Question

A waterfall is 145 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, cw = 4 186 J/kg⋅°C)

a. 0.16°C
b. 0.34°C
c. 0.69°C
d. 1.04°C

Answer 1

Answer:
`\Delta T=0.339^(\circ)C`

Step-by-step explanation:

Given

height from which water is falling
`h=145 m`

heat capacity of water
`c_w=4186 J/kg-^(\circ)C`

here Potential Energy is converted to heat the water

i.e.
`P.E.=mc_w\Delta T`

`mgh=mc_w\Delta T`

`9.8* 145=4186* \Delta T`

`\Delta T=0.339^(\circ)C`

Answer 2

option B

Step-by-step explanation:

height, h = 145 m

cw = 4186 J/kg °C

g = 9.8 m/s^2

According to the conservation of energy

Potential energy = thermal energy

m x g x h = m x c x ΔT

where, ΔT is the rise in temperature

9.8 x 145 = 4186 x ΔT

ΔT = 0.34°C