A heat engine working on carnot cycle receives heat at the rate of 40 kW from a source at 1200 K rejects to a sink at 300 K. The heat rejected is _______. 

This question was previously asked in
JKSSB JE Mechanical 02 Nov 2021 Shift 1 Official Paper
View all JKSSB JE Papers >
  1. 5 kW 
  2. 10 kW
  3. 20 kW
  4. 30 kW

Answer (Detailed Solution Below)

Option 2 : 10 kW
Free
ST 1: JKSSB JE - Surveying
20 Qs. 20 Marks 20 Mins

Detailed Solution

Download Solution PDF

Concept:

Clausius inequality for a reversible engine

For irreversible engine

The carnot cycle is a reversible cycle

Calculation:

Given:

Q1 = 40 kW, T1 = 1200 K, T2 =  300 K

For reversible cycle

Latest JKSSB JE Updates

Last updated on Jun 23, 2025

-> JKSSB JE application form correction facility has been started. Candidates can make corrections in the JKSSB recruitment 2025 form from June 23 to 27. 

-> JKSSB Junior Engineer recruitment exam date 2025 for Civil and Electrical Engineering has been released on its official website. 

-> JKSSB JE exam will be conducted on 10th August (Civil), and on 27th July 2025 (Electrical).

-> JKSSB JE recruitment 2025 notification has been released for Civil Engineering. 

-> A total of 508 vacancies has been announced for JKSSB JE Civil Engineering recruitment 2025. 

-> JKSSB JE Online Application form will be activated from 18th May 2025 to 16th June 2025 

-> Candidates who are preparing for the exam can access the JKSSB JE syllabus PDF from official website of JKSSB.

-> The candidates can check the JKSSB JE Previous Year Papers to understand the difficulty level of the exam.

-> Candidates also attempt the JKSSB JE Mock Test which gives you an experience of the actual exam.

More Carnot engine Questions

More Thermodynamics Questions

Hot Links: teen patti real cash withdrawal teen patti master list teen patti gold apk teen patti master 2024