If 500 calorie of heat energy is added to the system and the system does 350 calorie of work on the surroundings, the internal energy change of the system is
A
650 cal
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B
150 cal
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C
-150 cal
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D
-650 cal
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Solution
The correct option is B 150 cal Heat absosrbed by the system (q) = 500 cal Work done(w) = -350 cal According to the first law of thermodynamics dU = q + w = 500 + (-350) = 150 cal.