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

Consider the given diagram,

Here, ${\mathrm{r}}_1=10\times {10}^3\mathrm{Pa}\text{ and }{\mathrm{r}}_2=10\times {10}^{-3}{\mathrm{m}}^3$

 we know that for a complete cyclic process, change in internal energy, $\left(\mathrm{\Delta U}\right)=0$ According to 1st law of thermodynamics,

 $\mathrm{\Delta Q}=\mathrm{\Delta U}+\mathrm{W}$ From 

$\begin{array}{l}\Rightarrow \mathrm{\Delta Q}=0+\mathrm{W}\\ \mathrm{\Delta Q}=\mathrm{W}\\ \mathrm{W}=\text{ Area }\\ ={\mathrm{\pi r}}_1{\mathrm{r}}_2\\ =\mathrm{\pi }\times \left(10\times {10}^3\right)\times \left(10\times {10}^{-3}\right)\\ \mathrm{W}=100\mathrm{\pi J}\\ \\ \mathrm{\Delta Q}=100\mathrm{\pi J}\end{array}$