If 500cal of heat energy is added to a system and the system does 350cal of work on the surroundings, the internal energy change of the system is:
A
150 cal
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B
850 cal
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C
-150 cal
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D
-850 cal
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Solution
The correct option is A 150 cal Heat absorbed, q = 500 cal;
Work done by the system, w = -350 cal
From the first law of thermodynamics, △U=q+w=500+(−350)=150cal