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:

option B

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

Answer 2

Answer:$\Delta T=0.339^{\circ}C$                    

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. $P.E.=mc_w\Delta T$

$mgh=mc_w\Delta T$

$9.8\times 145=4186\times \Delta T$

$\Delta T=0.339^{\circ}C$