Chemical Thermodynamics Question 31

Question: An engine operating between 150oC and 25oC takes 500 J heat from a higher temperature reservoir if there are no frictional losses, then work done by engine is [MH CET 1999]

Options:

A) 147.7 J

B) 157.75 J

C) 165.85 J

D) 169.95 J

Show Answer

Answer:

Correct Answer: A

Solution:

T2=150+273=423,K

T1=25+273=298,K

Q=500,K

WQ=T2−T1T2 ; W=500(423−298423)=147.7,J .

Engineering Preparation

Chemistry 11th

Chemistry 12th

Mathematics 11th

Mathematics 12th

Physics 11th

Physics 12th

JEE Previous Year Questions

JEE PYQ - Chapterwise

JEE Tutorial Sessions

Medical Preparation

Biology 11th

Biology 12th

Chemistry 11th

Chemistry 12th

Physics 11th

Physics 12th

NEET Previous Year Questions

NEET PYQ - Chapterwise

NEET Tutorial Sessions

NCERT Books & Solutions

NCERT Books for JEE

NCERT Books for NEET

NCERT Solutions for JEE

NCERT Solutions for NEET

NCERT Exemplar for JEE

NCERT Exemplar for NEET

NCERT Books for 9th-10th

NCERT Exemplar for 9th-10th

Important Resources

Physics Formulas

Chemistry Formulas

Mathematics Formulas

Useful Articles

Sample Mock Tests

Other Resources

Media Coverage

Help Videos

Educational Games

CBSE Board Physics Previous Year Questions Chapterwise

Bank PO Previous Year Questions Chapterwise

For 10th and 12th Board Exams

GK

Syllabus

JEE Syllabus

NEET Syllabus

Test Platform

JEE / Engineering Test

NEET / Medical Test

Sample Mock Test

Mock Test for JEE

Additional problems chapterwise

Mentorship

Mentorship for JEE

Mentorship for NEET

NCERT Exercise Video Solution

NCERT Chapter Video Solution

Dual Pane

© 2024 Copyright SATHEE

Privacy Policy

Powered by Prutor@IITK

sathee@iitk.ac.in
Media coverage of SATHEE
goolge

Play Store
goolge

App Store