \mathrm{C}\) that cools to \(34.0^{\circ} \mathrm{C} ?\) (c) What do these results tell you about the relative severity of steam and hot-water bums?

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