temperature increase prob.?

Question

Water falls a distance of 50 meters. The specific heat of water is 1.0 cal/gm C. What is the increase in temperature of the water after the fall in degree C??

please help, thanks

Answer 1

As the water falls, it loses potential energy. In this question we will assume that this energy goes into heating the water. (This is somewhat wrong, as the energy will also be converted into KE)

specific heat water = 4184 J kg-1 K-1

PE for a height of 50 m = mgh = m*9.8*50 = 490m

energy required to heat water through increase of deltaT = m*c*deltaT

490m = m*c*deltaT

c*deltaT = 490

deltaT = 490/4184 = 0.1 K = 0.1 C

Answer 2

answer 1: Zero. This is the case when the water falls without stopping. (You didn't SAY it stopped!)

2: The kinetic energy is converted to heat energy. For each gram, the KE for a 50 m drop is ½(1/1000)(2g*50) = .49 J
.49 J * .24 = .1176 cal→ .1176°C increase in temp.