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

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+1kJ

-5kJ

+5kJ

+3kJ

answer is D.

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Detailed Solution

Change in internal energy is given by the relation ΔF=q+wΔE=−q+w (as heat is given out by the system) ΔE=−1+4ΔE=+3kJ

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