Convert the following Celsius temperatures to Kelvin and to Fahrenheit degrees. a. the temperature of someone with a fever, \(39.2^{\circ} \mathrm{C}\) b. a cold wintery day, \(-25^{\circ} \mathrm{C}\) c. the lowest possible temperature, \(-273^{\circ} \mathrm{C}\) d. the melting-point temperature of sodium chloride, \(801^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The converted temperatures are as follows: a. Fever: \(312.35 \ \mathrm{K}\), \(102.56 ^{\circ} \mathrm{F}\) b. Cold wintery day: \(248.15 \ \mathrm{K}\), \(-13 ^{\circ} \mathrm{F}\) c. Lowest possible temperature: \(0.15 \ \mathrm{K}\), \(-459.4 ^{\circ} \mathrm{F}\) d. Melting-point of sodium chloride: \(1074.15 \ \mathrm{K}\), \(1473.8 ^{\circ} \mathrm{F}\)

Step by step solution

01

Convert to Kelvin

Use the formula: \( K = C + 273.15 \) For \(39.2^{\circ} \mathrm{C}\), the Kelvin will be: \(K = 39.2 + 273.15 = 312.35 \) Kelvin
02

Convert to Fahrenheit

Use the formula: \( F = \frac{9}{5}C + 32 \) For \(39.2^{\circ} \mathrm{C}\), the Fahrenheit will be: \(F = \frac{9}{5}(39.2) + 32 = 102.56^{\circ} \mathrm{F}\) For b. a cold wintery day, \(-25^{\circ} \mathrm{C}\)
03

Convert to Kelvin

Use the formula: \( K = C + 273.15 \) For \(-25^{\circ} \mathrm{C}\), the Kelvin will be: \(K = -25 + 273.15 = 248.15 \) Kelvin
04

Convert to Fahrenheit

Use the formula: \( F = \frac{9}{5}C + 32 \) For \(-25^{\circ} \mathrm{C}\), the Fahrenheit will be: \(F = \frac{9}{5}(-25) + 32 = -13^{\circ} \mathrm{F}\) For c. the lowest possible temperature, \( -273^{\circ} \mathrm{C}\)
05

Convert to Kelvin

Use the formula: \( K = C + 273.15 \) For \(-273^{\circ} \mathrm{C}\), the Kelvin will be: \(K = -273 + 273.15 = 0.15 \) Kelvin
06

Convert to Fahrenheit

Use the formula: \( F = \frac{9}{5}C + 32 \) For \(-273^{\circ} \mathrm{C}\), the Fahrenheit will be: \(F = \frac{9}{5}(-273) + 32 = -459.4^{\circ} \mathrm{F}\) For d. the melting-point temperature of sodium chloride, \(801^{\circ} \mathrm{C}\)
07

Convert to Kelvin

Use the formula: \( K = C + 273.15 \) For \(801^{\circ} \mathrm{C}\), the Kelvin will be: \(K = 801 + 273.15 = 1074.15 \) Kelvin
08

Convert to Fahrenheit

Use the formula: \( F = \frac{9}{5}C + 32 \) For \(801^{\circ} \mathrm{C}\), the Fahrenheit will be: \(F = \frac{9}{5}(801) + 32 = 1473.8^{\circ} \mathrm{F}\)

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 Vaia!

Key Concepts

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

Celsius to Kelvin Conversion
The Celsius to Kelvin conversion is essential in various scientific fields as both scales are widely used. To convert temperature from Celsius to Kelvin, you use the simple formula:
\[\begin{equation} K = C + 273.15 \end{equation}\]
This equation highlights that the Kelvin scale and the Celsius scale are offset by 273.15. The Kelvin scale is an absolute temperature scale where zero Kelvin (\[\begin{equation} 0 K \end{equation}\]) is absolute zero, the point at which the fundamental particles of nature have minimal thermal motion.
To understand this concept better, consider a real-life scenario such as the temperature of boiling water. At sea level, water boils at \[\begin{equation}100^{\text{\circ} \text{C}}\end{equation}\]. Using the conversion formula, you end up with a temperature of \[\begin{equation}373.15 K\end{equation}\]. It's important to pay attention to the decimal places and use the formula accurately to ensure precision in scientific calculations.
Celsius to Fahrenheit Conversion
Converting temperatures from Celsius to Fahrenheit may seem complex at first, but it is actually quite straightforward with the correct formula. When you want to convert Celsius to Fahrenheit, you use the following equation:
\[\begin{equation} F = \frac{9}{5}C + 32 \end{equation}\]
This equation takes the Celsius temperature, multiplies it by \[\begin{equation}9/5\end{equation}\] or 1.8, and then adds 32 to adjust for the differences in the starting points of the two scales.
For example, the average human body temperature is approximately \[\begin{equation}37^{\text{\circ} \text{C}}\end{equation}\] which is equivalent to \[\begin{equation}98.6^{\text{\circ} \text{F}}\end{equation}\] when converted. This conversion is particularly useful in everyday life, especially when you are dealing with weather temperature reports or cooking recipes that prefer one scale over the other.
Temperature Scales
Temperature scales provide a consistent way to measure thermal energy or heat. The Celsius (\[\begin{equation}^{\text{\circ} \text{C}}\end{equation}\]), Fahrenheit (\[\begin{equation}^{\text{\circ} \text{F}}\end{equation}\]), and Kelvin (K) scales are the most prevalent. Each of these has its own place and purpose.
The Celsius scale is used worldwide for most temperature measurements. It has a freezing point of water at \[\begin{equation}0^{\text{\circ} \text{C}}\end{equation}\] and a boiling point at \[\begin{equation}100^{\text{\circ} \text{C}}\end{equation}\] at standard atmospheric pressure. In contrast, the Fahrenheit scale is mainly used in the United States and has water freezing at \[\begin{equation}32^{\text{\circ} \text{F}}\end{equation}\] and boiling at \[\begin{equation}212^{\text{\circ} \text{F}}\end{equation}\].
The Kelvin scale is the base unit of temperature in the International System of Units (SI). It is widely used in the scientific community because it is an absolute scale with its zero point at absolute zero, the coldest possible temperature where thermal motion ceases. The different scales serve various purposes, from day-to-day weather forecasts in Celsius or Fahrenheit to scientific research which typically resorts to Kelvin.

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

Evaluate each of the following, and write the answer to the appropriate number of significant figures. a. \(212.2+26.7+402.09\) b. \(1.0028+0.221+0.10337\) c. \(52.331+26.01-0.9981\) d. \(2.01 \times 10^{2}+3.014 \times 10^{3}\) e. \(7.255-6.8350\)

Define the following terms: solid, liquid, gas, pure substance, element, compound, homogeneous mixture, heterogeneous mixture, solution, chemical change, physical change.

The properties of a mixture are typically averages of the properties of its components. The properties of a compound may differ dramatically from the properties of the elements that combine to produce the compound. For each process described below, state whether the material being discussed is most likely a mixture or a compound, and state whether the process is a chemical change or a physical change. a. An orange liquid is distilled, resulting in the collection of a yellow liquid and a red solid. b. A colorless, crystalline solid is decomposed, yielding a pale yellow-green gas and a soft, shiny metal. c. A cup of tea becomes sweeter as sugar is added to it.

A copper wire (density \(=8.96 \mathrm{~g} / \mathrm{cm}^{3}\) ) has a diameter of \(0.25 \mathrm{~mm}\). If a sample of this copper wire has a mass of \(22 \mathrm{~g}\), how long is the wire?

The U.S. trade deficit at the beginning of 2005 was $$\$ 475,000,000$$. If the wealthiest \(1.00 \%\) of the U.S. population \((297,000,000)\) contributed an equal amount of money to bring the trade deficit to $$\$ 0$$, how many dollars would each person contribute? If one of these people were to pay his or her share in nickels only, how many nickels are needed? Another person living abroad at the time decides to pay in pounds sterling ( \(\mathrm{f}\) ). How many pounds sterling does this person contribute (assume a conversion rate of \(1 \mathrm{f}=\$ 1.869)\) ?

See all solutions

Recommended explanations on Chemistry 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.

Sign-up for free