Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :&

A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :         
                 

Option: 1

300 K


Option: 2

900 K


Option: 3

1000 K


Option: 4

360 K


Answers (1)

best_answer

\begin{aligned} & \eta=1-\frac{T_L}{T_H} \\ & 0.5=1-\frac{T_L}{600} \Rightarrow T_L=(1-0.5) \times 600 \mathrm{~K}=300 \mathrm{~K} \\ & \text { Now } 0.7=1-\frac{300}{T_2} \\ & \frac{300}{T_2}=0.3 \Rightarrow T_2=\frac{300}{0.3}=1000 \mathrm{~K} \end{aligned}

Posted by

Gautam harsolia

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: NTA exam city intimation slip link available; admit card date
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads
anam-khan

Bihar Board extends BSEB Super 50 free coaching application last date to March 26

Latest Question