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NEET PHYSICSEasy

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=ΔQ−ΔW=8400 J−500 J=7900 J\Delta U = \Delta Q - \Delta W = 8400\text{ J} - 500\text{ J} = 7900\text{ J}.

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