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In a non-flow process there is heat transfer loss of $1055 \mathrm{~kJ}$ and an internal energy increase of 210 $\mathrm{kJ}$. Determine the work transfer and state whether the process is an expansion or compression.
Solution:
Closed system for which the first law of Thermodynamics applies,
$$
\mathrm{Q}-\mathrm{W}=\Delta \mathrm{U}
$$
$$
1055-\mathrm{W}=210
$$
Hence the work done can be found as:
$$
\mathrm{W}=-1265 \mathrm{~kJ}
$$
Since negative, it must be work input, ie compression.
In a non-flow process carried out on $5.4 \mathrm{~kg}$ of a substance, there was a specific internal energy decrease of $50 \mathrm{~kJ} / \mathrm{kg}$ and a work transfer from the substance of $85 \mathrm{~kJ} / \mathrm{kg}$. Determine the heat transfer and state whether it is gain or loss.
[Ans: $189 \mathrm{~kJ}$, gain]
During the working stroke of an engine the heat transferred out of the system was $150 \mathrm{~kJ} / \mathrm{kg}$ of the working substance. If the work done by the engine is $250 \mathrm{~kJ} / \mathrm{kg}$, determine the change in internal energy and state whether it is decrease or increase.
Solution:
Closed system for which the first law of Thermodynamics applies,
$$
\mathrm{Q}-\mathrm{W}=\Delta \mathrm{U}
$$
Hence
$$
\begin{aligned}
\Delta \mathrm{U} & =\mathrm{Q}-\mathrm{W} \
& =(-150)-250 \
& =-400 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$
Since the sign is negative, there is a decrease in internal energy.
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