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**Hint:**When water hits the falls from a height, it has gravitational potential energy stored in it. When it hits the ground, the stored potential energy is converted to heat energy due to the impact. This is in accordance with the law of conservation of energy and will help us determine the rise in temperature of the water.

**Formulae used:**

Law of conservation of energy between gravitational potential energy and heat energy: $mgh = mc\Delta t$

Where $m$ is the mass of water and is expressed in kilogram $(kg)$, $g$ is the acceleration due to gravity and is approximately equal to $10m/s$, $h$ is the height of fall and is expressed in meter $(m)$, $c$ is the specific heat of the water and is expressed in Joules per kilogram per degree Celsius $(J/kg/^\circ C)$ and $\Delta t$ is the rise in temperature of the water and is expressed in degree Celsius $(^\circ C)$.

**Complete step by step answer:**

We have the following information from the question:

Height of fall $h = 50m$, $g = 10m/s$, Specific heat capacity of water $c = 4.2 \times {10^3}J/kg/^\circ C$

Let,

The rise in temperature of the falling water $ = \Delta t^\circ C$

When the water falls to the ground under the influence of only gravity, potential energy I stored in it. This potential energy is known as gravitational potential energy. When this falling water hits the ground below, the impact causes molecular change and therefore, gets heated up. This change is brought about by the previously stored gravitational potential energy. Using this in the equation for law f conservation of energy we get,

$mgh = mc\Delta t$

Substituting the known data and variables we get,

$

mgh = mc\Delta t \\

\Rightarrow m \times 10 \times 50 = m \times 4200 \times \Delta t \\

\Rightarrow \dfrac{{10 \times 50}}{{4200}} = \Delta t \\

\therefore\Delta t = 0.119^\circ C \\

$

**Therefore, the rise in temperature of the water is $0.119^\circ C$.**

**Note:**For the ease of calculation, it is considered that the entirety of the stored gravitational potential energy is converted into heat energy. There are some losses in real conditions, but we ignore them in our calculations.

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