Question

A water fall is $126$ meters high. If half of the potential energy of the falling water gets converted to heat , the rise in temperature of water will be

Answer 1

Assume $M$ kg of water is falling per second from the top of the fall. The potential energy of M kg of water at top of the fall $=M\times g\times h=M\times 10\times 126\left(kg\right)\times \left(m/s²\right)\times \left(m\right)=1260M$ Joule.

Half of this potential energy is converted to heat when water falls from top of the fall. Therefore energy converted to heat per second$=630M$Joule.

The mechanical equivalent of heat = 4.2 Joule/calorie.

So amount of heat generated per second $=\frac{\left(630MJoule\right)}{\left(4.2Joule/calorie\right)}=150M$ calorie

This heat will raise the temperature of $M$ kg of water by temperature $\Delta T$.

Specific heat of water $=1000$ Calorie per kg of water per one degree centigrade rise of temperature.

So

$150M$ Calorie $=M\times \left(1000Calorie/degreeC\right)\times \Delta T$ (in Centigrade)

$\Delta T=\frac{150}{1000}=0.15°$ C.