Question bank/JEE Main/Chemistry

JEE MainChemistryModeratePhysical Chemistry

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

Show answer and solution
T2=150+273=423KT_{2} = 150 + 273 = 423K
T1=25+273=298KT_{1} = 25 + 273 = 298K
Q=500KQ = 500K
WQ=T2−T1T2;W=500(423−298423)=147.7J\frac{W}{Q} = \frac{T_{2} - T_{1}}{T_{2}};W = 500\left( \frac{423 - 298}{423} \right) = 147.7J

Source: Grass-G / CH-11 Chemistry Paper-2 1Nov New Jee main.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans