A) The terminal velocity of a raindrop is dependent upon the size of the raindrop, but most sources list \sim 10 \frac{\mathrm{m}}{\mathrm{sec}} as a typical value.
Find Kinetic Energy of rainwater
\begin{aligned}& \mathrm{K} . \mathrm{E}=\operatorname{mass} \times \frac{\text { velocity }^{2}}{2}\\& \mathrm{K} . \mathrm{E}=(200 \mathrm{~kg})\left(\frac{1}{2}\right)\left(10 \frac{\mathrm{m}}{\mathrm{sec}}\right)^{2}\left(\frac{\mathrm{J}~ \mathrm{\textrm {sec } ^ { 2 }}}{\mathrm{kg}~ \mathrm{m}^{2}}\right)=10000 \mathrm{~J}=\bf 10 \,\mathbf{k J}\end{aligned}
Comparing kinetic energy to enthalpy directly is complicated by the fact that the kinetic energy is calculated on an absolute basis while enthalpy is known only relative to some reference state. However, if we assume the rain is saturated liquid at 25°C, and use the steam table values, the enthalpy is 104.8 kJ/kg. By contrast the kinetic energy of the rain is, per kilogram, is 0.05 (10 kJ/200 kg); by comparison it is negligible.
B) The better estimate is saturated water vapor at 20°C because the enthalpy of an ideal gas is a function of temperature but not of pressure. At atmospheric pressure and below we expect the ideal gas model to well approximate the properties of water vapor. Consequently, we will use the data at the correct temperature of 20°C, and assume the effect of pressure is negligible.
C) Set up an energy balance around the lake. We will use the assumption stated in part B) and the assumption that the velocity of the rainwater is negligible.
\begin{aligned}\Delta\left\{\mathrm { M } \left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2} +\mathrm{gh}\right)\right\} \\& =\mathrm{m}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\mathrm{m}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}+\mathrm{gh}_{\mathrm{out}}\right)+\mathrm{W}_{\mathrm{EC}}+\mathrm{W}_{\mathrm{S}}+\mathrm{Q}\end{aligned}
The left hand side is 0 because the final state of the lake is identical to the initial state of the lake- same temperature and mass of water. Cancelling terms:
0=m_{\rm {rain }} \widehat{H}_{25^{\circ} \mathrm{C}, 1 \text { bar }}-m_{\rm v a p} \widehat{H}_{20^{\circ} \mathrm{C}, 1 \text { bar }}+\mathrm{Q}_{\text {net }}
Liquid water at 25°C and 1 bar ≈ Saturated liquid water at 25^{\circ} \mathrm{C} \rightarrow 104.8 \frac{\mathrm{kJ}}{\mathrm{kg}}
Water vapor at 20°C and 1 bar ≈ Saturated water vapor at 20^{\circ} \mathrm{C} \rightarrow 2537.4 \frac{\mathrm{kJ}}{\mathrm{kg}}
\begin{aligned}& 0=(200 \mathrm{~kg})\left(104.8 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)-(200 \mathrm{~kg})\left(2537.4 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)+\mathrm{Q}_{\mathrm{net}}\\& Q_{\rm n e t}=\bf 486,520 \mathbf{~kJ}\end{aligned}
D) Set up an energy balance around the lake, using the same assumptions as in part (B).
\begin{aligned}\Delta\left\{\mathrm { M } \left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2}+\mathrm{gh}\right)\right\} \\& =\mathrm{m}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\mathrm{m}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}+\mathrm{gh}_{\mathrm{out}}\right)+\mathrm{W}_{\mathrm{EC}}+\mathrm{W}_{\mathrm{S}}+\mathrm{Q}\end{aligned}
Cancelling terms
M_{\text {lake }}\left(\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}\right)=m_{\text {rain }} \widehat{H}_{25^{\circ} \mathrm{C}, 1 \text { bar }}-m_{\text {vap }} \widehat{H}_{21^{\circ} \mathrm{C}, 1 \text { bar }}+\mathrm{Q}_{\text {net }}
Enthalpies
Liquid water at 25°C and 1 bar ≈ Saturated liquid water at 25^{\circ} \mathrm{C} \rightarrow 104.8 \frac{\mathrm{kJ}}{\mathrm{kg}}
Water vapor at 20°C and 1 bar ≈ Saturated water vapor at 20^{\circ} \mathrm{C} \rightarrow 2537.4 \frac{\mathrm{kJ}}{\mathrm{kg}}
\begin{aligned}& \mathrm{M}_{\text {lake }}\left(\widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }}\right)=\mathrm{M}_{\text {lake }} \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}=20000 \mathrm{~kg}\left(4.184 \frac{\mathrm{J}}{\mathrm{kg}^{\circ} \mathrm{C}}\right)\left(21^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}\right) \\& \mathrm{M}_{\text {lake }} \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}=\mathrm{m}_{\text {rain }} \widehat{\mathrm{H}}_{25^{\circ} \mathrm{C}, 1 \text { bar }}-\mathrm{m}_{\text {vap }} \widehat{\mathrm{H}}_{21^{\circ} \mathrm{C}, 1 \text { bar }}+\mathrm{Q}_{\text {net }} \\& 20000 \mathrm{~kg}\left(4.184 \frac{\mathrm{J}}{\mathrm{kg}^{\circ} \mathrm{C}}\right)\left(21^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}\right) \\& =(200 \mathrm{~kg})\left(104.8 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)-(200 \mathrm{~kg})\left(2537.4 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)+\mathrm{Q}_{\mathrm{net}}\\ & \rm Q_{n e t}=\bf 570,200 \mathbf{~kJ}\end{aligned}
E) Set up an energy balance around the lake. We will use the assumption from part B) and the assumption that the velocity of the rainwater is negligible.
\begin{aligned}\Delta\left\{\mathrm { M } \left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2}+\mathrm{gh}\right)\right\} \\& =\mathrm{m}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\mathrm{m}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}+\mathrm{gh}_{\mathrm{out}}\right)+\mathrm{W}_{\mathrm{EC}}+\mathrm{W}_{\mathrm{S}}+\mathrm{Q}\end{aligned}
Cancelling terms
\mathrm{M}_{\text {final }} \widehat{U}_{\text {final }}-\mathrm{M}_{\text {initial }} \widehat{U}_{\text {initial }}=\mathrm{m}_{\text {rain }} \widehat{H}_{25^{\circ} \mathrm{C}, 1 \text { bar }}-\mathrm{m}_{\text {vap }} \widehat{H}_{20^{\circ} \mathrm{C}, 1 \text { bar }}+\mathrm{Q}_{\text {net }}
Enthalpies
\widehat{\mathrm{H}}_{25^{\circ} \mathrm{C}, 1 \text { bar }}= Liquid water at 25°C and 1 bar ≈ Saturated liquid water at 25^{\circ} \mathrm{C} \rightarrow 104.8 \frac{\mathrm{kJ}}{\mathrm{kg}}
\widehat{\mathrm{H}}_{20^{\circ} \mathrm{C}, 1 \text { bar }}= Water vapor at 20°C and 1 bar ≈ Saturated water vapor at 20^{\circ} \mathrm{C} \rightarrow 2537.4 \frac{\mathrm{kJ}}{\mathrm{kg}}
\widehat{\mathrm{U}}_{\text {initial }}= Liquid water at 20°C and 1 bar ≈ Saturated liquid water at 20^{\circ} \mathrm{C} \rightarrow 83.9 \frac{\mathrm{kJ}}{\mathrm{kg}}
\widehat{\mathrm{U}}_{\text {final }}= Liquid water at 21°C and 1 bar ≈ Saturated liquid water at 21^{\circ} \mathrm{C} \rightarrow Must be interpolated
Interpolate
\begin{aligned}& y=\frac{\left(y_{2}-y_{1}\right)\left(x-x_{1}\right)}{\left(x_{2}-x_{1}\right)}+y_{1}\\& y=\frac{(104.8-83.9)(21-20)}{(25-20)}+83.9=\bf 92.4 \frac{\mathbf{k J}}{\mathbf{k g}}\end{aligned}
\begin{array}{|c|c|c|c|}\hline&1&&2 \\\hline \text{Specific Energy (y) }&83.9&???&104.8\\\hline\text{Temperature (x)}&20&21&25\\\hline\end{array}
Set up a mass balance to solve for M_{\text {final }}
\begin{aligned}& M_{\text {final }}-M_{\text {initial }}=m_{\text {in }}-m_{\text {out }}\\& \mathrm{M}_{\text {final }}-20000 \mathrm{~kg}=200 \mathrm{~kg}-50 \mathrm{~kg}\\& \mathrm{M}_{\text {final }}=20150 \mathrm{~kg}\\ \end{aligned}
Plugging into the energy balance for the lake
\begin{aligned}& (20150 \mathrm{kg})\left(92.4 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)-(20000 \mathrm{kg})\left(83.9 \frac{\mathrm{kJ}}{\mathrm{kg}}\right) \\& =(200 \mathrm{kg})\left(104.8 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)-(50 \mathrm{kg})\left(2537.4 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)+Q_{n e t}\\& \mathrm{Q}_{\text {net }}=\bf 289,770 \mathbf{~kJ}\end{aligned}