The internal energy change in a system that has absorbed $2 \;k cal$ of heat and done $500 \;J $ of work is ...... $J$
AIPMT 2009, Medium
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(a)$\Delta Q = 2k\,cal = 2 \times {10^3} \times 4.2J$$ = 8400J$ and $\Delta W = 500J.$
Hence from $\Delta Q = \Delta U + \Delta W$
$\Delta U = \Delta Q - \Delta W$
$= 8400 -500 = 7900 J$
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