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A system given out $\mathrm{x} \mathrm{J}$ of heat and does $\mathrm{y} \mathrm{J}$ of work on it's surrounding. What is the internal energy change?
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Verified Answer
The correct answer is:
$-x-y ~ J$
$\mathrm{q}=-\mathrm{xJ}$ Heat released by system $(-)$ ve $\mathrm{w}=-\mathrm{yJ}$ Work done by system $(-)$ ve From first law of thermodynamics.
$$
\begin{aligned}
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \\
& =-\mathrm{x}-\mathrm{y} \mathrm{J}
\end{aligned}
$$
$$
\begin{aligned}
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \\
& =-\mathrm{x}-\mathrm{y} \mathrm{J}
\end{aligned}
$$
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