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$2 \cdot 5 \mathrm{~kJ}$ of work is done on the system and it releases $1500 \mathrm{~J}$ of heat. What is the change in internal energy?
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The correct answer is:
$1000 \mathrm{~J}$
$\Delta U=q+w = -1500 + 2500 = 1000 ~J$
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