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Change in internal energy, when 4 kJ of work is done on the system and 1 kJ of heat is given out by the system is ……(a) +1 kJ(b) -5 kJ (c) +3 kJ(d) -3 kJ |
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Answer» (c) +3 kJ ∆U = q + w ∆U = – 1 kJ + 4 kJ ∆U = + 3 kJ |
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