JEE PHYSICS MCQ: THERMODYNAMICS

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When 20 K cal of heat is supplied to the system and the increase in the internal energy is 8400 J, then the work done by the system is

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Correct option: B

18 K cal

The correct answer is 18 K cal because it is calculated by applying the first law of thermodynamics:

\Delta W = \Delta Q - \Delta U

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