Question

Water falls from a height of 800 m from a water fall. If the initial temperature of water at top of water fall is 10°C, what is its temperature at the bottom? Assume all the energy of the falling water changes to heat energy.

[Take g = 10 ms⁻² and sp. heat capacity of water 4200 J kg⁻¹K⁻¹]

Answer 1

K.E. of falling water = P.E. of water at the top of water = mgh

K.E. of falling water at the bottom = Heat energy produced in the water

⇒ mgh = mсθ_R

∴ \[ \theta_{R}=\frac{g h}{c}\]

\[=\frac{10 \mathrm{~ms}^{-2} \times 800 \mathrm{~m}}{4200 \mathrm{Jkg}^{-1}{ }^{\circ} \mathrm{C}^{-1}}\]

= 1.90°C

Thus, temperature of water at bottom of water fall = 10°C + 1.9°C = 11.9°C