Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A Carnot engine whose sink is at has an efficiency of <img alt="40 \%" src="https://ent

A Carnot engine whose sink is at 300 \mathrm{~K} has an efficiency of 40 \%. By how much should the temperature of source be increased so as to increase its efficiency by 50 \% of original efficiency?

Option: 1

275 \mathrm{~K}


Option: 2

325 \mathrm{~K}


Option: 3

250 \mathrm{~K}


Option: 4

380 \mathrm{~K}


Answers (1)

best_answer

The efficiency of Carnot engine \mathrm{=1-\frac{T_{2}}{T_{1}}}.

Here, \mathrm{T_{1}} is the temperature of source and \mathrm{T_{2}} is the temperature of sink.
As given, \mathrm{\eta=40 \%=\frac{40}{100}=0.4}

and
\mathrm{T_{2}=300 \mathrm{~K}}
So,
\mathrm{0.4=1-\frac{300}{T_{1}}}
\mathrm{\Rightarrow \quad T_{1}=\frac{300}{1-0.4}=500 \mathrm{~K}}
Let temperature of the source be increased by \mathrm{x}, then efficiency becomes
\mathrm{\eta^{\prime} =40 \%+50 \% \text { of } \eta}
\mathrm{=\frac{40}{100}+\frac{50}{100} \times 0.4=0.4+0.5 \times 0.4=0.6}
Hence,
\mathrm{0.6 =1-\frac{300}{500+x}}
\mathrm{\Rightarrow \quad \frac{300}{500+x} =0.4}
\mathrm{\therefore \quad x =750-500=250 \mathrm{~K}}


 

Posted by

jitender.kumar

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

Jharkhand NEET UG counselling registration 2025 begins; round 1 choice filing from July 30
anam-khan

Chandigarh NEET counselling registration 2025 begins for MBBS, BDS, BHMS admission; apply by July 24
anam-khan

Karnataka NEET UG Counselling 2025: KEA begins round 1 option entry at cetonline.karnataka.gov.in

Latest Question