Phy-Heat

The answer should be C.) 63 m.

Let m be the mass of water.
Let h be the height of the waterfall

Accleration due to gravity, g = 10 ms^-2
Specific heat capacity of water, c = 4200 Jkg^-1°C^-1
Temperature change of water, △T = 0.15 °C

Assume no heat is lost to the surroundings,
Loss of P.E. = Gain of internal energy
mgh = mc△T
gh = c△T
(10)h = (4200)(0.15)
h = 63 m

So, the height of the waterfall is 63 m.

Assume all potential energy lost by water (of mass m) is converted to heat

mgh = mc dt

h = c dt / g = 4200 x 0.15 / 10 = 63 m