Water initially at `20^@`C at a height of 1.68 km above the ground, falls down on ground. Taking the specific heat capacity of water to be 4200 J `kg^(-1) K^(-1)`, find the final temperature of water on reaching the ground. Take g = 10 m `s^(-2)`.

Explanation

Given, initial temperature = `20^@`C, g = 10 m `s^(-2)`, h = 1.68 km = `1.68 xx 10^3` m and c = 4200 J `kg^(-1) K^(-1)`.
Initially, water at a height h has the potential energy mgh stored in it which changes into the kinetic energy during the fall. On reaching the ground, water has the kinetic energy equal to the initial potential energy (equal to mgh) which on striking the ground changes into the heat energy mc`Delta`t, if `Delta`t is the rise in temperature of water. Assuming that whole of the kinetic energy of water changes into the heat energy and there is no loss of energy,
mgh = mc`Delta`t
or rise in temperature `Deltat = (gn)/(c)`
= `(10×x(1-68x10^3))/(4200)=4^@` C
Hence final temperature of water on reaching the ground = initial temperature + rise in temperature = `20^2C + 4^@C = 24^@`C.