Question

Water falls from a height $20\;m$ at a rate of $100\;kgs^{-1}$. How many calories of heat will be produced per second on striking the earth? Assume that the whole energy is converted to heat.

Answer 1

Hint: Water at a height possesses a potential energy, which gets converted to kinetic energy due to the falling motion of water, following which it strikes the earth to generate heat. Given the assumption in the question, calculate the energy in joules, and use the appropriate conversion to obtain the same in calories.

Formula Used:
Potential energy $PE = mgh$
$1\;calorie = 4.184\;joules$

Complete Step-by-Step Solution:
We are given that water falls from a height of $h=20\;m$. The water possesses a potential energy by virtue of the height from which it falls. The water falls under the influence of the acceleration due to gravity $g = 9.8\;ms^{-2}$, and we are given that this fluidic motion is at a rate of $m = 100\;kgs^{-1}$.

The potential energy possessed by the water at a height can be quantized as:
$PE = mgh = (100 \times 9.8 \times 20) \; (kgs^{-1}. ms^{-2}. m)= 19600\;kgm^2s^{-2}.s^{-1} = 19600\;Js^{-1}$

Now, the water falls by converting the potential energy into kinetic energy to facilitate its vertical motion. As the water strikes the earth, usually a part of this kinetic energy gets dissipated in the form of heat. But we are given that the entire energy gets dissipated as heat, i.e.,
$H = 19600\;J$

We know the empirical relation between calories and joules is given as: $1 cal = 4.184\;J$
$\Rightarrow H = \dfrac{19600}{4.184}$
$\Rightarrow 4684.5\;cals^{-1}$.

Thus, the amount of heat produced per second is $\approx 4685\;cal$.

Note:
The calorie as a unit in general, is ill-defined since the energy required to raise the temperature of 1 gram of water by $1^{\circ}$C is not always constant! In fact, it varies as a function of water temperature.
The amount of energy required to warm 1 gram of air-free water:
From $3.5^{\circ}C$ to $4.5^{\circ}C$ is called the $4^{\circ}$ calorie = $4.204\;J$
From $14.5^{\circ}C$ to $15.5^{\circ}C$ is called the $15^{\circ}$ calorie = $4.1855\;J$
From $19.5^{\circ}C$ to $20.5^{\circ}$C is called the $20^{\circ}$ calorie = $4.204\;J$
So the value that we have chosen for our evaluation is another type called the Thermochemical calorie.