Calculate the difference in temperature between the water at the top and bottom of a waterfall 25m high assuming that 15% of the energy of fall is spent in heating the water $\left(\mathrm{J}=4.18{\mathrm{JCal}}^{-1}\right)$

Potential energy available from ‘m’ kg of water$=\mathrm{mgh} \mathrm{J}.$

Where h=height of water fall in meters and 15percent of this converted into heat. 
Therefore,  (15/100) mgh is converted into heat:
$\to$Let $∆\theta$ be the raise in temperature
Therefore, heat taken up by ‘m’ kg of water
$=\mathrm{ms\Delta \theta } \mathrm{Cal}$
$\to \mathrm{ms\Delta \theta }\times 4.18\mathrm{J}$
Where s =specific heat of water$=1000\mathrm{cal}/{\mathrm{kg}}^0\mathrm{c}$
Therefore, $\left(15/100\right)$ $\mathrm{mgh}=\mathrm{ms\Delta \theta }\times 4.18$
Or $\mathrm{\Delta }\theta =\frac{15\times 9.8\times 25}{1000\times 4.18}={0.0088}^0\mathrm{C}$