Question

A water fall is $84m$ high. Assuming that half the kinetic energy of the falling water gets converted into heat, the rise in temperature of water is:

• A. ${0.98}^{0}C$
• B. ${9.8}^{0}C$
• C. ${0.098}^{0}C$
• D. ${0.0098}^{0}C$

Answer 1

The correct option is C ${0.098}^{0}C$

Let $m$ gram be the mass of water.

Then potential energy of water is:

$P.E.=mgh=m\times 980\times 8400$

Since potential energy is converted to kinetic energy

$P.E.=K.E.$

$K.E.=m\times 980\times 8400$

Half of kinetic energy is converted into heat.

Then heat energy:

$Heatenergy=\frac{1}{2}\times m\times 980\times 8400$

Convert joule into calories

$heat=\frac{m\times 980\times 4200}{4.2{\times 10}^7}=m\times 0.098$

But heat is:

$Heat=ms\Delta T$

$\therefore ms\Delta T=m\times 0.098m\times 1\times \Delta T=m\times 0.098\Delta T=0.098C$

Rise in temperature of water is $={0.098}^{\circ }C$ .