Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 1183

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 \(300 \mathrm{~K}\)

Short Answer

Expert verified
The temperatures T1 and T2 are 500 K and 300 K, respectively (Option D).

Step by step solution

01

Write down the formula for Carnot engine efficiency

For a Carnot engine operating between two temperatures T1 (higher) and T2 (lower), the efficiency is given by: Efficiency = \(1 - \frac{T_2}{T_1}\)
02

Equate efficiencies

Given efficiency 1 = 0.4, and efficiency 2 = 0.5. We also know that in the second case, T2 is lowered by 50 K. Efficiency 1 = \(1 - \frac{T_2}{T_1}\) = 0.4 And Efficiency 2 = \(1 - \frac{T_2 - 50}{T_1}\) = 0.5
03

Solve for T1 and T2

Now, we need to solve the two equations to find T1 and T2. From Efficiency 1 equation, we get: T2 = T1 (1 - 0.4) = 0.6 * T1 From Efficiency 2 equation, we get: T2 - 50 = T1 (1 - 0.5) = 0.5 * T1 Substituting T2 from the first equation into the second equation: 0.6 * T1 - 50 = 0.5 * T1 Solving for T1: 0.1 * T1 = 50 T1 = 500 K Now, we can find T2 using T1: T2 = 0.6 * T1 = 0.6 * 500 = 300 K
04

Compare with given options

Comparing our found values of T1 and T2 with the given options, we can conclude the correct answer is: (D) T1 = \(500 \mathrm{~K}\) and T2 = \(300 \mathrm{~K}\)

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

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) \(27 \mathrm{~K}\)

Which of the following is not a thermodynamic co-ordinate. (A) \(\mathrm{R}\) (B) \(\mathrm{P}\) (C) \(\mathrm{T}\) (D) \(\mathrm{V}\)

In Column I different Process is given match corresponding option of column \(\mathrm{I}_{1}\) Column - I Columm - II (A) adiabatic process (p) \(\Delta \mathrm{p}=0\) (B) Isobaric process (q) \(\Delta \mathrm{u}=0\) (C) Isochroic process (r) \(\Delta Q=0\) (D) Isothermal process (s) \(\Delta \mathrm{W}=0\) (A) A-p, B-s, C-r, D-q (B) (B) A-s, B-q, C-p, D-r (C) $\mathrm{A}-\mathrm{r}, \mathrm{B}-\mathrm{p}, \mathrm{C}-\mathrm{s}, \mathrm{D}-\mathrm{q}$ (D) $\mathrm{A}-\mathrm{q}, \mathrm{B}-\mathrm{r}, \mathrm{C}-\mathrm{q}, \mathrm{D}-\mathrm{p}$

If for a gas $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=1.67$, this gas is made up to molecules which are (A) diatomic (B) Polytomic (C) monoatomic (D) mixnese of diatomic and polytomic molecules

What is the value of absolute temperature on the Celsius Scale? (A) \(-273.15^{\circ} \mathrm{C}\) (B) \(100^{\circ} \mathrm{C}\) (C) \(-32^{\circ} \mathrm{C}\) (D) \(0^{\circ} \mathrm{C}\)
See all solutions

Recommended explanations on English Textbooks

Free Response Essay

Read Explanation

Graphology

Read Explanation

Discourse

Read Explanation

Sociolinguistics

Read Explanation

Listening and Speaking

Read Explanation

Linguistic Terms

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