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(C) For a cyclic process,the net heat absorbed $Q$ is equal to the net work done $W$,which is equal to the area enclosed by the $p-V$ diagram.From the

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$10^2 \pi ~J$

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(C) For a cyclic process,the net heat absorbed $Q$ is equal to the net work done $W$,which is equal to the area enclosed by the $p-V$ diagram.From the figure,the process is an ellipse in the $p-V$ plane.The pressure axis $(p)$ ranges from $10 	ext{ kPa}$ to $30 	ext{ kPa}$,so the major/minor axis length is $\Delta p = 30 - 10 = 20 	ext{ kPa} = 20 	imes 10^3 	ext{ Pa}$.The volume axis $(V)$ ranges from $10 	ext{ lt}$ to $30 	ext{ lt}$,so the major/minor axis length is $\Delta V = 30 - 10 = 20 	ext{ lt} = 20 	imes 10^{-3} 	ext{ m}^3$.The area of an ellipse is given by $A = \pi 	imes a 	imes b$,where $a$ and $b$ are the semi-axes.Here,$a = \frac{\Delta p}{2} = 10 	imes 10^3 	ext{ Pa}$ and $b = \frac{\Delta V}{2} = 10 	imes 10^{-3} 	ext{ m}^3$.Therefore,$W = \pi 	imes (10 	imes 10^3) 	imes (10 	imes 10^{-3}) = 100 \pi 	ext{ J} = 10^2 \pi 	ext{ J}$.

The heat energy absorbed by a system undergoing the cyclic process shown in the figure is:

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