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NEET 1993 (AIPMT) · Heat, Work & First Law of Thermodynamics

Heat, Work & First Law of Thermodynamics question from NEET 1993 (AIPMT)

Q23NEET 1993 · AIPMTEasy

Heat, Work & First Law of Thermodynamics

110 joule of heat is added to a gaseous system whose internal energy is 40 J, then the amount of external work done is

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Answer: (b)

Take the 40 J as the increase in internal energy, $\Delta U = 40$ J.
$W = Q - \Delta U = 110 - 40 = 70$ J
So the answer is (b) 70 J.

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