Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The temperature on Celsius scale is \(25^{\circ} \mathrm{C}\). What is the corresponding temperature on the Fahrenheit Scale? (A) \(40^{\circ} \mathrm{F}\) (B) \(45^{\circ} \mathrm{F}\) (C) \(50^{\circ} \mathrm{F}\) (D) \(77^{\circ} \mathrm{F}\)

Short Answer

Expert verified
The corresponding temperature on the Fahrenheit scale is \(77^{\circ} \mathrm{F}\). (D)

Step by step solution

01

Write down the formula to convert Celsius to Fahrenheit

We begin by writing down the formula needed for this conversion: \(F = \frac{9}{5}C + 32\)
02

Substitute the given value for C

Given that the temperature in Celsius is 25ºC, we replace C with 25 in the formula: \(F = \frac{9}{5}(25) + 32\)
03

Calculate the temperature in Fahrenheit

Now, we perform the calculations to find the temperature in Fahrenheit: \(F = \frac{9}{5}(25) + 32 = \frac{9(25)}{5} + 32 = 45 + 32\) \(F = 77\)
04

Find the corresponding answer

With the knowledge that the temperature in Fahrenheit is 77ºF, we look at the answer choices and find that: (D) \(77^{\circ} \mathrm{F}\) So, the correct answer for the problem is choice (D).

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!

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

The temperature of sink of Carnot engine is \(27^{\circ} \mathrm{C}\). Efficiency of engine is \(25 \%\) Then find the temperature of source. (A) \(227^{\circ} \mathrm{C}\) (B) \(327^{\circ} \mathrm{C}\) (C) \(27^{\circ} \mathrm{C}\) (D) \(127^{\circ} \mathrm{C}\)

In an isothermal reversible expansion, if the volume of \(96 \mathrm{~J}\) of oxygen at \(27^{\circ} \mathrm{C}\) is increased from 70 liter to 140 liter, then the work done by the gas will be (A) \(300 \mathrm{R} \log _{\mathrm{e}}^{(2)}\) (B) \(81 \mathrm{R} \log _{\mathrm{e}}^{(2)}\) (C) \(2.3 \times 900 \mathrm{R} \log _{10} 2\) (D) \(100 \mathrm{R} \log _{10}^{(2)}\)

A gas mixture consists of 2 mole of oxygen and 4 mole of argon at temperature \(\mathrm{T}\). Neglecting all vibrational modes, the total internal energy of the system is (A) \(11 \mathrm{RT}\) (B) \(9 \mathrm{RT}\) (C) \(15 \mathrm{RT}\) (D) \(4 \mathrm{RT}\)

An ideal gas at \(27 \mathrm{C}\) is Compressed adiabatically, to \(\\{8 / 27\\}\) of its original Volume. If \(\mathrm{v}=(5 / 3)\), then the rise in temperature is (A) \(225 \mathrm{k}\) (B) \(450 \mathrm{~K}\) (C) \(375 \mathrm{~K}\) (D) \(405 \mathrm{~K}\)

70 calorie of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from \(30^{\circ} \mathrm{C}\) to $35^{\circ} \mathrm{C}$ The amount of heat required to raise the temperature of the same gas through the same range at constant volume is $\ldots \ldots \ldots \ldots \ldots .$ calorie. (A) 50 (B) 30 (C) 70 (D) 90

See all solutions

Recommended explanations on English 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