# Thermodynamics: First Law Application ## Problem Restatement - **Step 1:** 200 J heat supplied **to** system, 80 J work **done by** system. - **Step 2:** 150 J work **done on** system, 120 J heat **released by** system. - **Task:** Find the **total change in internal energy** (\(\Delta U\)) for both steps together. --- ## First Law of Thermodynamics \[ \Delta U = Q - W \] Where: - \( \Delta U \) = Change in internal energy - \( Q \) = Heat added to system (**positive** if to system, **negative** if released) - \( W \) = Work done **by** system (**positive** if by system, **negative** if done on system) --- ## Step 1 Calculations - \( Q_1 = +200 \) 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 Calculations - \( Q_2 = -120 \) 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 \[ \Delta U_{\text{total}} = \Delta U_1 + \Delta U_2 = 120\,\text{J} + 30\,\text{J} = 150\,\text{J} \] --- ## **Final Answer** \[ \boxed{150\ \text{J}} \] The **total change in internal energy** for both steps combined is **150 J**.