/*! 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 25 Convert the following temperatur... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Convert the following temperatures to degrees Celsius or Fahrenheit: (a) \(95^{\circ} \mathrm{F}\), the temperature on a hot summer day; (b) \(12^{\circ} \mathrm{F},\) the temperature on a cold winter day; (c) a \(102^{\circ} \mathrm{F}\) fever; (d) a furnace operating at \(1852^{\circ} \mathrm{F} ;\) (e) \(-273.15^{\circ} \mathrm{C}\) (theoretically the lowest attainable temperature).

Short Answer

Expert verified
(a) \(35^{\circ} C\), (b) \(-11.1^{\circ} C\), (c) \(38.9^{\circ} C\), (d) \(1011.1^{\circ} C\), (e) \(-459.7 ^{\circ} F\)

Step by step solution

01

Convert Fahrenheit to Celsius

For converting Fahrenheit to Celsius, the formula is \[ C = (F - 32) \times \frac{5}{9} \]. Applying the formula, \[ 95^{\circ} F = (95 - 32) \times \frac{5}{9} = 35^{\circ} C\]; \[ 12^{\circ} F = (12 - 32) \times \frac{5}{9} = -11.1^{\circ} C \]; \[ 102^{\circ} F = (102 - 32) \times \frac{5}{9} = 38.9^{\circ} C \]; \[ 1852^{\circ} F = (1852 - 32) \times \frac{5}{9} = 1011.1^{\circ} C \].
02

Convert Celsius to Fahrenheit

For converting Celsius to Fahrenheit, the formula is \[ F = C \times \frac{9}{5} + 32\]. Applying the formula, \[ -273.15^{\circ} C = (-273.15 \times \frac{9}{5}) + 32 = -459.7 ^{\circ} F \]

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 91Ó°ÊÓ!

Key Concepts

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

Fahrenheit to Celsius formula
When converting temperatures from Fahrenheit to Celsius, there's a straightforward formula to guide you. The formula is:
  • \[ C = (F - 32) \times \frac{5}{9} \]
This formula translates a temperature value from Fahrenheit (denoted as \( F \)) to Celsius (denoted as \( C \)).

To use the formula, subtract 32 from the Fahrenheit temperature. This adjustment accounts for the different starting points of the two temperature scales. Then, multiply the result by \( \frac{5}{9} \). This multiplier aligns with the difference between the two scales in terms of the temperature increment between degrees.

For example, if you want to convert \( 95^{\circ} F \), a hot summer day temperature, into Celsius, the calculation would be:
  • \[ 95^{\circ} F = (95 - 32) \times \frac{5}{9} = 35^{\circ} C \]
Celsius to Fahrenheit formula
Converting temperatures from Celsius to Fahrenheit is equally straightforward with its own formula:
  • \[ F = C \times \frac{9}{5} + 32 \]
This formula translates a temperature value from Celsius (\( C \)) to Fahrenheit (\( F \)).

To apply this formula, you first multiply the Celsius temperature by \( \frac{9}{5} \). This factor increases the Celsius value to match the larger Fahrenheit increments.

Then, add 32 to the result. The addition compensates for the differing zero points on the two scales.

Consider converting \( -273.15^{\circ} C \) (the temperature known as absolute zero) to Fahrenheit:
  • \[ -273.15^{\circ} C = (-273.15 \times \frac{9}{5}) + 32 = -459.7^{\circ} F \]
absolute zero
Absolute zero is a fascinating concept in thermodynamics. It represents the theoretical lowest temperature possible.

Measured as \( -273.15^{\circ} C \) or \( -459.7^{\circ} F \), at this temperature, the particles constituting matter would be at their minimal kinetic energy state. In other words, the motion that constitutes heat ceases almost entirely.

Absolute zero is more than a theoretical curiosity; it forms a foundational element of thermodynamics and helps scientists understand how temperature affects the physical properties of materials.
  • It's essential in determining the laws of thermodynamics, particularly the Third Law, which states that entropy approaches a minimum as temperature approaches absolute zero.
  • Though it cannot be reached physically, scientists have achieved temperatures just above absolute zero using advanced cooling techniques.
By digitizing our understanding of absolute zero, learners can appreciate the depth of temperature's role in scientific and practical scenarios.

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

Classify each of the following as an element, a compound, a homogeneous mixture, or a heterogeneous mixture: (a) water from a well, (b) argon gas, (c) sucrose, (d) a bottle of red wine, (e) chicken noodle soup, (f) blood flowing in a capillary, (g) ozone.

How many seconds are there in a solar year (365.24 days)?

Convert the following temperatures to degrees Celsius: (a) \(77 \mathrm{~K}\), the boiling point of liguid nitrogen, (b) \(4.2 \mathrm{~K},\) the boiling point of liquid helium, (c) \(601 \mathrm{~K},\) the melting point of lead.

Percent error is often expressed as the absolute value of the difference between the true value and the experimental value, divided by the true value: percent error \(=\frac{\mid \text { true value }-\text { experimental value } \mid}{\mid \text { true value } \mid} \times 100 \%\) The vertical lines indicate absolute value. Calculate the percent error for the following measurements: (a) The density of alcohol (ethanol) is found to be \(0.802 \mathrm{~g} / \mathrm{mL}\). (True value: \(0.798 \mathrm{~g} / \mathrm{mL}\).) (b) The mass of gold in an earring is analyzed to be \(0.837 \mathrm{~g}\). (True value: \(0.864 \mathrm{~g}\).)

The following procedure was used to determine the volume of a flask. The flask was weighed dry and then filled with water. If the masses of the empty flask and filled flask were \(56.12 \mathrm{~g}\) and \(87.39 \mathrm{~g},\) respectively, and the density of water is \(0.9976 \mathrm{~g} / \mathrm{cm}^{3}\), calculate the volume of the flask in \(\mathrm{cm}^{3}\).

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.