Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$2 \cdot 5 \mathrm{~kJ}$ of work is done on the system and it releases $1500 \mathrm{~J}$ of heat. What is the change in internal energy?
ChemistryThermodynamics (C)MHT CETMHT CET 2020 (20 Oct Shift 2)
Options:
  • A $1000 \mathrm{~J}$
  • B $4000 \mathrm{~J}$
  • C $2500 \mathrm{~J}$
  • D $1500 \mathrm{~J}$
Solution:
1039 Upvotes Verified Answer
The correct answer is: $1000 \mathrm{~J}$
$\Delta U=q+w = -1500 + 2500 = 1000 ~J$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.