• Heat Transfer Area Calculation in Counterflow Exchanger:
• Heat Duty Absorbed by Water:

$$Q = \dot{m}_w C_{p,w} (T_{w,\text{out}} - T_{w,\text{in}}) = 3\text{ kg/s} \times 4,187\text{ J/(kg}\cdot\text{K)} \times (80 - 30)\text{ K} = 628,050\text{ W}$$

• Temperature Differences:

$$\Delta T_1 = T_{s,\text{in}} - T_{w,\text{out}} = 180^\circ\text{C} - 80^\circ\text{C} = 100^\circ\text{C}$$ $$\Delta T_2 = T_{s,\text{out}} - T_{w,\text{in}} = 130^\circ\text{C} - 30^\circ\text{C} = 100^\circ\text{C}$$ $$\text{LMTD} = \Delta T_1 = \Delta T_2 = 100^\circ\text{C} = 100\text{ K}$$

• Required Heat Transfer Surface Area ($A$):

$$A = \frac{Q}{U \times \text{LMTD}} = \frac{628,050\text{ W}}{820\text{ W/(m}^2\cdot\text{K)} \times 100\text{ K}} = 7.659\text{ m}^2 \approx 7.66\text{ m}^2$$

• Conclusion: 7.66 m² (Option 1).

Correct option: (1) 7.66 m?