Answer
$q_{\text {regen }}=1070 \text{ kJ/kg}$
Work Step by Step
Applying the ideal gas equation to the isothermal process 3-4 gives $$
P_4=P_3 \frac{v_3}{v_4}=(50 \mathrm{kPa})(12)=600\ \mathrm{kPa}
$$ Since process 4-1 is one of constant volume,
$$ T_1=T_4\left(\frac{P_1}{P_4}\right)=(298 \mathrm{~K})\left(\frac{3600 \mathrm{kPa}}{600 \mathrm{kPa}}\right)=1788 \mathrm{~K}
$$ Application of the first law to process 4-1 gives $$
q_{\text {regen }}=c_v\left(T_1-T_4\right)=(0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(1788-298) \mathrm{K}=1070 \mathrm{~kJ} / \mathbf{k g}
$$
