In the reported figure,the heat energy absorb | vedclass

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(C) For a complete cyclic process,the change in internal energy is $\Delta U = 0$.From the first law of thermodynamics,$\Delta Q = \Delta U + W$.Since

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$100$

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(C) For a complete cyclic process,the change in internal energy is $\Delta U = 0$.From the first law of thermodynamics,$\Delta Q = \Delta U + W$.Since $\Delta U = 0$,the heat absorbed $\Delta Q$ is equal to the work done $W$ by the system.The work done $W$ in a cyclic process is equal to the area enclosed by the $P-V$ curve.The area of the circle is $A = \pi \cdot r_P \cdot r_V$,where $r_P$ is the radius along the pressure axis and $r_V$ is the radius along the volume axis.From the figure,the diameter along the pressure axis is $40 	ext{ kPa} - 20 	ext{ kPa} = 20 	ext{ kPa}$,so $r_P = 10 	ext{ kPa} = 10 	imes 10^3 	ext{ Pa}$.The diameter along the volume axis is $40 	ext{ L} - 20 	ext{ L} = 20 	ext{ L}$,so $r_V = 10 	ext{ L} = 10 	imes 10^{-3} 	ext{ m}^3$.Therefore,$\Delta Q = W = \pi 	imes (10 	imes 10^3 	ext{ Pa}) 	imes (10 	imes 10^{-3} 	ext{ m}^3) = 100 \pi 	ext{ J}$.Thus,the value is $100$.

In the reported figure,the heat energy absorbed by a system in going through a cyclic process is $......\,\pi \text{ J}$.

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