Question 13.4: For the data in question 3, use the Penman-Monteith Equation...

When air temperature T_{a} = 22 °C, \Delta = 162\;  Pa\; K^{-1} and \gamma = 66.3 \; Pa \; K^{-1} . The saturation deficit δ is 2643 − 1000 = 1643 Pa. Substituting values into the Penman-Monteith equation gives λE = [162 × 300 + (1.2 × 10³ × 1643/40)]/ [162 + 66.3(110/40)] = 284 W m^{−2}. Then C = R_{n} − λE = 16 W m^{−2}. Since C = \rho c_{p} (T_{l} -T_{a} ) / r_{H} ,(T_{l} -T_{a} ) = 40 \times 16 / (1.2\times 10^{3}) = 0.53 \; ^{\circ } C , so T_{l} = 22.5 °C.