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NEET 2009 (AIPMT) · Heat, Work & First Law of Thermodynamics

Heat, Work & First Law of Thermodynamics question from NEET 2009 (AIPMT)

Q20NEET 2009 · AIPMTEasy

Heat, Work & First Law of Thermodynamics

The internal energy change in a system that has absorbed 2 kcal of heat and done 500 J of work is

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Answer: (c)

  1. Q=2Q = 2 kcal =2000×4.2=8400= 2000 \times 4.2 = 8400 J
  2. ΔU=Q−W=8400−500=7900\Delta U = Q - W = 8400 - 500 = 7900 J
  3. So the answer is (c) 7900 J.
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