According to the first law of thermodynamics, $\Delta U = Q - W$. For the path $ABC$, the total heat added is $Q_{ABC} = Q_{AB} + Q_{BC} = 400\text{ J} + 100\text{ J} = 500\text{ J}$. From the standard $P$-$V$ diagram given in this PYQ, the coordinates are: $A \equiv (2 \times 10^{-3}\text{ m}^3, 2 \times 10^4\text{ Pa})$ $B \equiv (2 \times 10^{-3}\text{ m}^3, 6 \times 10^4\text{ Pa})$ $C \equiv (4 \times 10^{-3}\text{ m}^3, 6 \times 10^4\text{ Pa})$ Work done in path $AB$ (isochoric process, $\Delta V = 0$) is $W_{AB} = 0$. Work done in path $BC$ (isobaric process) is $W_{BC} = P_B(V_C - V_B) = 6 \times 10^4 \times (4 - 2) \times 10^{-3} = 120\text{ J}$. Total work done along path $ABC$ is $W_{ABC} = W_{AB} + W_{BC} = 120\text{ J}$. Change in internal energy is $\Delta U = Q_{ABC} - W_{ABC} = 500\text{ J} - 120\text{ J} = 380\text{ J}$. Since internal energy is a state function, $\Delta U$ for path $AC$ is also $380\text{ J}$. Work done along path $AC$ is the area of the trapezium under the curve $AC$: $W_{AC} = \frac{1}{2} (P_A + P_C)(V_C - V_A) = \frac{1}{2}(2 \times 10^4 + 6 \times 10^4)(4 - 2) \times 10^{-3} = \frac{1}{2}(8 \times 10^4)(2 \times 10^{-3}) = 80\text{ J}$. Therefore, heat absorbed in path $AC$ is $Q_{AC} = \Delta U + W_{AC} = 380\text{ J} + 80\text{ J} = 460\text{ J}$.