Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 34

(a) The temperature on a warm summer day is \(87^{\circ} \mathrm{F}\) . What is the temperature in \(^{\circ} \mathrm{C} ?\) (b) Many scientific data are reported at \(25^{\circ} \mathrm{C}\) . What is this temperature in kelvins and in degrees Fahrenheit? (c) Suppose that a recipe calls for an oven temperature of \(400^{\circ} \mathrm{F}\) . Convert this temperature to degrees Celsius and to kelvins. (d) Liquid nitrogen boils at 77 \(\mathrm{K}\) . Convert this temperature to degrees Fahrenheit and to degrees Celsius.

Short Answer

Expert verified
(a) 87°F is approximately \(30.56^{\circ} \mathrm{C}\). (b) 25°C is equivalent to 298.15K and \(77^{\circ} \mathrm{F}\). (c) 400°F is approximately \(204.44^{\circ} \mathrm{C}\) and 477.59K. (d) 77K is equivalent to \(-196.15^{\circ} \mathrm{C}\) and approximately \(-320.87^{\circ} \mathrm{F}\).

Step by step solution

01

(a) Convert 87°F to Celsius

First, apply the Fahrenheit to Celsius formula: \(C = (F - 32) \times \frac{5}{9}\) Substitute F with 87: \(C = (87 - 32) \times \frac{5}{9}\) Now, calculate the Celsius temperature: \(C = 55 \times \frac{5}{9} \approx 30.56^{\circ}\)
02

(b) Convert 25°C to Kelvin and Fahrenheit

Converting Celsius to Kelvin: \(K = C + 273.15\) Substitute C with 25: \(K = 25 + 273.15 = 298.15K\) Converting Celsius to Fahrenheit: \(F = (C \times \frac{9}{5}) + 32\) Substitute C with 25: \(F = (25 \times \frac{9}{5}) + 32 \approx 77^{\circ}\)
03

(c) Convert 400°F to Celsius and Kelvin

Converting Fahrenheit to Celsius: \(C = (F - 32) \times \frac{5}{9}\) Substitute F with 400: \(C = (400 - 32) \times \frac{5}{9} \approx 204.44^{\circ}\) Now convert Celsius to Kelvin: \(K = C + 273.15\) Substitute C with 204.44: \(K = 204.44 + 273.15 = 477.59K\)
04

(d) Convert 77K to Fahrenheit and Celsius

Converting Kelvin to Celsius: \(C = K - 273.15\) Substitute K with 77: \(C = 77 - 273.15 = -196.15^{\circ}\) Now convert Celsius to Fahrenheit: \(F = (C \times \frac{9}{5}) + 32\) Substitute C with -196.15: \(F = (-196.15 \times \frac{9}{5}) + 32 \approx -320.87^{\circ}\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Carry out the following operations and express the answers with the appropriate number of significant figures. $$ \begin{array}{ll}{\text { (a) } 14.3505+2.65} & {\text { (b) } 952.7-140.7389} \\\ {\text { (c) }\left(3.29 \times 10^{4}\right)(0.2501)} & {\text { (d) } 0.0588 / 0.677}\end{array} $$

The total rate at which power is used by humans world wide is approximately 15 TW (terawatts). The solar flux averaged over the sunlit half of Earth is 680 \(\mathrm{W} / \mathrm{m}^{2}\) (assuming no clouds). The area of Earth's disc as seen from the Sun is \(1.28 \times 10^{14} \mathrm{m}^{2} .\) The surface area of Earth is approximately \(197,000,000\) square miles. How much of Earth's surface would we need to cover with solar energy collectors to power the planet for use by all humans? Assume that the solar energy collectors can convert only 10\(\%\) of the available sun light into useful power.

What type of quantity (for example, length, volume, density) do the following units indicate? (a) \(\mathrm{mL},(\mathbf{b}) \mathrm{cm}^{2}\) ,\((\mathbf{c}) \mathrm{mm}^{3},(\mathbf{d}) \mathrm{mg} / \mathrm{L},(\mathbf{e}) \mathrm{ps},(\mathbf{f}) \mathrm{nm},(\mathrm{g}) \mathrm{K}\)

Consider the two spheres shown here, one made of silver and the other of aluminum.(a) What is the mass of each sphere in \(\mathrm{kg}\) ( b) The force of gravity acting on an object is \(F=m g\) where \(m\) is the mass of an object and \(g\) is the acceleration of gravity \(\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right) .\) How much work do you do on each sphere it you raise it from the floor to a height of 2.2 \(\mathrm{m} ?(\mathrm{c})\) Does the act of lifting the sphere off the ground increase the potential energy of the aluminum sphere by a larger, smaller, or same amount as the silver sphere? (d) If you release the spheres simultaneously, they will have the same velocity when they hit the ground. Will they have the same kinetic energy? If not, which sphere will have more kinetic energy? \([\) Section 1.4\(]\)

A 32.65 -g sample of a solid is placed in a flask. Toluene, in which the solid is insoluble, is added to the flask so that the total volume of solid and liquid together is 50.00 \(\mathrm{mL} .\) The solid and toluene together weigh 58.58 \(\mathrm{g}\) . The density of toluene at the temperature of the experiment is 0.864 \(\mathrm{g} / \mathrm{mL}\) . What is the density of the solid?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation
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