Lake loss from monthly pan evaporation

Lake evaporation $= C_p \times$ pan evaporation, and the volume lost $=$ depth $\times$ mean water-spread area.

Step 1: Annual pan evaporation

$16.7+14.3+17.8+25.0+28.6+21.4+16.7+16.7+16.7+16.7+21.4+16.7 = 228.7$ cm

Step 2: Lake evaporation depth

$$E_L = 0.7\times228.7 = 160.09\ \text{cm} = 1.6009\ \text{m}$$

Step 3: Mean water-spread area

$$A_{avg} = \frac{2.80 + 2.55}{2} = 2.675\ \text{km}^2 = 2.675\times10^6\ \text{m}^2$$

Step 4: Volume of evaporation loss

$$V = 1.6009\times 2.675\times10^6 = 4,282,407\ \text{m}^3$$

Answer: Evaporation depth = 160.1 cm; volume lost = 4.28 × 10⁶ m³ (about 4.28 million m³) in the year.