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NEET PHYSICSThermodynamicsEasy

Question

The internal energy change in a system that has absorbed 2 kcal2\text{ kcal} of heat and done 500 J500\text{ J} of work is

A

8900 J8900\text{ J}

B

6400 J6400\text{ J}

C

5400 J5400\text{ J}

D

7900 J7900\text{ J}

Step-by-Step Solution

According to the first law of thermodynamics, ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W. Given: Heat absorbed by the system, ΔQ=2 kcal=2×1000×4.2 J=8400 J\Delta Q = 2\text{ kcal} = 2 \times 1000 \times 4.2\text{ J} = 8400\text{ J} Work done by the system, ΔW=500 J\Delta W = 500\text{ J} Therefore, the change in internal energy is: ΔU=ΔQΔW=8400 J500 J=7900 J\Delta U = \Delta Q - \Delta W = 8400\text{ J} - 500\text{ J} = 7900\text{ J}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermodynamicsinternalenergychangesystemabsorbed

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