[tex]21^{\circ} \mathrm{C}[/tex] is the final river temperature after the cooling water and river have mixed.
Explanation:
Final river temperature after mixing using energy energy balance equation
[tex]\Delta E_{\text {River }}+\Delta E_{\text {Cooling water}}=0 \ldots \ldots(1)[/tex]
[tex]\Delta E_{\text {River }}=Q_{\text {River}} \rho C_{P}\left(T-T_{\text {River }}\right)[/tex]
[tex]$\Delta E_{\text {cooling water }}=Q_{\text {cooling water }} \rho C_{P}\left(T-T_{\text {Cooling water }}\right)$[/tex]
[tex]Q_{\text {River}} \rho C_{P}\left(T-T_{\text {River }}\right)+Q_{\text {Cooling water }} \rho C_{P}\left(T-T_{\text {Cooling water }}\right)=0 \ldots \ldots .(2)[/tex]
Where,[tex]C_{P}[/tex] is specific heat at constant pressure,[tex]\Delta T[/tex] change in temperature,[tex]Q_{\text {River}}[/tex] is flow in the river,[tex]$Q_{\text {cooling water }}$[/tex]is the flow of cooling water from plant,[tex]T[/tex] final required temperature after mixing cooling water and river water,and,[tex]\rho[/tex] is density of water.
From Diagram
While substituting,
[tex]40 \mathrm{m}^{3} / \mathrm{s} \text { for } Q_{\text {River }}[/tex]
[tex]$2 \mathrm{m}^{3} / \mathrm{s} \quad for\quad Q_{\text {cooling water }}[/tex]
[tex]80^{\circ} \mathrm{C} \text { for } T_{\text {cooling water }}[/tex] and
[tex]18^{\circ} \mathrm{C} \text { for } T_{\mathrm{River}}[/tex]
[tex]Q_{\text {River}} \rho C_{P}\left(T-T_{\text {River }}\right)+Q_{\text {Cooling water }} \rho C_{P}\left(T-T_{\text {Cooling water }}\right)=0[/tex]
[tex]40 \times \rho C_{P}(T-18)+2 \times \rho C_{P}(T-40)=0[/tex]
[tex]40 \times(T-18)+2 \times(T-40)=0[/tex]
[tex]T=21^{\circ} \mathrm{C}[/tex]
Finally, the temperature after the mixing of cool water and river is [tex]21^{\circ} \mathrm{C}[/tex] .
