Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 1086

The temperature of a substance increases by \(27^{\circ} \mathrm{C}\) What is the value of this increase of Kelvin scale? (A) \(300 \mathrm{~K}\) (B) \(2-46 \mathrm{~K}\) (C) \(7 \mathrm{~K}\) (D) \(27 \mathrm{~K}\)

Short Answer

Expert verified
The value of the temperature increase on the Kelvin scale is \(27 \mathrm{~K}\).

Step by step solution

01

Understand the relationship between Celsius and Kelvin scales

The Celsius and Kelvin scales have the same interval or "size" between their degrees. The difference between them is that the Kelvin scale starts at absolute zero (-273.15°C), while the Celsius scale starts at the freezing point of water (0°C).
02

Convert the temperature increase from Celsius to Kelvin

Since both Celsius and Kelvin scales have the same interval, a temperature change of 27°C is equal to a temperature change of 27K.
03

Identify the correct answer

Comparing our result with the given options, we can see that the correct answer is: (D) \(27 \mathrm{~K}\)

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

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

A Carnot engine operating between temperature \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) has efficiency \(0.4\), when \(\mathrm{T}_{2}\) lowered by $50 \mathrm{~K}\(, its efficiency increases to \)0.5\(. Then \)\mathrm{T}_{1}$ and \(\mathrm{T}_{2}\) are respectively. (A) \(300 \mathrm{~K}\) and \(100 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) and \(200 \mathrm{~K}\) (C) \(600 \mathrm{~K}\) and \(400 \mathrm{~K}\) (D) \(500 \mathrm{~K}\) and \(300 \mathrm{~K}\)

The work of \(62.25 \mathrm{KJ}\) is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by \(5^{\circ} \mathrm{C}\) The gas is $\mathrm{R}=8.3\\{\mathrm{~J} /(\mathrm{mole})\\}$ (A) triatomic (B) diatomic (C) monoatomic (D) a mixture of monoatomic and diatomic

An ideal gas heat engine is operating between \(227^{\circ} \mathrm{C}\) and \(127^{\circ} \mathrm{C}\). It absorbs \(10^{4} \mathrm{~J}\) Of heat at the higher temperature. The amount of heat Converted into. work is \(\ldots \ldots\) J. (A)2000 (B) 4000 (C) 5600 (D) 8000

One mole of a monoatomic gas is heat at a constant pressure of 1 atmosphere from \(0 \mathrm{k}\) to \(100 \mathrm{k}\). If the gas constant $\mathrm{R}=8.32 \mathrm{~J} / \mathrm{mol} \mathrm{k}$ the change in internal energy of the gas is approximate ? (A) \(23 \mathrm{~J}\) (B) \(1.25 \times 10^{3} \mathrm{~J}\) (C) \(8.67 \times 10^{3} \mathrm{~J}\) (D) \(46 \mathrm{~J}\)
See all solutions

Recommended explanations on English Textbooks

English Language Study

Read Explanation

Creative Story

Read Explanation

Syntax

Read Explanation

Essay Plan

Read Explanation

English Grammar Summary

Read Explanation

Essay Prompts

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