Question

Water falls from a height of$800m$ from a water fall. If the initial temperature of water at top of water fall is $10°C$, what is its temperature at the bottom? Assume all the energy of falling water changes to heat energy.
[Take $g=10m{s}^{-2}$and sp. heat capacity of water$4200Jk{g}^{-1}{K}^{-1}$ ]

Answer 1

Step 1: Given data

height $\left(h\right)=800m$

acceleration due to gravity$\left(g\right)=10m{s}^{-2}$

Sp. heat capacity$\left(c\right)=4200Jk{g}^{-1}{K}^{-1}$

$m$ is the mass of the object

Temperature change$\left({\theta }_R\right)=?$

Step 2: equating the kinetic energy to the potential energy of the water

The kinetic energy of falling water=Potential energy of water at the top of water$=mgh$

The kinetic energy of falling water= heat produced in water

$$\begin{gathered} \Rightarrow mgh=mc{\theta }_R \\ \Rightarrow {\theta }_R=\frac{gh}{c} \\ \Rightarrow {\theta }_R=\frac{10m{s}^{-2}\times 800m}{4200Jk{g}^{-1}{K}^{-1}}=1.9°C \end{gathered}$$

So, the temperature of water at bottom$=10°C+1.9°C=11.9°C$

The temperature at the bottom of the water is $11.9°C$.