Question

Water falls from height $500m.$ The rise in temperature of water at bottom if the whole of energy remains in water, will be: [sp. Heat of water $4.2kJ/kg$]
A) ${0.23^\circ}C$
B) ${1.16^\circ}C$
C) ${27^\circ}C$
D) ${1.02^\circ}C$

Answer 1

Hint: Energy is conserved, that is the total initial and final energies are equal. Although energy can change forms.

Complete step by step answer:
Specific Heat: It is the amount of energy needed to raise the temperature of $1kg$ water by ${1^\circ}C$. The formula to calculate specific heat is,
$S = \dfrac{{\Delta Q}}{{m\Delta T}}$
Where, $S = $Specific heat of water
$\Delta Q = $ Change in energy
$m = $Mass of water
$\Delta T = $ Change in temperature
When the water was held at a height, it had energy stored in it in the form of potential energy. This potential energy gets converted into heat energy as soon as the water hits the ground. The conversion does not include any kind of energy loss. Hence, energy is conserved.
So, the potential energy when water is held up at a height,
$PE = mgh$
And, the heat energy which forms when water hits the ground, $Q = mS\Delta T$
Since,
$
  PE = Q \\
   \Rightarrow mgh = mS\Delta T \\
   \Rightarrow gh = S\Delta T \\
   \Rightarrow \Delta T = \dfrac{{gh}}{S} \\
   \Rightarrow \Delta T = \dfrac{{9.8 \times 500}}{{4.2 \times {{10}^3}}} \\
  \therefore \Delta T = {1.16^\circ}C $

Hence, option B is the correct answer.

Note: The law of conservation of energy is applicable in all cases of Newtonian physics, but it has limitations when it comes to nuclear physics. Since there is a whole lot of unaccounted energy and mass loss, the law of conservation of energy tends to fail in that particular area.