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