Falling Water A water parcel of mass 1 kg and initial temperature 15 degree C fa

Question

Falling Water A water parcel of mass 1 kg and initial temperature 15 degree C falls adiabatically over a waterfall that is 100 m high. What is the change in potential energy of the parcel? What is the change in temperature of the water parcel? Assume all the potential energy lost is converted into internal energy of the parcel. What is the change in its entropy? Calculate the amount of energy degraded during the fall. Compare your answer to part d with the work required to put the parcel back at the top of the falls.

Explanation / Answer

Given:

Water parcel mass (m) = 1 Kg

Initial temperature (T) = 150 C

Height (h) = 100 m

Adiabatic process: Without transfer of heat between a thermodynamic system and its surroundings.

a.

Potential energy = mgh

g - Acceleration due to gravity [g = 9.8 m/s2]

Initial potential energy = mgh = 1(9.8)(100) = 980 J

Final potential energy = 0 [since h becomes zero]

Change in potential energy = initial potential energy - final potential energy = 980 - 0 = 980 J

c.

Change in entropy dS = dQ/T = 980/15 = 65.333 J/K

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