If 500 J of heat leaves a body at 250 K reversibly, its entropy change is:

  1. -0.5 J/K
  2. +2 J/K
  3. -2 J/K
  4. +125,000 J/K
Answer

C) -2 J/K

Delta S = Q/T = -500/250. SLO P-11-C-22 · Understanding

Physics MCQ · More Heat Engines, Second Law and Entropy MCQs · Practise