# Step-by-Step Solution: Change in Internal Energy (ΔU) We will use the **First Law of Thermodynamics**: \[ \Delta U = Q - W \] where: - \(\Delta U\) = Change in internal energy - \(Q\) = Heat supplied **to** the system (positive if added, negative if released) - \(W\) = Work done **by** the system (positive if by the system, negative if done on the system) --- ## **Step 1: First Process** - **Heat supplied:** \(Q_1 = +200\,\text{J}\) (to the system, positive) - **Work done by the system:** \(W_1 = +80\,\text{J}\) (by the system, positive) \[ \Delta U_1 = Q_1 - W_1 = 200\,\text{J} - 80\,\text{J} = 120\,\text{J} \] --- ## **Step 2: Second Process** - **Work done on the system:** \(W_2 = -150\,\text{J}\) (on the system, negative) - **Heat released:** \(Q_2 = -120\,\text{J}\) (released, negative) \[ \Delta U_2 = Q_2 - W_2 = (-120\,\text{J}) - (-150\,\text{J}) = -120\,\text{J} + 150\,\text{J} = 30\,\text{J} \] --- ## **Step 3: Total Change in Internal Energy** \[ \Delta U_{\text{total}} = \Delta U_1 + \Delta U_2 = 120\,\text{J} + 30\,\text{J} = 150\,\text{J} \] --- ## **Final Answer** \[ \boxed{150\,\text{J}} \] **The total change in internal energy for both steps combined is \(150\,\text{J}\).**