Question

In a water-fall, the water falls from a height of $100$m. If the entire K.E. of water is converted into heat, the rise in temperature of water will be
A. ${0.23^ \circ }C$
B. ${0.46^ \circ }C$
C. ${2.3^ \circ }C$
D. ${0.023^ \circ }C$

Answer 1

Hint: To calculate the rise of temperature of water, we need to use the concept of kinetic energy converted into heat. Also, the velocity of water falling freely under gravity depends on the mass of the body, acceleration due the gravity of the earth and height from which it is falling.

Formulae used:
Kinetic energy $K = \dfrac{1}{2}m{v^2}$
Velocity of water falling freely under gravity $v = \sqrt {2gh} $
Where, $m$ - mass of water, $h$ - height of the water fall and $g$ - acceleration due to gravity.

Complete step by step answer:
We are given $h = 100$ m.
Let us consider $m = 1$ kg of water converting into heat.
So, velocity of the water in waterfall, $v = \sqrt {2gh} $
$v = \sqrt {2 \times 9.8 \times 100} $
\[\Rightarrow v = 44.27\,m{s^{ - 1}} - - - - - - - (1)\]
The kinetic energy of the water is given by $K = \dfrac{1}{2}m{v^2}$
$K = \dfrac{1}{2} \times 1 \times {\left( {44.27} \right)^2}$
$\Rightarrow K = 980\,J - - - - - - - - - - (2)$
Now, the whole kinetic energy is converted into heat.So the heat capacity of the water is $4184Jk{g^{ - 1}}{K^{ - 1}}$ . The $980\,J$ of water raise the temperature of each $1\,kg$ of water by \[\left( {\dfrac{{980}}{{4184}}} \right)\]$ ^\circ C$. So, the rise in temperature is ${0.23^ \circ }C$.

Hence, option A is correct.

Note: Heat capacity is the quantity of the heat required to raise the temperature of the substance by one degree. If the mass of the substance is unity then the heat capacity is called Specific heat capacity. The motion of water from the waterfall is under the freely falling bodies.