A waterfall is 70 m high. What is the increase in water temperature at the bottom of the falls if all the initial potential energy goes into heating the water? (g = 9.8 m/s², c_w = 4186 J/kg·$\degree$C) A) 0.16$\degree$C B) 0.34$\degree$C C) 0.69$\degree$C D) 1.04$\degree$C

Question

Quizplus · Accepted Answer

The potential energy converted to heat is mgh = mcΔT. Solving for ΔT gives ΔT = gh/c. Substituting the values: ΔT = (9.8 m/s² * 70 m) / 4186 J/kg°C ≈ 0.16°C.

A Waterfall Is 70 M High °\degree° C) A) 0 °\degree° C B) 0

A Waterfall Is 70 M High $\degree$ C)
A) 0 $\degree$ C
B) 0