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Q.
What is $\Delta E$ for system that does $500\, cal$-of work on surrounding and $300\, cal$ of heat is absorbed by the system ?
Bihar CECEBihar CECE 2003Thermodynamics
Solution:
From first law of thermodynamic.
$\Delta E=q+ W$
Given $q=+300\, cal$ ($\because$ heat is absorbed)
$W-500\, cal$
$(\because$ work is done on surroundings).
$\therefore \Delta E=q+ W$
$=300+(-500)$
$=-200\, cal.$